Tag: Class 11 Physics

Question 7.1

Answer the following:

Question 7.1 (a)

You can shield a charge from electrical forces by putting it inside a hollow conductor.
Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?

Answer:
No, gravitational shielding is not possible. Although the gravitational force due to a hollow spherical shell on a body inside it is zero, external masses still exert gravitational force on the body. Unlike electric charges, there is no negative mass and no gravitational analogue of a conductor. Hence, gravity cannot be shielded by any means.

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Question 7.1 (b)

An astronaut inside a small spaceship orbiting around the Earth cannot detect gravity. If the space station orbiting around the Earth has a large size, can he hope to detect gravity?

Answer:
No, even in a large space station the astronaut cannot directly detect gravity because both the astronaut and the station are in free fall under Earth’s gravity. However, in a very large station, extremely small tidal effects (due to variation of gravity from one point to another) may be detectable with sensitive instruments.

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Question 7.1 (c)

If you compare the gravitational force on the Earth due to the Sun to that due to the Moon, you would find that the Sun’s pull is greater than the Moon’s pull. However, the tidal effect of the Moon’s pull is greater than the tidal effect of the Sun. Why?

Answer:
Tidal effects depend on the variation (difference) of gravitational force across the Earth, which is proportional to mass divided by the cube of distance. Although the Sun is much more massive, it is very far away. The Moon, being much closer to the Earth, produces a larger variation in gravitational pull across the Earth and hence causes stronger tides.

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Exercise 7.2

Choose the correct alternative 

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(a) Acceleration due to gravity decreases with increasing altitude.

✔ As altitude increases, the distance from the Earth’s centre increases, so gravitational pull decreases.

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(b) Acceleration due to gravity decreases with increasing depth

(assuming the Earth to be a sphere of uniform density).

✔ Inside the Earth, only the mass enclosed within the given radius contributes to gravity, so $g$ decreases as depth increases.

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(c) Acceleration due to gravity is independent of mass of the body.

✔ $g={\displaystyle \frac{GM}{R^2}}$ depends on the mass of the Earth, not on the mass of the falling body.

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(d) The formula

$$-GMm(\frac{1}{r_2}-\frac{1}{r_1})$$

is more accurate than the formula

$$mg(r_2-r_1)$$

✔ The $mg(r_2-r_1)$ formula is only an approximation valid near the Earth’s surface, whereas the gravitational potential energy formula is exact.

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Question 7.3

Suppose there existed a planet that went around the Sun twice as fast as the Earth. What would be its orbital size as compared to that of the Earth?

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Answer:

Using Kepler’s Third Law,

$$T^2\propto R^3$$

Let

• $T_E$ = time period of Earth

• $R_E$ = orbital radius of Earth

The new planet goes twice as fast, so its time period is half:

$$T_P=\frac{T_E}{2}$$

Applying Kepler’s law:

$${\left(\frac{T_P}{T_E}\right)}^2={\left(\frac{R_P}{R_E}\right)}^3$$

$${\left(\frac{1}{2}\right)}^2={\left(\frac{R_P}{R_E}\right)}^3$$

$$\frac{1}{4}={\left(\frac{R_P}{R_E}\right)}^3$$

$$\frac{R_P}{R_E}={\left(\frac{1}{4}\right)}^{1\mathrm{/}3}$$

$$\boxed{{\displaystyle \frac{R_P}{R_E}=\frac{1}{\sqrt[3]{4}}\approx 0.63}}$$

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Question 7.4

Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is
$4.22\times {10}^8\text{m}$. Show that the mass of Jupiter is about one–thousandth that of the Sun.

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Answer:

For a satellite moving in a circular orbit, by Kepler’s third law:

$$M=\frac{4{\pi }^2{R}^3}{G{T}^2}$$where
$M$ = mass of Jupiter,
$R=4.22\times {10}^8\text{m}$
$T=1.769\text{days}=1.769\times 24\times 3600\text{s}$
$G=6.67\times {10}^{-11}{\text{N m}}^2{\text{kg}}^{-2}$

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Step 1: Convert time period into seconds

$$T=1.769\times 24\times 3600\approx 1.53\times {10}^5\text{s}$$

Step 2: Substitute values

$$M_J=\frac{4{\pi }^2(4.22\times {10}^8{)}^3}{6.67\times {10}^{-11}(1.53\times {10}^5{)}^2}$$

$$M_J\approx 1.9\times {10}^{27}\text{kg}$$

Step 3: Compare with mass of the Sun

$$M_{\text{Sun}}\approx 2.0\times {10}^{30}\text{kg}$$

$$\frac{M_J}{M_{\text{Sun}}}=\frac{1.9\times {10}^{27}}{2.0\times {10}^{30}}\approx {10}^{-3}$$

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Question 7.5

Let us assume that our galaxy consists of $2.5\times {10}^{11}$ stars, each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution?
Take the diameter of the Milky Way to be ${10}^5$ ly.

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Answer:

The star revolves around the galactic centre under the gravitational attraction of the mass enclosed within its orbit.

Step 1: Mass of the galaxy

Each star has one solar mass:

$$M_{\odot }=2\times {10}^{30}\text{kg}$$

Total mass of galaxy:

$$M=2.5\times {10}^{11}\times 2\times {10}^{30}=5\times {10}^{41}\text{kg}$$

Step 2: Radius of orbit

Given distance of star from galactic centre:

$$R=\mathrm{50,000}\text{ly}$$

Convert light year to metre:

$$1\text{ly}=9.46\times {10}^{15}\text{m}$$

$$R=5\times {10}^4\times 9.46\times {10}^{15}=4.73\times {10}^{20}\text{m}$$

Step 3: Time period using circular orbit relation

$$T=2\pi \sqrt{\frac{R^3}{GM}}$$

Substitute values:

$$T=2\pi \sqrt{\frac{(4.73\times {10}^{20}{)}^3}{6.67\times {10}^{-11}\times 5\times {10}^{41}}}$$

$$T\approx 3.5\times {10}^{15}\text{s}$$

Step 4: Convert seconds to years

$$1\text{year}=3.15\times {10}^7\text{s}$$

$$T=\frac{3.5\times {10}^{15}}{3.15\times {10}^7}\approx 1.1\times {10}^8\text{years}$$

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Exercise 7.6

Choose the correct alternative

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(a)

If the zero of potential energy is at infinity, the total energy of an orbiting satellite is the negative of its kinetic energy.

✔ For a satellite in circular orbit:

$$E=-\frac{1}{2}m{v}^2$$

So, total energy = negative of kinetic energy.

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(b)

The energy required to launch an orbiting satellite out of Earth’s gravitational influence is less than the energy required to project a stationary object at the same height out of Earth’s influence.

✔ An orbiting satellite already has kinetic energy, so additional energy needed is less.

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Question 7.7

Does the escape speed of a body from the Earth depend on
(a) the mass of the body,
(b) the location from where it is projected,
(c) the direction of projection,
(d) the height of the location from where the body is launched?

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Answer:

(a) Mass of the body:
❌ No
Escape speed is independent of the mass of the body.

(b) Location from where it is projected:
✔ Yes
Escape speed depends on the distance from the centre of the Earth.

(c) Direction of projection:
❌ No
Escape speed is independent of the direction of projection.

(d) Height of the point of projection:
✔ Yes
Escape speed decreases with increase in height above the Earth’s surface.

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Question 7.8

A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant
(a) linear speed,
(b) angular speed,
(c) angular momentum,
(d) kinetic energy,
(e) potential energy,
(f) total energy
throughout its orbit?
(Neglect any mass loss of the comet when it comes very close to the Sun.)

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Answer:

(a) Linear speed:
❌ No
The speed is maximum at perihelion and minimum at aphelion.

(b) Angular speed:
❌ No
Angular speed changes; it is higher when the comet is closer to the Sun.

(c) Angular momentum:
✔ Yes
Angular momentum remains constant because gravitational force is a central force.

(d) Kinetic energy:
❌ No
Kinetic energy varies with speed, so it is not constant.

(e) Potential energy:
❌ No
Gravitational potential energy depends on distance from the Sun and hence changes.

(f) Total energy:
✔ Yes
Total mechanical energy remains constant for an orbiting comet (bound system).

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Question 7.9

Which of the following symptoms is likely to afflict an astronaut in space?
(a) swollen feet
(b) swollen face
(c) headache
(d) orientational problem

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Answer:

✔ (b) Swollen face
✔ (d) Orientational problem

❌ (a) Swollen feet – Not likely, because in weightlessness body fluids move upward.
❌ (c) Headache – Not a typical direct effect of weightlessness.

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Question 7.10

In the following two exercises, choose the correct answer from among the given ones:

The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig. 7.11):

(i) a  (ii) b  (iii) c  (iv) 0

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Answer:

✔ (ii) b

Explanation (exam-oriented):

• For a complete spherical shell, gravitational intensity at the centre is zero.

• In a hemispherical shell, mass is present only on one side.

• The gravitational pulls do not cancel completely.

• The resultant gravitational intensity is directed towards the curved surface of the hemisphere, along its axis of symmetry.

Hence, the correct direction is arrow b.

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Question 7.11

For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow:
(i) d  (ii) e  (iii) f  (iv) g

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Answer:

✔ (iii) f

Explanation (brief):

• Gravitational intensity at any point is the resultant of attractions due to all mass elements.

• At an arbitrary point P near a hemispherical shell, components perpendicular to the axis partially cancel, while components towards the mass distribution dominate.

• The net gravitational intensity therefore points in the direction shown by arrow f.

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Question 7.12

A rocket is fired from the Earth towards the Sun. At what distance from the Earth’s centre is the gravitational force on the rocket zero?
Mass of the Sun $=2\times {10}^{30}\text{kg}$,
Mass of the Earth $=6\times {10}^{24}\text{kg}$.
Neglect the effect of other planets.
Distance between Earth and Sun $=1.5\times {10}^{11}\text{m}$.

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Answer:

Let the point where the net gravitational force is zero be at a distance $x$ from the centre of the Earth along the line joining Earth and Sun. At this point, gravitational attraction due to Earth equals that due to the Sun.

Step 1: Write force balance condition

$$\frac{G{M}_E}{x^2}=\frac{G{M}_S}{(R-x{)}^2}$$

Cancel $G$:

$$\frac{M_E}{x^2}=\frac{M_S}{(R-x{)}^2}$$

Step 2: Substitute values

$$\frac{6\times {10}^{24}}{x^2}=\frac{2\times {10}^{30}}{(1.5\times {10}^{11}-x{)}^2}$$

Taking square root on both sides:

$$\frac{\sqrt{6\times {10}^{24}}}{x}=\frac{\sqrt{2\times {10}^{30}}}{1.5\times {10}^{11}-x}$$

Step 3: Simplify

$$\frac{1.5\times {10}^{11}-x}{x}=\sqrt{\frac{2\times {10}^{30}}{6\times {10}^{24}}}=\sqrt{3.33\times {10}^5}\approx 577$$

$$1.5\times {10}^{11}-x=577x$$

$$1.5\times {10}^{11}=578x$$

$$x\approx \frac{1.5\times {10}^{11}}{578}\approx 2.6\times {10}^8\text{m}$$

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Question 7.13

How will you “weigh the Sun”, that is, estimate its mass?
The mean orbital radius of the Earth around the Sun is
$1.5\times {10}^8\text{km}$.

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Answer:

The mass of the Sun can be estimated using the motion of the Earth around the Sun and Newton’s law of gravitation.

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Step 1: Use centripetal force condition

The gravitational force provided by the Sun acts as the centripetal force for Earth’s circular motion:

$$\frac{G{M}_{S}m}{R^2}=\frac{m{v}^2}{R}$$

where
$M_S$ = mass of the Sun,
$R$ = orbital radius of Earth,
$m$ = mass of Earth,
$v$ = orbital speed of Earth.

Step 2: Express velocity in terms of time period

$$v=\frac{2\pi R}{T}$$

Substitute in the force equation:

$$\frac{G{M}_S}{R^2}=\frac{4{\pi }^{2}R}{T^2}$$

$$M_S=\frac{4{\pi }^2{R}^3}{G{T}^2}$$

Step 3: Substitute numerical values

Mean orbital radius:

$$R=1.5\times {10}^8\text{km}=1.5\times {10}^{11}\text{m}$$

Time period of Earth:

$$T=1\text{year}=3.15\times {10}^7\text{s}$$

Gravitational constant:

$$G=6.67\times {10}^{-11}{\text{N m}}^2{\text{kg}}^{-2}$$

$$M_S=\frac{4{\pi }^2(1.5\times {10}^{11}{)}^3}{6.67\times {10}^{-11}(3.15\times {10}^7{)}^2}$$

$$M_S\approx 2.0\times {10}^{30}\text{kg}$$

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Question 7.14

A Saturn year is 29.5 times the Earth year. How far is Saturn from the Sun if the Earth is
$1.50\times {10}^8\text{km}$ away from the Sun?

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Answer:

Using Kepler’s Third Law:

$$\frac{T_S^2}{T_E^2}=\frac{R_S^3}{R_E^3}$$

where
$T_S=29.5T_E$,
$R_E=1.50\times {10}^8\text{km}$.

Step 1: Substitute values

$$(29.5{)}^2={\left(\frac{R_S}{R_E}\right)}^3$$

$$\frac{R_S}{R_E}=(29.5{)}^{2\mathrm{/}3}$$

Step 2: Calculate

$$(29.5{)}^{2\mathrm{/}3}\approx 9.55$$

Step 3: Find Saturn’s distance

$$R_S=9.55\times 1.50\times {10}^8$$

$$R_S\approx 1.43\times {10}^9\text{km}$$

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Question 7.15

A body weighs 63 N on the surface of the Earth.
What is the gravitational force on it due to the Earth at a height equal to half the radius of the Earth?

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Answer:

Weight on the surface of the Earth:

$$W=63\text{N}$$

Let the radius of Earth be $R$.
Given height:

$$h=\frac{R}{2}$$

Step 1: Use variation of gravity with height

Acceleration due to gravity at height $h$ is:

$$g_h=g{\left(\frac{R}{R+h}\right)}^2$$

Substitute $h=\frac{R}{2}$:

$$g_h=g{\left(\frac{R}{R+\frac{R}{2}}\right)}^2=g{\left(\frac{R}{\frac{3R}{2}}\right)}^2=g{\left(\frac{2}{3}\right)}^2=\frac{4g}{9}$$

Step 2: Find weight at height $h$

Since weight $W=mg$,

$$W_h=m{g}_h=m\left(\frac{4g}{9}\right)$$

But $mg=63\text{N}$, so:

$$W_h=\frac{4}{9}\times 63$$

$$W_h=28\text{N}$$

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Question 7.16

Assuming the Earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the Earth if it weighed 250 N on the surface?

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Answer:

Weight on the surface of the Earth:

$$W=250\text{N}$$

Let the radius of the Earth be $R$.

Halfway down to the centre means depth:

$$d=\frac{R}{2}$$

Step 1: Variation of gravity with depth

For a uniform-density Earth, acceleration due to gravity at depth $d$ is:

$$g_d=g(1-\frac{d}{R})$$

Substitute $d=\frac{R}{2}$:

$$g_d=g(1-\frac{1}{2})=\frac{g}{2}$$

Step 2: Find the weight at depth

Since weight $W=mg$,

$$W_d=m{g}_d=m\left(\frac{g}{2}\right)$$But $mg=250\text{N}$, therefore:

$$W_d=\frac{250}{2}=125\text{N}$$

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Question 7.17

A rocket is fired vertically with a speed of 5 km s⁻¹ from the Earth’s surface. How far from the Earth does the rocket go before returning to the Earth?

Given:
Mass of Earth $M=6.0\times {10}^{24}\text{kg}$
Radius of Earth $R=6.4\times {10}^6\text{m}$
$G=6.67\times {10}^{-11}{\text{N m}}^2{\text{kg}}^{-2}$

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Answer:

Initial speed of the rocket:

$$v=5{\text{km s}}^{-1}=5\times {10}^3{\text{m s}}^{-1}$$

At the highest point, the speed becomes zero.

Step 1: Use conservation of mechanical energy

Initial energy at Earth’s surface:

$$E_i=\frac{1}{2}m{v}^2-\frac{GMm}{R}$$

Final energy at maximum distance $r$:

$$E_f=-\frac{GMm}{r}$$

By energy conservation:

$$\frac{1}{2}m{v}^2-\frac{GMm}{R}=-\frac{GMm}{r}$$

Cancel $m$:

$$\frac{1}{2}{v}^2=GM(\frac{1}{R}-\frac{1}{r})$$

Step 2: Substitute values

$$\frac{1}{2}(5\times {10}^3{)}^2=(6.67\times {10}^{-11})(6.0\times {10}^{24})(\frac{1}{6.4\times {10}^6}-\frac{1}{r})$$
$$1.25\times {10}^7=4.002\times {10}^{14}(\frac{1}{6.4\times {10}^6}-\frac{1}{r})$$

$$\frac{1}{6.4\times {10}^6}-\frac{1}{r}=3.12\times {10}^{-8}$$

$$\frac{1}{r}=1.56\times {10}^{-7}-3.12\times {10}^{-8}=1.25\times {10}^{-7}$$

$$r=8.0\times {10}^6\text{m}$$

Step 3: Find height above Earth’s surface

$$h=r-R=(8.0-6.4)\times {10}^6$$

$$h=1.6\times {10}^6\text{m}$$

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Question 7.18

The escape speed of a projectile on the Earth’s surface is 11.2 km s⁻¹. A body is projected with thrice this speed. What is the speed of the body far away from the Earth?
(Neglect the presence of the Sun and other planets.)

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Answer:

Escape speed from Earth:

$$v_e=11.2{\text{km s}}^{-1}$$

Given initial speed:

$$v=3v_e=33.6{\text{km s}}^{-1}$$

Step 1: Use energy conservation

At Earth’s surface:

$$E_i=\frac{1}{2}m{v}^2-\frac{GMm}{R}$$

At infinity, gravitational potential energy is zero:

$$E_f=\frac{1}{2}m{v}_{\mathrm{\infty }}^2$$

For escape speed:

$$\frac{1}{2}m{v}_e^2=\frac{GMm}{R}$$

Step 2: Substitute $\frac{GM}{R}$

$$\frac{1}{2}m{v}^2-\frac{1}{2}m{v}_e^2=\frac{1}{2}m{v}_{\mathrm{\infty }}^2$$

Cancel $\frac{1}{2}m$:

$$v_{\mathrm{\infty }}^2=v^2-v_e^2$$

Step 3: Substitute values

$$v_{\mathrm{\infty }}^2=(3v_e{)}^2-v_e^2=9v_e^2-v_e^2=8v_e^2$$

$$v_{\mathrm{\infty }}=\sqrt{8}{v}_e=2\sqrt{2}{v}_e$$

$$v_{\mathrm{\infty }}=2\sqrt{2}\times 11.2\approx 31.7{\text{km s}}^{-1}$$

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Question 7.19

A satellite orbits the Earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the Earth’s gravitational influence?

Given:
Mass of satellite $m=200\text{kg}$
Mass of Earth $M=6.0\times {10}^{24}\text{kg}$
Radius of Earth $R=6.4\times {10}^6\text{m}$
Height $h=400\text{km}=4.0\times {10}^5\text{m}$
$G=6.67\times {10}^{-11}{\text{N m}}^2{\text{kg}}^{-2}$

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Answer:

For a satellite in circular orbit, the total mechanical energy is:

$$E=-\frac{GMm}{2r}$$

where $r=R+h$

To escape Earth’s gravitational field, the total energy must be raised to zero.
Hence, energy required:

$$\mathrm{\Delta }E=0-E=\frac{GMm}{2r}$$

Step 1: Calculate orbital radius

$$r=R+h=6.4\times {10}^6+4.0\times {10}^5=6.8\times {10}^6\text{m}$$

Step 2: Substitute values

$$\mathrm{\Delta }E=\frac{(6.67\times {10}^{-11})(6.0\times {10}^{24})(200)}{2(6.8\times {10}^6)}$$

$$\mathrm{\Delta }E=\frac{8.004\times {10}^{16}}{1.36\times {10}^7}$$

$$\mathrm{\Delta }E\approx 5.9\times {10}^9\text{J}$$

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Question 7.20

Two stars, each of one solar mass $(=2\times {10}^{30}\text{kg})$, are approaching each other for a head-on collision.
When they are at a distance of ${10}^9\text{km}$, their speeds are negligible. What is the speed with which they collide?

Radius of each star $={10}^4\text{km}$.(Assume the stars remain undistorted. Use the known value of $G$.)

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Answer:

Given:
Mass of each star

$$M=2\times {10}^{30}\text{kg}$$

Initial separation

$$r_i={10}^9\text{km}={10}^{12}\text{m}$$

At collision, distance between centres

$$r_f=2R=2\times {10}^4\text{km}=2\times {10}^7\text{m}$$

Step 1: Use conservation of mechanical energy

Initially, kinetic energy is negligible, so total energy is purely gravitational potential energy:

$$E_i=-\frac{G{M}^2}{r_i}$$

At collision, both stars move with equal speed $v$.
Total kinetic energy:

$$K=2\times \frac{1}{2}M{v}^2=M{v}^2$$

Final potential energy:

$$E_f=-\frac{G{M}^2}{r_f}$$

Energy conservation:

$$M{v}^2=G{M}^2(\frac{1}{r_f}-\frac{1}{r_i})$$

Cancel $M$:

$$v^2=GM(\frac{1}{r_f}-\frac{1}{r_i})$$

Step 2: Substitute values

$$v^2=(6.67\times {10}^{-11})(2\times {10}^{30})(\frac{1}{2\times {10}^7}-\frac{1}{{10}^{12}})$$

Since

$$\frac{1}{{10}^{12}}\ll \frac{1}{2\times {10}^7},$$

we neglect the second term:

$$v^2\approx (6.67\times {10}^{-11})(2\times {10}^{30})\left(\frac{1}{2\times {10}^7}\right)$$

$$v^2\approx 6.67\times {10}^{12}$$

$$v\approx 2.58\times {10}^6{\text{m s}}^{-1}$$

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Question 7.21

Two heavy spheres, each of mass 100 kg and radius 0.10 m, are placed 1.0 m apart on a horizontal table.
What is the gravitational force and gravitational potential at the mid-point of the line joining the centres of the spheres?
Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

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Answer

Given

• Mass of each sphere: $M=100\text{kg}$

• Distance between centres: $1.0\text{m}$

• Distance of midpoint from each centre:

  $$r=0.5\text{m}$$

• Gravitational constant:

  $$G=6.67\times {10}^{-11}{\text{N m}}^2{\text{kg}}^{-2}$$

(Outside a uniform sphere, gravity acts as if all mass were concentrated at the centre.)

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(a) Gravitational force at the midpoint

Gravitational force on a test mass $m$ at the midpoint due to one sphere:

$$F_1=\frac{GMm}{r^2}$$

Both spheres exert equal forces in opposite directions.

$$F_{\text{net}}=F_1-F_1=0$$

$$\boxed{{\displaystyle F_{\text{mid}}=0}}$$

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(b) Gravitational potential at the midpoint

Gravitational potential due to one sphere:

$$V_1=-\frac{GM}{\mathrm{r<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"></span>}}$$

Total potential at midpoint (scalar addition):

$$V=V_1+V_1=-\frac{2GM}{r}$$

Substitute values:

$$V=-\frac{2(6.67\times {10}^{-11})(100)}{0.5}$$$$V=-2.67\times {10}^{-8}{\text{J kg}}^{-1}$$
$$\boxed{{\displaystyle V_{\text{mid}}=-2.67\times {10}^{-8}{\text{J kg}}^{-1}}}$$

(c) Nature of equilibrium

• Net gravitational force at the midpoint is zero, so the object is in equilibrium.

• If the object is displaced slightly towards one sphere, the nearer sphere pulls it more strongly.

• This causes the object to move further away from the midpoint.

$$\boxed{{\displaystyle \text{Equilibrium is unstable}}}$$

 

 

Question 6.1

Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body ?

Answer:

For bodies having uniform mass density, the centre of mass coincides with their geometric centre due to symmetry.

1. Sphere:
   The centre of mass lies at the geometric centre of the sphere.

2. Cylinder:
   The centre of mass lies at the midpoint of its axis, i.e., the geometric centre of the cylinder.

3. Ring:
   The centre of mass lies at the centre of the ring (centre of the circular shape), even though there is no material present at that point.

4. Cube:
   The centre of mass lies at the geometric centre of the cube (intersection of its body diagonals).

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Question 6.2

In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10⁻¹⁰ m). Find the approximate location of the centre of mass (CM) of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Answer:

Let the hydrogen nucleus (H) be at the origin.

• Mass of hydrogen nucleus = $m$

• Mass of chlorine nucleus = $35.5m$

• Distance between H and Cl nuclei = $1.27\mathring{\text{A}}$

So,

• Position of H nucleus: $x_1=0$

• Position of Cl nucleus: $x_2=1.27\mathring{\text{A}}$

The centre of mass is given by:

$$x_{\text{CM}}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$$

Substitute values:

$$x_{\text{CM}}=\frac{m(0)+35.5m(1.27)}{m+35.5m}$$

$$\frac{\mathrm{<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"></span>}}{}$$$$x_{\text{CM}}=\frac{35.5\times 1.27}{36.5}$$

$$x_{\text{CM}}\approx 1.24\mathring{\text{A}}$$

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Question 6.3

A child sits stationary at one end of a long trolley moving uniformly with a speed $V$ on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the centre of mass (CM) of the (trolley + child) system?

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Answer :

Since the trolley is moving on a smooth horizontal floor, there is no external horizontal force acting on the trolley + child system.

According to the law of conservation of linear momentum:

If no external force acts on a system, the velocity of its centre of mass remains constant.

Initially, the entire system (trolley + child) is moving with speed $V$.
When the child runs about on the trolley, all the forces involved are internal forces between the child and the trolley.

Internal forces cannot change the motion of the centre of mass.

Final Answer

$$\boxed{{\displaystyle \text{The speed of the centre of mass remains }V\mathrm{.}}}$$

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Question 6.4

Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.

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Proof :

Let a and b be two vectors with an angle θ between them.

Step 1: Area using basic geometry

The area of a triangle formed by two sides a and b with included angle θ is:

$$\text{Area of triangle}=\frac{1}{2}ab\sin \theta$$

Step 2: Magnitude of vector product

By definition of the vector (cross) product:

$$\mathrm{\mid }\mathbf{a}\times \mathbf{b}\mathrm{\mid }=ab\sin \theta$$

Step 3: Compare the two results

From above,$$\text{Area of triangle}=\frac{1}{2}\mathrm{\mid }\mathbf{a}\times \mathbf{b}\mathrm{\mid }$$

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Question 6.5

Show that $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$ is equal in magnitude to the volume of the parallelepiped formed on the three vectors $\mathbf{a},\mathbf{b}$and $\mathbf{c}$.

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Proof :

Consider a parallelepiped formed by the three vectors
$\mathbf{a},\mathbf{b}$ and $\mathbf{c}$.

Step 1: Area of the base

The base of the parallelepiped is the parallelogram formed by vectors
$\mathbf{b}$ and $\mathbf{c}$.

The area of this base is given by the magnitude of their vector product:

$$\text{Area of base}=\mathrm{\mid }\mathbf{b}\times \mathbf{c}\mathrm{\mid }$$

Step 2: Height of the parallelepiped

Let θ be the angle between vector $\mathbf{a}$ and the vector $\mathbf{b}\times \mathbf{c}$.

The height (h) of the parallelepiped is the component of $\mathbf{a}$ along the direction perpendicular to the base:

$$h=\mathrm{\mid }\mathbf{a}\mathrm{\mid }\cos \theta$$

Step 3: Volume of the parallelepiped

$$\text{Volume}=(\text{Area of base})\times (\text{Height})$$

$$\text{Volume}=\mathrm{\mid }\mathbf{b}\times \mathbf{c}\mathrm{\mid }\mathrm{\mid }\mathbf{a}\mathrm{\mid }\cos \theta$$

Step 4: Use scalar triple product

By definition of the scalar (triple) product:

$$\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=\mathrm{\mid }\mathbf{a}\mathrm{\mid }\mathrm{\mid }\mathbf{b}\times \mathbf{c}\mathrm{\mid }\cos \theta$$

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Question 6.6

Find the components along the x, y and z axes of the angular momentum l of a particle whose position vector is r with components (x, y, z) and momentum p with components $(p_x,p_y,p_z)$. Show that if the particle moves only in the x–y plane, the angular momentum has only a z-component.

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Solution :

The angular momentum of a particle is defined as:

$$\mathbf{l}=\mathbf{r}\times \mathbf{p}$$

Step 1: Write vectors in component form

$$\mathbf{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}$$

$$\mathbf{p}=p_x\widehat{i}+p_y\widehat{j}+p_z\widehat{k}$$

Step 2: Find the cross product using determinant

$$\mathbf{l}=\mid \begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ x& y& z\\ p_x& p_y& p_z\end{array}\mid$$

Expanding:

$$\mathbf{l}=\widehat{i}(y{p}_z-z{p}_y)-\widehat{j}(x{p}_z-z{p}_x)+\widehat{k}(x{p}_y-y{p}_x)$$

Step 3: Components of angular momentum

$$l_x=y{p}_z-z{p}_y$$

$$l_y=z{p}_x-x{p}_z$$

$$l_z=x{p}_y-y{p}_x$$

Step 4: Particle moving only in the x–y plane

For motion in the x–y plane:

$$z=0\text{and}{p}_z=0$$

Substitute into the components:

$$l_x=y(0)-0(p_y)=0$$

$$l_y=0(p_x)-x(0)=0$$

$$l_z=x{p}_y-y{p}_x\mathrm{\ne }0$$

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Question 6.7

Two particles, each of mass $m$ and speed $v$, travel in opposite directions along parallel lines separated by a distance $d$. Show that the angular momentum vector of the two-particle system is the same whatever be the point about which the angular momentum is taken.

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Solution :

Let the two particles move along parallel straight lines, separated by a perpendicular distance $d$.

• Mass of each particle = $m$

• Speed of each particle = $v$

• Momenta:

  $${\mathbf{p}}_1=m\mathbf{v},{\mathbf{p}}_2=-m\mathbf{v}$$

Step 1: Choose any arbitrary origin O

Let the perpendicular distances of the two lines of motion from O be $r_1$ and $r_2$, such that:

$$r_1+r_2=d$$

Step 2: Angular momentum of each particle

Angular momentum magnitude of a particle moving in a straight line:

$$L=(\text{momentum})\times (\text{perpendicular distance})$$

For particle 1:

$$L_1=mv{r}_1$$

For particle 2:

$$L_2=mv{r}_2$$

Step 3: Total angular momentum of the system

Both angular momentum vectors point in the same direction (given by the right-hand rule), so they add:

$$L_{\text{total}}=L_1+L_2=mv(r_1+r_2)$$

$$L_{\text{total}}=mvd$$

Step 4: Independence from choice of origin

The result depends only on $d$ (the separation between the lines) and not on $r_1$ or $r_2$ individually.

Hence, no matter where the origin is chosen, the total angular momentum remains the same.

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Question 6.8

A non-uniform bar of weight $W$ is suspended at rest by two strings of negligible weight as shown in Fig. 6.33. The angles made by the strings with the vertical are ${36.9}^{\circ }$ and ${53.1}^{\circ }$respectively. The bar is 2 m long. Calculate the distance $d$ of the centre of gravity of the bar from its left end.

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Solution :

Let the tensions in the left and right strings be $T_1$ and $T_2$ respectively.

Given:

$$\sin {36.9}^{\circ }=0.6,\cos {36.9}^{\circ }=0.8$$

$$\sin {53.1}^{\circ }=0.8,\cos {53.1}^{\circ }=0.6$$

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Step 1: Horizontal equilibrium

Since the bar is at rest, horizontal components must balance:

$$T_1\sin {36.9}^{\circ }=T_2\sin {53.1}^{\circ }$$

$$T_1(0.6)=T_2(0.8)$$

$$T_1=\frac{4}{3}{T}_2$$

Step 2: Vertical equilibrium

Sum of vertical components equals weight $W$:

$$T_1\cos {36.9}^{\circ }+T_2\cos {53.1}^{\circ }=W$$

$$\frac{4}{3}{T}_2(0.8)+T_2(0.6)=W$$

$$(1.0667+0.6)T_2=W$$

$$T_2=0.6W,T_1=0.8W$$

Step 3: Take moments about the left end

• Vertical component of $T_1$ acts at the left end → no moment

• Vertical component of $T_2$:

$$T_2\cos {53.1}^{\circ }=0.6W\times 0.6=0.36W$$Taking moments about the left end:

$$(0.36W)(2)=W\cdot d$$

$$d=0.72\text{ m}$$

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Question 6.9

A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

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Given

• Mass of car, $m=1800\text{kg}$

• Weight of car,

  $$W=mg=1800\times 9.8=17640\text{N}$$

• Distance between axles = $1.8\text{m}$

• Distance of centre of gravity from front axle = $1.05\text{m}$

So, distance of CG from rear axle:

$$1.8-1.05=0.75\text{m}$$

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Step 1: Let reactions be

• Reaction at front axle = $R_f$

• Reaction at rear axle = $R_r$

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Step 2: Vertical equilibrium

$$R_f+R_r=W=17640$$

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Step 3: Take moments about the front axle

Moment due to rear reaction = moment due to weight

$$R_r\times 1.8=17640\times 1.05$$

$$R_r=\frac{17640\times 1.05}{1.8}$$

$$R_r=10290\text{N}$$

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Step 4: Find reaction at front axle

$$R_f=17640-10290=7350\text{N}$$

Step 5: Force on each wheel

Each axle has two wheels.

• Force on each back wheel:

$$\frac{R_r}{2}=\frac{10290}{2}=5145\text{N}$$

• Force on each front wheel:

$$\frac{R_f}{2}=\frac{7350}{2}=3675\text{N}$$

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Question 6.10

Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

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Solution :

Given:

• Same torque $\tau$

• Same mass $M$

• Same radius $R$

• Same time of application

Angular acceleration is given by:

$$\alpha =\frac{\tau }{I}$$

where $I$ is the moment of inertia.

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Step 1: Moments of inertia

• Hollow cylinder (about its symmetry axis):

$$I_{\text{cyl}}=M{R}^2$$

• Solid sphere (about diameter):

$$I_{\text{sphere}}=\frac{2}{5}M{R}^2$$

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Step 2: Compare angular accelerations

$${\alpha }_{\text{cyl}}=\frac{\tau }{M{R}^2}$$

$${\alpha }_{\text{sphere}}=\frac{\tau }{\frac{2}{5}M{R}^2}=\frac{5\tau }{2M{R}^2}$$

Clearly,

$${\alpha }_{\text{sphere}}>{\alpha }_{\text{cyl}}$$

Step 3: Angular speed after given time

Angular speed acquired after time $t$:

$$\omega =\alpha t$$

Since both start from rest and torque acts for the same time:

$${\omega }_{\text{sphere}}>{\omega }_{\text{cyl}}$$

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Question 6.11

A solid cylinder of mass 20 kg rotates about its axis with angular speed $100{\text{rad s}}^{-1}$
The radius of the cylinder is 0.25 m.
(i) What is the kinetic energy associated with the rotation of the cylinder?
(ii) What is the magnitude of angular momentum of the cylinder about its axis?

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Given

• Mass, $M=20\text{kg}$

• Radius, $R=0.25\text{m}$

• Angular speed, $\omega =100{\text{rad s}}^{-1}$

For a solid cylinder about its axis:

$$I=\frac{1}{2}M{R}^2$$

Step 1: Moment of inertia

$$I=\frac{1}{2}(20)(0.25{)}^2=10\times 0.0625=0.625{\text{kg m}}^2$$

(i) Rotational kinetic energy

$$K=\frac{1}{2}I{\omega }^2$$

$$K=\frac{1}{2}(0.625)(100{)}^2$$

$$K=0.3125\times 10000=3125\text{J}$$

(ii) Angular momentum

$$L=I\omega$$

$$L=0.625\times 100=62.5{\text{kg m}}^2{\text{s}}^{-1}$$

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Question 6.12

(a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to $\frac{2}{5}$ times the initial value? Assume that the turntable rotates without friction.

(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

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Solution

(a) New angular speed

Since the turntable rotates without friction, there is no external torque on the system (child + turntable).

Hence, angular momentum is conserved.

$$I_1{\omega }_1=I_2{\omega }_2$$

Given:

$${\omega }_1=40\text{rev/min}$$

$$I_2=\frac{2}{5}{I}_1$$

Substitute:

$$I_1(40)=\frac{2}{5}{I}_1{\omega }_2$$

Cancel $I_1$:

$$40=\frac{2}{5}{\omega }_2$$

$${\omega }_2=40\times \frac{5}{2}=100\text{rev/min}$$

Answer (a)

$$\boxed{{\displaystyle {\omega }_2=100\text{rev/min}}}$$

(b) Change in kinetic energy

Rotational kinetic energy:

$$K=\frac{1}{2}I{\omega }^2$$

Initial kinetic energy

$$K_1=\frac{1}{2}{I}_1{\omega }_1^2$$

Final kinetic energy

$$K_2=\frac{1}{2}{I}_2{\omega }_2^2$$

Substitute $I_2=\frac{2}{5}{I}_1$ and ${\omega }_2=\frac{5}{2}{\omega }_1$:

$$K_2=\frac{1}{2}\left(\frac{2}{5}{I}_1\right){\left(\frac{5}{2}{\omega }_1\right)}^2$$

$$K_2=\frac{1}{2}{I}_1{\omega }_1^2\times \frac{25}{10}$$

$$K_2=\frac{5}{2}{K}_1$$

Conclusion (b)

$$\boxed{{\displaystyle K_2>K_1}}$$

The kinetic energy of rotation increases when the child folds his arms.

Explanation for increase in kinetic energy

The increase in kinetic energy comes from the work done by the child while pulling his arms inward against centrifugal effects.
This work is converted into rotational kinetic energy.

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Question 6.13

A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.

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Given

• Mass of hollow cylinder, $M=3\text{kg}$

• Radius, $R=40\text{cm}=0.40\text{m}$

• Pulling force, $F=30\text{N}$

For a hollow cylinder about its axis:

$$I=M{R}^2$$

Step 1: Torque on the cylinder

The force applied on the rope produces a torque:

$$\tau =FR=30\times 0.40=12\text{N m}$$

Step 2: Angular acceleration

Using rotational equation of motion:

$$\tau =I\alpha$$

$$12=(M{R}^2)\alpha$$

$$12=(3)(0.40{)}^2\alpha$$

$$12=3\times 0.16\alpha =0.48\alpha$$

$$\alpha =\frac{12}{0.48}=25{\text{rad s}}^{-2}$$

Angular acceleration

$$\boxed{{\displaystyle \alpha =25{\text{rad s}}^{-2}}}$$

Step 3: Linear acceleration of the rope

Since there is no slipping:

$$a=\alpha R$$
$$a=25\times 0.40=10{\text{m s}}^{-2}$$

Linear acceleration of the rope

$$\boxed{{\displaystyle a=10{\text{m s}}^{-2}}}$$

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Question 6.14

To maintain a rotor at a uniform angular speed of $200{\text{rad s}}^{-1}$, an engine needs to transmit a torque of $180\text{N m}$. What is the power required by the engine?
(Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient.

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Solution :

Given:

• Angular speed, $\omega =200{\text{rad s}}^{-1}$

• Torque, $\tau =180\text{N m}$

The power delivered by a rotating engine is given by:

$$P=\tau \omega$$

Substitute the given values:

$$P=180\times 200$$
$$P=36000\text{W}$$

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Question 6.15

From a uniform disc of radius $R$, a circular hole of radius $R\mathrm{/}2$ is cut out. The centre of the hole is at a distance $R\mathrm{/}2$from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

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Solution :

Step 1: Use method of superposition

Treat the removed hole as a negative mass.

Let:

• Surface mass density = $\sigma$ (uniform)

• Mass of full disc:

$$M=\sigma \pi R^2$$

• Mass of removed disc (hole):

$$m=\sigma \pi {\left(\frac{R}{2}\right)}^2=\frac{1}{4}M$$

Step 2: Choose coordinate system

• Take the centre of the original disc as origin $O$

• Let the centre of the hole be along the +x-axis at distance $R\mathrm{/}2$

So:

• Position of full disc CM: $x_1=0$

• Position of hole CM: $x_2=R\mathrm{/}2$

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Step 3: Centre of mass formula

$$x_{\text{CM}}=\frac{M{x}_1-m{x}_2}{M-m}$$

(Subtract because the hole represents negative mass.)

Substitute values:

$$x_{\text{CM}}=\frac{M(0)-\frac{1}{4}M\left(\frac{R}{2}\right)}{M-\frac{1}{4}M}$$

$$x_{\text{CM}}=\frac{-\frac{MR}{8}}{\frac{3M}{4}}$$

$$x_{\text{CM}}=-\frac{R}{6}$$

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Question 6.16

A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g, are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

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Solution :

Given

• Length of metre stick = 100 cm

• Initial balance point (centre of stick) = 50 cm

• New balance point = 45 cm

• Mass of each coin = 5 g

• Total mass of coins = $10\text{g}$

• Position of coins = 12 

Let the mass of the metre stick be $M$ grams.

Principle Used

For equilibrium about the knife edge,

$$\text{Sum of clockwise moments}=\text{Sum of anticlockwise moments}$$

Step 1: Distances from the new balance point (45 cm)

• Distance of coins from knife edge:

$$45-12=33\text{cm}$$

• Distance of stick’s centre of gravity from knife edge:

$$50-45=5\text{cm}$$

Step 2: Take moments about the knife edge

Moment due to coins = Moment due to metre stick

$$(10)(33)=M(5)$$

Step 3: Solve for $M$

$$M=\frac{10\times 33}{5}=66\text{g}$$

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Question 6.17

The oxygen molecule has a mass of $5.30\times {10}^{-26}\text{kg}$and a moment of inertia of   
$1.94\times {10}^{-46}{\text{kg m}}^2$about an axis through its centre perpendicular to the line joining the two atoms. Suppose the mean speed of such a molecule in a gas is $500{\text{m s}}^{-1}$ and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

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Solution :

Given

• Mass of molecule,

  $$m=5.30\times {10}^{-26}\text{kg}$$

• Moment of inertia,

  $$I=1.94\times {10}^{-46}{\text{kg m}}^2$$

• Mean speed,

  $$v=500{\text{m s}}^{-1}$$

• Rotational KE $=\frac{2}{3}$ of translational KE

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Step 1: Translational kinetic energy

$$K_{\text{trans}}=\frac{1}{2}m{v}^2$$

$$K_{\text{trans}}=\frac{1}{2}(5.30\times {10}^{-26})(500{)}^2$$

$$K_{\text{trans}}=\frac{1}{2}(5.30\times {10}^{-26})(2.5\times {10}^5)$$

$$K_{\text{trans}}=6.63\times {10}^{-21}\text{J}$$

Step 2: Rotational kinetic energy

$$K_{\text{rot}}=\frac{2}{3}{K}_{\text{trans}}$$$$K_{\text{rot}}=\frac{2}{3}(6.63\times {10}^{-21})=4.42\times {10}^{-21}\text{J}$$

Step 3: Use rotational KE formula

$$K_{\text{rot}}=\frac{1}{2}I{\omega }^2$$

$$\frac{1}{2}(1.94\times {10}^{-46}){\omega }^2=4.42\times {10}^{-21}$$

$${\omega }^2=\frac{2(4.42\times {10}^{-21})}{1.94\times {10}^{-46}}$$

$${\omega }^2=4.56\times {10}^{25}$$

$$\omega =\sqrt{4.56\times {10}^{25}}\approx 6.75\times {10}^{12}{\text{rad s}}^{-1}$$