WBCHSE Class 11 Physics Circular Motion Short Answer Questions

Circular Motion Short Answer Type Questions

Question 1. When a car takes a circular turn on a level road, which force acts as the centripetal force?
Answer:

When a vehicle takes a turn on a road, the frictional force acting between the tyres of the vehicle and the road supplies the necessary centripetal free.

Question 2.

1. A particle is moving in a circle of radius r with constant angular velocity ω. At any point (r, θ) on its path, its position vector is \(\vec{r}=r \cos \theta \hat{i}+r \sin \theta \hat{j}\). Show that the velocity of the particle has no component along the radius.
2. Find an expression for the acceleration of the particle and hence indicate its direction.

Answer:

1. \(\vec{r}=r \cos \theta \hat{i}+r \sin \theta \hat{j}\)

∴ \(\frac{d \vec{r}}{d t}=-r \sin \theta \frac{d \theta}{d t} \hat{i}+r \cos \theta \frac{d \theta}{d t} \hat{j}\)

= \(r \omega(-\sin \theta \hat{i}+\cos \theta \hat{j})\)

or, \(\nu=\left|\frac{d \vec{r}}{d t}\right|=\sqrt{r^2 \omega^2}=r \omega\)

So, \(\vec{r} \cdot \frac{d \vec{r}}{d t}=r^2 \omega(-\cos \theta \sin \theta+\cos \theta \sin \theta)=0\)

As, \(\vec{r}\) and \(\frac{d \vec{r}}{d t}\) are mutually perpendicular, the component of \(\frac{d \vec{r}}{d t}\) along \(\vec{r}\) is zero.

2. Acceleration of the particle, \(\vec{a} =\frac{d \vec{v}}{d t}=r \omega(-\cos \theta \hat{i}-\sin \theta \hat{j}) \frac{d \theta}{d t}\)

= \(-r \omega^2(\cos \theta \hat{i}+\sin \theta \hat{j})\) [because ω = constant]

= \(-\omega^2 \vec{r}\)

∴ \(\vec{a}\) and \(\vec{r}\) are acted in opposite direction.

Question 3. Two particles having masses M and m are moving in a circular path having radii R and r respectively. If their periods are the same, then the ratio of their angular velocities will be

1. \(\frac{R}{r}\)
2. \(\sqrt{\frac{R}{r}}\)
3. \(\frac{r}{R}\)
4. 1

Answer:

Time period = \(\frac{2 \pi R}{v_1}=\frac{2 \pi r}{v_2}\)

or, \(\frac{v_1}{v_2}=\frac{R}{r}\) [where \(v_1\) and \(v_2\) are linear velocities]

If the angular velocities are \(\omega_1\) and \(\omega_2\) respectively, \(v_1=\omega_1 R \quad \text { and } v_2=\omega_2 r\)

∴ \(\frac{\omega_1}{\omega_2}=\frac{v_1}{R} \cdot \frac{r}{v_2}=\frac{R}{r} \cdot \frac{r}{R}=1\)

The option 4 is correct.

Question 4. Express in diagram, angular velocity, angular acceleration, linear velocity and linear acceleration as vector quantity.
Answer:

Here, \(\vec{v}\) = linear velocity ; \(\vec{\omega}\) = angular velocity ;\(\vec{a}\) = linear acceleration and \(\vec{\alpha}\) = angular acceleration

When velocity gradually increases, both \(\vec{\omega}\) and \(\vec{\alpha}\) act in the same direction. On the other hand, if the velocity gradually decreases, \(\vec{\alpha}\) acts in the direction opposite to that of \(\vec{\omega}\).

Question 5. The maximum velocity of the car which is moving in a circular path of radius 150m in the horizontal plane during banking is (coefficient of friction 0.6)

1. 60 m/s
2. 30 m/s
3. 15 m/s
4. 25 m/s

Answer:

⇒ \(\nu_{\max }=\sqrt{r \mu g}=\sqrt{150 \times 0.6 \times 9.8} \approx 30 \mathrm{~m} / \mathrm{s}\)

The option 2 is correct

Question 6. If the radii of circular paths of two particles of the same masses are in the ratio 1:2, then, in order to have the same centripetal force, their velocities should be in the ratio of

1. 1:√2
2. √2:1
3. 4:1
4. 1:4

Answer:

Centripetal force = \(\frac{m v^2}{r}\) as the masses of the two particles are equal.

\(\frac{m v_1^2}{r_1}=\frac{m v_2^2}{r_2}\)

or, \(\frac{v_1}{v_2}=\sqrt{\frac{r_1}{r_2}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}\)

The option 1 is correct.

Question 7. Calculate the angular speed of a car which rounds a curve of radius 8 m at a speed of 50 km/h.
Answer:

v = \(50 \mathrm{~km} / \mathrm{h}=\frac{50 \times 1000}{60 \times 60} \mathrm{~m} / \mathrm{s}=\frac{125}{9} \mathrm{~m} / \mathrm{s}\)

∴ Angular velocity, \(\omega=\frac{\nu}{r}=\frac{125 / 9}{8}=\frac{125}{72}=1.736 \mathrm{~s}^{-1}\)

Question 8. A particle is moving uniformly in a circular path of radius r. When it moves through an angular displacement θ, then the magnitude of the corresponding linear displacement will be

1. \(2 r \cos \left(\frac{\theta}{2}\right)\)
2. \(2 r \cot \left(\frac{\theta}{2}\right)\)
3. \(2 r \tan \left(\frac{\theta}{2}\right)\)
4. \(2 r \sin \left(\frac{\theta}{2}\right)\)

Answer:

Here, \(\angle A O D=\frac{\theta}{2}=\angle B O D, A O=r, A D=B D\)

From \(\triangle A O D\)

∴ \(\sin \frac{\theta}{2}=\frac{A D}{O A} \quad \text { or, } A D=r \sin \frac{\theta}{2}\)

Linear displacement, \(A B=2 \times A D=2 r \sin \frac{\theta}{2}\)

The option 4 is correct

Question 9. A circular disc rolls on a horizontal floor without slipping and the centre of the disc moves with a uniform velocity v. Which of the following values the velocity at a point on the rim of the disc can have?

1. v
2. -v
3. 2v
4. Zero

Answer:

Velocity of the point of contact of the circular disc with respect to the horizontal floor = v – v = 0

Velocity of the farthest point = v + v = 2v

Horizontal velocity of the left or right endpoint

= velocity of the centre of the disc

= v

Option 1, 3 and 4 is correct.

Question 10. A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then

1. \(T \propto R^{(n+1) / 2}\)
2. \(T \propto R^{n / 2}\)
3. \(T \propto R^{3 / 2}\) for any n
4. \(T \propto R^{\frac{n}{2}+1}\)

Answer:

According to the given condition, \(F \propto \frac{1}{R^n} or, F=\frac{k}{R^n} \)

or, \(m \omega^2 R=\frac{k}{R^n} or, m\left(\frac{2 \pi}{T}\right)^2=\frac{k}{R^{n+1}} \quad or, \frac{4 \pi^2 m}{T^2}=\frac{k}{R^{n+1}}\)

So, \(\frac{1}{T^2} \propto \frac{1}{R^{n+1}} \quad or, T \propto R^{\frac{n+1}{2}}\)

The option 1 is correct

Question 11. A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis. A massless string is wound around the cylinder with one end attached to it and the other hanging freely. Tension in the string required to produce an angular acceleration of 2 revolutions s^-2 is

1. 25 N
2. 50N
3. 78.5 N
4. 157N

Answer:

Here, \(\alpha=2 \mathrm{rev} \cdot \mathrm{s}^{-2}=4 \pi \mathrm{rad} \cdot \mathrm{s}^{-2}\)

I = \(\frac{1}{2} M R^2=\frac{1}{2}(50)(0.5)=\frac{25}{4} \mathrm{~kg} \cdot \mathrm{m}^2\)

As, \(\tau=I \alpha, \text { so, } T R=I \alpha\)

∴ T = \(\frac{I \alpha}{R}=\frac{\left(\frac{25}{4}\right)(4 \pi)}{0.5}=50 \pi=157 \mathrm{~N}\)

The option 4 is correct

Question 12. A car is negotiating a curved road of radius R. The road is banked at an angle θ. The coefficient of friction between the tyres of the car and the road is μ_s. The maximum safe velocity on this road is

1. \(\sqrt{g R\left(\frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}\right)}\)
2. \(\sqrt{\frac{g}{R}\left(\frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}\right)}\)
3. \(\sqrt{\frac{g}{R^2}\left(\frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}\right)}\)
4. \(\sqrt{g R^2\left(\frac{\mu_s+\tan \theta}{1-\mu_s \tan \theta}\right)}\)

Answer: Option 1 is correct

Question 13. A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad/s². Its net acceleration in m/s² at the end of 2.0 s is approximately

1. 7.0
2. 6.0
3. 3.0
4. 8.0

Answer:

Angular acceleration, a = 2 rad/s²

diameter of the disc, R = 50 cm = 0.5 m

The angular speed of the disc after 2 s,

ω = ω₀ + αt

= (0 + 2×2) [ω₀ = 0]

= 4 rad/s

At that instant, radial acceleration, a_r = Rω² = 0.5 x (4)² = 8 m/s²

and tangential acceleration, a_t = Rα = 0.5 x 2 = 1 m/s²

∴ Resultant acceleration a = \(\sqrt{a_r^2+a_t^2}=\sqrt{8^2+1^2} \approx 8 \mathrm{~m} / \mathrm{s}^2\)

The option 4 is correct

Question 14. A cyclist on a level road takes a sharp circular turn of radius 3m (g = 10 m · s^-2). If the coefficient of static friction between the cycle tyres and the road is 0.2, at which of the following speeds will the cyclist not skid while taking the turn?

1. 14.4 km · h^-1
2. 7.2 km · h^-1
3. 9 km · h^-1
4. 10.8 km · h^-1

Answer:

The maximum speed at which the cyclist will not skid is,

∴ \(v_m =\sqrt{\mu r g}=\sqrt{0.2 \times 3 \times 10}=2.45 \mathrm{~m} / \mathrm{s} \)

= \(\frac{2.45 \times 60 \times 60}{1000} \mathrm{~km} / \mathrm{h}=8.82 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

Among the given options, 7.2 km · h^-1 is less than this.

The option 2 is correct.

Question 15. Derive an expression for the acceleration of a body of mass m moving with a uniform speed v in a circular path of radius r.
Answer:

Let A = initial position of the body;

B = position of the body after a small time interval t.

So, the angular displacement θ = ∠AOB is also small, and the displacement x effectively coincides with the arc AB of the circular path. Then,

Now at A, the velocity v is horizontal; at B, the velocity makes the same angle θ with the horizontal direction, keeping its magnitude v to be the same, as the body moves with a uniform speed v.

Again, as θ is small, the change of velocity u, effectively coincides with a circular arc.

So, \(\theta=\frac{u}{v}\)

Then we get, \(\frac{x}{r}=\frac{u}{v}\) or, \(u=\frac{v}{r} x\)

Hence, the acceleration is, a = \(\frac{\text { change of velocity }}{\text { time }}=\frac{u}{t}=\frac{v x}{r t}\)

= \(\frac{v}{r} v=\frac{v^2}{r}\)

Shows that, as θ is small, u is normal to the velocity v. As v is tangential at every point on the circular path, the acceleration a is radial. This is the centripetal acceleration of a uniform circular motion.

Question 16. Why are circular roads banked? Deduce an expression for maximum speed of a vehicle which can be achieved while taking a turn on the banked curved road neglecting friction.
Answer:

The breadths of circular roads are sometimes kept slanted with the horizontal. This is called banking. In a banked road, the normal reaction of a vehicle would have a horizontal component. This contributes to the centripetal force of circular motion and thus helps a vehicle to have a greater safe speed.

θ = banking angle; N = normal reaction; Ncosθ = vertical component of N, which balances the weight mg of the vehicle; Nsinθ = horizontal component of N, which supplies the centripetal force \(\frac{m v^2}{r}\) for the vehicle moving with velocity v in a path of radius r.

∴ \(N \cos \theta=m g\) and \(N \sin \theta=\frac{m v^2}{r}\)

Then, \(\frac{N \sin \theta}{N \cos \theta}=\frac{\left(m v^2\right) / r}{m g}\)

or, \(\tan \theta=\frac{v^2}{r g} or, v=\sqrt{r g \tan \theta}\)

This is the maximum velocity that a vehicle may achieve in the curved path.

In this treatment, we neglected the effect of friction [However, friction plays a very important role in motions along curved paths. For example, if the road is not banked, θ = 0; then our formula gives v = 0. But in practice, vehicles can turn in curved paths even in the absence of banking. In that case, the entire centripetal force is provided by friction].

Question 17. What is the need for the banking of tracks?
Answer:

A vehicle moving along a curved path needs some centripetal force. If the path is banked, i.e., inclined with the horizontal, then the horizontal components of friction and normal reaction help to provide this centripetal force. The vehicle may attain a velocity greater than that if the path has been horizontal along the curve.

Question 18. Why are the spokes fitted in a cycle wheel?
Answer:

To preserve the circular structure of the rim of the wheel.

To withstand the centripetal force, acting radially inwards, when the wheel is in circular motion during running along a road.

Question 19. An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes revolutions in 100 s.

1. What is the angular speed and the linear speed of the motion?
2. Is the acceleration vector a constant vector? What is its magnitude?

Answer:

1. Here, the radius of the circular groove, r = 12 cm and

time period, T = \(\frac{100}{7}\)s

Hence, angular speed, \(\omega=\frac{2 \pi}{T}=\frac{2 \times 3.14 \times 7}{100}=0.44 \mathrm{rad} \cdot \mathrm{s}^{-1}\)

and linear speed, \(\nu=r \omega=12 \times 0.44=5.28 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

2. The acceleration is directed radially towards the centre of the circular groove. As the insect moves along the groove, its direction of acceleration changes but its magnitude remains constant.

Its magnitude is \(\frac{v^2}{r}=\frac{(5.28)^2}{12}=2.32 \mathrm{~cm} \cdot \mathrm{s}^{-2}\)

Question 20. Calculate the maximum speed with which a vehicle can travel on a banked circular road without skidding. A cyclist at a 18 km/h on a level road takes a sharp turn of radius 3 m without reducing the speed. The coefficient of static friction between the tyres and the road is 1. Will the cyclist slip while taking the turn?
Answer:

Here, v = 18 km/h =5 m/s, r = 3 m and = 1

Only frictional force can provide the centripetal force on an unbanked road.

The condition for which the cyclist will not slip is \(\frac{m v^2}{r}\) ≤ \(F_s\)

or, \(\frac{m v^2}{r}\) ≤ \(\mu_s m g \quad \text { or, } v^2\) ≤ \(\mu_s r g\)

Here, \(v^2=25\)

and \(\mu_s r g=1 \times 3 \times 10=30\)

Since, 25 < 30, therefore it satisfies the condition v² ≤ \(\mu_s r g\)

Hence, the cyclist will not slip while taking the turn.

Question 21. A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s. what is the magnitude and direction of acceleration of the stone?
Answer:

According to the question,

r = 80 cm = 0.8 m

and ω = 2 x π X 14/25 rad/s

Now, the magnitude of acceleration produced,

∴ \(r \omega^2=0.8 \times\left(2 \times \pi \times \frac{14}{25}\right)^2=9.9 \mathrm{~m} / \mathrm{s}^2\)

The direction of this acceleration will be along the radius of the circle towards its centre.