CBSE Class 9 Physics Uniform Circular Motion MCQs Set 13

Answer: (d) \( vr^{-1} \)

Explanation:
1. The angle swept per second is the angular velocity \( \omega \).
2. \( \omega = \dfrac{v}{r} = vr^{-1} \).

Teacher's Note:
1. Use v = r omega.
2. The unit of omega is rad/s.

Answer: (c) 8 x 10⁵ m/s²

Explanation:
1. \( \omega = 2\pi \) rad/s and r = 2 x 10⁴ m.
2. \( a = \omega^2 r = 4\pi^2 \times 2 \times 10^4 \approx 7.9 \times 10^5 \approx 8 \times 10^5 \) m/s².

Teacher's Note:
1. Convert the radius to metres.
2. Centripetal acceleration = omega squared times r.

Answer: (b) 10.47 rad/s

Explanation:
1. Frequency = 100/60 rev/s.
2. \( \omega = 2\pi \times \dfrac{100}{60} = 10.47 \) rad/s.

Teacher's Note:
1. Each revolution is 2 pi radians.
2. Convert minutes to seconds.

Answer: (a) \( \dfrac{mv^2}{r} \)

Explanation:
1. Centripetal acceleration is \( \dfrac{v^2}{r} \); multiplying by the mass gives the force \( \dfrac{mv^2}{r} \).

Teacher's Note:
1. It is also written as m omega squared r.
2. It acts towards the centre.

Answer: (b) 20 N

Explanation:
1. \( F = m\omega^2 r = 5 \times 4 \times 1 = 20 \) N.

Teacher's Note:
1. Use the omega form of the formula.
2. Check units: kg x rad²/s² x m = N.

Answer: (a) radius of the path

Explanation:
1. \( a = \dfrac{v^2}{r} \), so for constant v it varies inversely as r.
2. It does not depend on the mass.

Teacher's Note:
1. Centripetal acceleration is independent of mass.
2. It is directly proportional to the square of the speed.

Answer: (a) It moves in a circle and its kinetic energy remains constant

Explanation:
1. A force perpendicular to the velocity changes only the direction of velocity, so the speed and kinetic energy stay constant and the path is a circle.
2. The acceleration has constant magnitude but its direction keeps changing.

Teacher's Note:
1. The force does no work because it is perpendicular to the displacement.
2. The acceleration vector rotates with the particle.

Answer: (c) 1000 N

Explanation:
1. 36 km/h = 10 m/s.
2. \( F = \dfrac{mv^2}{r} = \dfrac{500 \times 100}{50} = 1000 \) N.

Teacher's Note:
1. Convert the speed to m/s first.
2. On a level road this force is provided by friction.

Answer: (a) 1 : 1

Explanation:
1. Angular speed \( \omega = \dfrac{2\pi}{t} \), which depends only on the period.
2. Both cars have the same period, so the ratio is 1 : 1.

Teacher's Note:
1. Linear speeds are in the ratio r₁ : r₂.
2. Angular speed is independent of mass and radius for a given period.

Answer: (b) 3 pi rad/s

Explanation:
1. Frequency = 90/60 = 1.5 rev/s.
2. \( \omega = 2\pi \times 1.5 = 3\pi \) rad/s.

Teacher's Note:
1. Multiply the frequency by 2 pi.
2. 90 rpm is 1.5 revolutions per second.

Answer: (a) 0.2 rad/s

Explanation:
1. \( \omega = \dfrac{v}{r} = \dfrac{0.02}{0.10} = 0.2 \) rad/s.

Teacher's Note:
1. Use v = r omega.
2. Angular velocity is measured in rad/s.

Answer: (a) 13.2 m/s

Explanation:
1. \( \omega = 2\pi \times 3 = 18.85 \) rad/s.
2. \( v = r\omega = 0.7 \times 18.85 \approx 13.2 \) m/s, along the tangent.

Teacher's Note:
1. The stone flies off along the tangent with its speed at that instant.
2. Convert revolutions to radians.

Answer: (a) 4.12 rad/s

Explanation:
1. The tension is greatest at the bottom: \( T - mg = m\omega^2 r \).
2. \( 2.7 - 1 = 0.1\omega^2 (1) \Rightarrow \omega^2 = 17 \Rightarrow \omega = 4.12 \) rad/s.

Teacher's Note:
1. The string is most likely to snap at the lowest point.
2. Gravity adds to the required centripetal force there.

Answer: (a) 10 m

Explanation:
1. Speed 10 m/s. At the top of the loop the minimum speed is \( \sqrt{rg} \), so \( v^2 \geq rg \).
2. \( r \leq \dfrac{v^2}{g} = \dfrac{100}{10} = 10 \) m.

Teacher's Note:
1. He must keep at least the minimum speed at the top.
2. A bigger track needs a higher speed.

Answer: (a) 3.536 rad/s

Explanation:
1. At the top the water just stays if \( m\omega^2 r = mg \).
2. \( \omega = \sqrt{\dfrac{g}{r}} = \sqrt{\dfrac{10}{0.8}} = \sqrt{12.5} = 3.536 \) rad/s.

Teacher's Note:
1. Spilling begins when the required centripetal force falls below the weight.
2. The result is independent of the mass of water.