JEE Physics Circular Motion MCQs Set 01

Answer: (b) \( \dfrac{2m}{\sqrt r} \)

Explanation:
1. \( a_c = \dfrac{v^2}{r} = \dfrac{4}{r^2} \Rightarrow v^2 = \dfrac{4}{r} \), so \( v = \dfrac{2}{\sqrt r} \).
2. Momentum \( p = mv = \dfrac{2m}{\sqrt r} \).

Teacher's Note:
1. Write the centripetal acceleration as v²/r and solve for v. Then multiply by the mass.

Answer: (c) \( v\omega \)

Explanation:
1. The acceleration is centripetal, \( a = \omega^2 r \).
2. Since \( v = \omega r \), this equals \( v\omega \).

Teacher's Note:
1. Memorise the three forms: a = v²/r = ω²r = vω. Pick the one that matches the options.

Answer: (b) The magnitude of the velocity is constant but its direction changes

Explanation:
1. Equal angles in equal times means the angular speed, and so the linear speed v = ωr, is constant.
2. The velocity is tangential, so its direction changes continuously.

Teacher's Note:
1. Constant speed does not mean constant velocity. Velocity is a vector and turns as the body goes round.

Answer: (b) 5.3 cm/s

Explanation:
1. Distance in 100 s \( = 7\times 2\pi\times 12 = 528 \) cm.
2. \( v = 528/100 \approx 5.3 \) cm/s.

Teacher's Note:
1. Speed is distance per unit time. Use 2πr for the length of one revolution.

Answer: (c) \( 4\pi^2 \)

Explanation:
1. \( \omega = 2\pi f = 4\pi \) rad/s.
2. \( a = \omega^2 r = 16\pi^2\times 0.25 = 4\pi^2 \) m/s².

Teacher's Note:
1. Convert the radius to metres before using ω²r. The frequency gives ω through 2πf.

Answer: (a) 7 r.p.m.

Explanation:
1. Tension \( T = m\omega^2 r \propto \omega^2 \).
2. Doubling T multiplies ω by \( \sqrt{2} \): \( 5\sqrt{2} \approx 7 \) r.p.m.

Teacher's Note:
1. Tension varies as the square of the angular speed. Doubling the tension raises the speed by only √2.

Answer: (d) a, c, d

Explanation:
1. \( \vec\omega \) is along the axis, perpendicular to the plane of motion, so \( \vec\omega \perp \vec v \) and \( \vec\omega \perp \vec a_c \).
2. The centripetal acceleration is radial and the velocity is tangential, so \( \vec v \perp \vec a_c \).
3. For rotation about a fixed axis \( \vec\alpha \) is parallel to \( \vec\omega \), so (b) is wrong. Correct: a, c, d.

Teacher's Note:
1. Visualise the axis, radius and tangent as three mutually perpendicular directions. Angular velocity and angular acceleration both lie along the axis.

Answer: (b) Its kinetic energy is constant

Explanation:
1. A force perpendicular to velocity does no work, so the kinetic energy and the speed stay constant.
2. The direction of the velocity changes, so the particle moves in a circle with the acceleration changing in direction.

Teacher's Note:
1. Perpendicular force means circular motion at constant speed. The magnitude of acceleration is constant but its direction is not.

Answer: (c) 24 rad/s

Explanation:
1. \( \omega = \dfrac{d\theta}{dt} = 6t^2 \).
2. At t = 2 s, \( \omega = 6\times 4 = 24 \) rad/s.

Teacher's Note:
1. Angular velocity is the time derivative of the angular position. The constant 0.5 disappears on differentiation.

Answer: (c) Zero

Explanation:
1. \( \alpha = d\omega/dt = 0 \) because ω is constant.
2. The centripetal acceleration \( r\omega^2 \) is not the angular acceleration.

Teacher's Note:
1. Do not confuse angular acceleration α (rate of change of ω) with centripetal acceleration ω²r.

Answer: (d) The change in momentum is 2mv

Explanation:
1. The speed is unchanged, so the kinetic energy does not change.
2. The velocity reverses, so the change in momentum is \( mv - (-mv) = 2mv \).

Teacher's Note:
1. Momentum is a vector, so reversing the velocity gives a change of 2mv. Kinetic energy is a scalar and does not change.

Answer: (a) 493 cm/s²

Explanation:
1. Frequency \( f = 0.5 \) Hz, so \( \omega = 2\pi f = \pi \) rad/s.
2. \( a = \omega^2 r = \pi^2\times 50 \approx 493 \) cm/s².

Teacher's Note:
1. Frequency is the number of revolutions per second. Keep the radius in centimetres to get the answer in cm/s².

Answer: (d) Velocity is transverse and acceleration has both radial and transverse components

Explanation:
1. The velocity is always tangential (transverse).
2. The acceleration has a radial part v²/r (change of direction) and a tangential part dv/dt (change of speed).

Teacher's Note:
1. Non-uniform circular motion adds a tangential acceleration to the usual centripetal one. The velocity is still along the tangent.

Answer: (b) 10.47 rad/s

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

Teacher's Note:
1. Convert revolutions per minute to radians per second by multiplying by 2π/60.

Answer: (c) 1

Explanation:
1. \( \omega = 2\pi/T \).
2. Equal periods give equal angular velocities, so the ratio is 1.

Teacher's Note:
1. Angular velocity depends only on the time period. Neither the mass nor the radius matters.

Answer: (d) \( \dfrac{\pi}{1800} \) rad/s

Explanation:
1. The minute hand completes one revolution in 3600 s.
2. \( \omega = \dfrac{2\pi}{3600} = \dfrac{\pi}{1800} \) rad/s.

Teacher's Note:
1. Option (c) is twice the true value because it uses 1800 s as the period. The period of the minute hand is 3600 s.

Answer: (d) 2.7 m/s²

Explanation:
1. Centripetal acceleration \( = v^2/r = 900/500 = 1.8 \) m/s².
2. Tangential acceleration = 2 m/s².
3. \( a = \sqrt{1.8^2 + 2^2} = \sqrt{7.24} \approx 2.7 \) m/s².

Teacher's Note:
1. Add the radial and tangential accelerations as perpendicular vectors, using Pythagoras.

Answer: (a) 1.88 m/s, 35.5 m/s², 35.5 N

Explanation:
1. \( \omega = 2\pi\times 3 = 18.85 \) rad/s.
2. \( v = \omega r = 1.885 \) m/s; \( a = \omega^2 r = 35.5 \) m/s².
3. \( T = ma = 1\times 35.5 = 35.5 \) N.

Teacher's Note:
1. Tension supplies the centripetal force when gravity is ignored. Compute ω first and everything else follows.

Answer: (a) 40 m/s²

Explanation:
1. Distance for two revolutions \( s = 2\times 2\pi r = 4\pi\times\dfrac{20}{\pi} = 80 \) m.
2. \( v^2 = 2a_t s \Rightarrow 6400 = 2a_t(80) \Rightarrow a_t = 40 \) m/s².

Teacher's Note:
1. Use the linear equation v² = 2as along the arc. The radius cancels with π.

Answer: (b) 35 m/s

Explanation:
1. \( v = \omega r = 70\times 0.5 = 35 \) m/s.

Teacher's Note:
1. Direct use of v = ωr. Check that the angular velocity is in rad/s.

Answer: (b) \( \pi^2 \) m/s², along the radius towards the centre

Explanation:
1. \( f = 22/44 = 0.5 \) Hz, so \( \omega = \pi \) rad/s.
2. \( a = \omega^2 r = \pi^2\times 1 = \pi^2 \) m/s², directed towards the centre.

Teacher's Note:
1. In uniform circular motion the acceleration always points to the centre. There is no tangential part.

Answer: (a) \( \dfrac{\pi}{6} \)

Explanation:
1. \( \omega_0 = 600\times\dfrac{2\pi}{60} = 20\pi \) rad/s.
2. \( \alpha = \dfrac{\omega_0}{t} = \dfrac{20\pi}{120} = \dfrac{\pi}{6} \) rad/s².

Teacher's Note:
1. Convert rev/min to rad/s first. Time must also be in seconds (2 minutes = 120 s).

Answer: (d) 2 m

Explanation:
1. The tangential acceleration is \( a_t = \alpha r \).
2. \( r = \dfrac{a_t}{\alpha} = \dfrac{10}{5} = 2 \) m.

Teacher's Note:
1. Tangential acceleration equals αr. Here the linear acceleration is taken as the tangential one.

Answer: (b) Zero

Explanation:
1. Uniform speed means ω is constant.
2. \( \alpha = d\omega/dt = 0 \).

Teacher's Note:
1. Constant speed means no tangential acceleration, which is the same as zero angular acceleration.

Answer: (c) Tripled

Explanation:
1. \( F = \dfrac{mv^2}{r} \). With v → 3v and r → 3r: \( F' = \dfrac{m(9v^2)}{3r} = 3F \).

Teacher's Note:
1. The speed enters as a square and the radius in the denominator. Substitute both changes together.