JEE Physics Circular Motion MCQs Set 04

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

Explanation:
1. In uniform circular motion the acceleration is centripetal, \( a = \omega^2 r \).
2. Since \( v = \omega r \), this can be written \( a = v\omega \).

Teacher's Note:
1. Centripetal acceleration has three equivalent forms: \( v^2/r \), \( \omega^2 r \) and \( v\omega \).
2. Check the dimensions of every option before choosing.

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

Explanation:
1. Equal angles in equal times means constant angular speed, so the speed \( v = \omega r \) is constant.
2. The velocity is along the tangent, and the tangent direction keeps turning, so only the direction changes.

Teacher's Note:
1. Uniform circular motion has constant speed but changing velocity.
2. The acceleration, not the velocity, points to the centre.

Answer: (d) \( \dfrac{P^2}{mr} \)

Explanation:
1. \( v = P/m \).
2. \( F = \dfrac{mv^2}{r} = \dfrac{m}{r}\cdot\dfrac{P^2}{m^2} = \dfrac{P^2}{mr} \).

Teacher's Note:
1. Replace v by P/m in the centripetal force formula.
2. The result \( P^2/(mr) \) is a handy form when momentum is given.

Answer: (d) a, c, d

Explanation:
1. The angular velocity vector is along the axis, perpendicular to the plane of motion, so it is perpendicular to both \( \vec{v} \) and \( \vec{a}_c \): (a) and (d) hold.
2. The centripetal acceleration is radial and the velocity is tangential, so (c) holds.
3. For rotation about a fixed axis \( \vec{\alpha} \) is parallel to \( \vec{\omega} \), so (b) is false.

Teacher's Note:
1. Both \( \vec{\omega} \) and \( \vec{\alpha} \) lie along the axis for fixed-axis rotation.
2. Velocity and centripetal acceleration lie in the plane of the circle.

Answer: (a) 1.1 cm/s

Explanation:
1. Period \( T = 200/5 = 40 \) s, so \( \omega = 2\pi/40 \) rad/s.
2. \( v = \omega r = \dfrac{2\pi}{40}\times 7 = \dfrac{2\times 22\times 7}{7\times 40} = 1.1 \) cm/s.

Teacher's Note:
1. Find the period first, then use \( v = 2\pi r/T \).
2. Using \( \pi = 22/7 \) makes the arithmetic clean.

Answer: (d) \( 32\pi^2 \)

Explanation:
1. \( \omega = 2\pi f = 8\pi \) rad/s and \( r = 0.5 \) m.
2. \( a = \omega^2 r = 64\pi^2 \times 0.5 = 32\pi^2 \) m/s\( ^2 \).

Teacher's Note:
1. Always convert the radius to metres when the answer is asked in m/s\( ^2 \).
2. Frequency f in rev/s gives \( \omega = 2\pi f \).

Answer: (b) its K.E. is constant

Explanation:
1. A force perpendicular to the velocity does no work, so kinetic energy does not change.
2. The speed is constant, but the direction of velocity and of acceleration keeps changing, and the path is a circle.

Teacher's Note:
1. Zero work means constant kinetic energy and constant speed.
2. The acceleration has constant magnitude but not constant direction.

Answer: (c) Zero

Explanation:
1. Angular acceleration \( \alpha = \dfrac{d\omega}{dt} \).
2. The angular velocity is constant, so \( \alpha = 0 \).

Teacher's Note:
1. Do not confuse angular acceleration with centripetal acceleration \( r\omega^2 \).
2. Zero angular acceleration still has non-zero centripetal acceleration.

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

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

Teacher's Note:
1. Momentum is a vector, so reversal of direction doubles the change.
2. Kinetic energy is a scalar and depends only on speed.

Answer: (d) velocity is tangential and acceleration has both radial and tangential components

Explanation:
1. The velocity of a particle on a circle is always along the tangent.
2. When the speed changes there is a tangential component of acceleration in addition to the radial (centripetal) component.

Teacher's Note:
1. Uniform motion has only the radial component.
2. Non-uniform motion adds a tangential component.

Answer: (b) 5.23 rad/s

Explanation:
1. \( \omega = \dfrac{2\pi \times 50}{60} = 5.23 \) rad/s.

Teacher's Note:
1. Divide by 60 to convert minutes to seconds.
2. Do not mix radians and degrees.

Answer: (a) \( \dfrac{r}{R} \)

Explanation:
1. \( \omega = v/\text{radius} \).
2. \( \dfrac{\omega_M}{\omega_m} = \dfrac{v/R}{v/r} = \dfrac{r}{R} \).

Teacher's Note:
1. For equal linear speed, the smaller circle has the larger angular velocity.
2. The masses do not matter here.

Answer: (c) \( \dfrac{7\pi}{6} \) m/s

Explanation:
1. \( 60^\circ/\text{s} = \pi/3 \) rad/s.
2. \( v = \omega r = \dfrac{\pi}{3}\times 3.5 = \dfrac{7\pi}{6} \) m/s.

Teacher's Note:
1. Convert degrees to radians before using \( v = \omega r \).
2. \( 180^\circ = \pi \) rad.

Answer: (b) 39.4 cm/s\( ^2 \)

Explanation:
1. Period \( T = 5 \) s, so \( \omega = 2\pi/5 \) rad/s.
2. \( a = \omega^2 r = \dfrac{4\pi^2}{25}\times 25 = 4\pi^2 \approx 39.4 \) cm/s\( ^2 \).

Teacher's Note:
1. Work in cm throughout when the options are in cm/s\( ^2 \).
2. Constant speed means the acceleration is purely centripetal.

Answer: (c) \( \tan 30^\circ \)

Explanation:
1. The tangential acceleration lies along the velocity and the centripetal acceleration is perpendicular to it.
2. The angle between the total acceleration and the velocity satisfies \( \tan\phi = a_c/a_t \), so the ratio is \( \tan 30^\circ \).

Teacher's Note:
1. Resolve the total acceleration along and perpendicular to the velocity.
2. Opposite over adjacent gives the tangent.

Answer: (d) 2.2 m/s\( ^2 \)

Explanation:
1. \( a_c = v^2/r = 400/400 = 1 \) m/s\( ^2 \) and \( a_t = 2 \) m/s\( ^2 \).
2. \( a = \sqrt{1^2 + 2^2} = \sqrt{5} \approx 2.2 \) m/s\( ^2 \).

Teacher's Note:
1. The two components are at right angles, so add them as vectors.
2. Use Pythagoras for the resultant.

Answer: (a) \( \sqrt{257} \)

Explanation:
1. \( \omega = 2 + 1\times 2 = 4 \) rad/s.
2. \( a_c = \omega^2 r = 16 \), \( a_t = \alpha r = 1 \).
3. \( a = \sqrt{16^2 + 1^2} = \sqrt{257} \) m/s\( ^2 \).

Teacher's Note:
1. Find the angular velocity at that instant before the centripetal acceleration.
2. Add radial and tangential parts at right angles.

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

Explanation:
1. Period \( T = 44/22 = 2 \) s, so \( \omega = \pi \) rad/s.
2. \( a = \omega^2 r = \pi^2 \) m/s\( ^2 \).
3. At constant speed the acceleration is centripetal, towards the centre.

Teacher's Note:
1. Constant speed means no tangential acceleration.
2. The direction is always towards the centre.

Answer: (b) \( \dfrac{1}{2} \) m

Explanation:
1. \( a_t = \alpha r \).
2. \( r = \dfrac{20}{40} = \dfrac{1}{2} \) m.

Teacher's Note:
1. Tangential acceleration equals angular acceleration times radius.
2. Check units: m/s\( ^2 \) divided by rad/s\( ^2 \) gives metres.

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

Explanation:
1. \( \omega_0 = 1200\times\dfrac{2\pi}{60} = 40\pi \) rad/s and \( t = 240 \) s.
2. \( \alpha = \dfrac{\omega_0}{t} = \dfrac{40\pi}{240} = \dfrac{\pi}{6} \) rad/s\( ^2 \).

Teacher's Note:
1. Convert both rev/min and minutes to SI units.
2. Uniform deceleration: \( \omega = \omega_0 - \alpha t \).

Answer: (b) \( 4\sqrt{2} \) m/s\( ^2 \)

Explanation:
1. At t = 2 s, \( v = 4 \) m/s.
2. \( a_t = \dfrac{dv}{dt} = 2t = 4 \) m/s\( ^2 \) and \( a_c = \dfrac{v^2}{r} = \dfrac{16}{4} = 4 \) m/s\( ^2 \).
3. \( a = \sqrt{4^2+4^2} = 4\sqrt{2} \) m/s\( ^2 \).

Teacher's Note:
1. Tangential acceleration is the rate of change of speed.
2. Both components are equal here, so the angle with the velocity is 45°.

Answer: (b) 4 m/s\( ^2 \)

Explanation:
1. \( \alpha = \dfrac{d\omega}{dt} = 2t - 2 = 2 \) rad/s\( ^2 \) at t = 2 s.
2. \( a_t = \alpha r = 2\times 2 = 4 \) m/s\( ^2 \).

Teacher's Note:
1. Differentiate \( \omega \) to get \( \alpha \).
2. Then multiply by r for the tangential acceleration.

Answer: (d) \( \dfrac{V}{2d} \)

Explanation:
1. OA has length \( \sqrt{d^2+d^2} = \sqrt{2}\,d \) and makes 45° with the horizontal.
2. The component of V perpendicular to OA is \( V\sin 45^\circ = \dfrac{V}{\sqrt{2}} \).
3. \( \omega = \dfrac{V/\sqrt{2}}{\sqrt{2}\,d} = \dfrac{V}{2d} \).

Teacher's Note:
1. Use only the velocity component perpendicular to the line of sight.
2. Divide it by the distance OA, not by d.

Answer: (b) accelerated circular motion

Explanation:
1. The velocity is along \( \vec{p} \).
2. \( \vec{a}\cdot\vec{p} = 32 - 18 = 14 \gt 0 \), so the acceleration has a component along the velocity and the speed is increasing.
3. So it is accelerated circular motion.

Teacher's Note:
1. The dot product of a and v tells whether the speed rises or falls.
2. For uniform circular motion it would be zero.

Answer: (a) 9 rad

Explanation:
1. Angle in the n-th second from rest: \( \theta_n = \tfrac{1}{2}\alpha(2n-1) \).
2. \( \theta_5 = \tfrac{1}{2}\times 2\times 9 = 9 \) rad.

Teacher's Note:
1. Fifth-second angle is the angle in 5 s minus the angle in 4 s.
2. That is \( 25 - 16 = 9 \) rad.