JEE Physics Circular Motion MCQs Set 14

Answer: (b) 200 rad

Explanation:
1. \( \theta = \frac{\omega_0+\omega}{2}t = \frac{0+80}{2}\times5 = 200 \) rad.

Teacher's Note:
1. Average angular velocity is 40 rad/s.
2. Angular displacement = average angular velocity × time.

Answer: (a) \( 2\pi \) mm/s and \( 2\sqrt2\,\pi \) mm/s

Explanation:
1. \( \omega = \frac{2\pi}{60} \) rad/s, so \( v = \omega r = \frac{2\pi}{60}\times60 \text{ mm} = 2\pi \) mm/s.
2. At perpendicular positions the velocities are perpendicular, so \( |\Delta\vec v| = v\sqrt2 = 2\sqrt2\,\pi \) mm/s.

Teacher's Note:
1. Convert 6 cm to 60 mm.
2. Perpendicular vectors of equal magnitude differ by √2 times the magnitude.

Answer: (c) r

Explanation:
1. Chord \( = 2r\sin\frac{\theta}{2} = 2r\sin30^\circ = r \).

Teacher's Note:
1. The chord length for angle θ is 2r sin(θ/2).
2. For 60° the triangle formed is equilateral.

Answer: (d) \( 2a\sin\frac{\omega t}{2} \)

Explanation:
1. Angle turned \( = \omega t \), so chord \( = 2a\sin\frac{\omega t}{2} \).

Teacher's Note:
1. Displacement is the chord.
2. It is zero at t = 0 and maximum at half a period.

Answer: (b) 2/17 rad/s

Explanation:
1. \( \vec v = 2\hat i + 4t\hat j \), so \( \tan\theta = 2t \).
2. Differentiating: \( \sec^2\theta\,\dot\theta = 2 \), so \( \dot\theta = \frac{2}{1+4t^2} = \frac{2}{17} \) rad/s at t = 2 s.

Teacher's Note:
1. Find tan θ as a function of t.
2. Differentiate with respect to time.

Answer: (a) 1 rad/s

Explanation:
1. Line of sight length \( r = \frac{30}{\cos30^\circ} = 20\sqrt3 \) m.
2. Component of the car's velocity perpendicular to the line of sight \( = 40\cos30^\circ = 20\sqrt3 \) m/s, so \( \omega = \frac{20\sqrt3}{20\sqrt3} = 1 \) rad/s.

Teacher's Note:
1. Angular speed = perpendicular velocity / distance.
2. Resolve the velocity perpendicular to the line of sight.

Answer: (b) perpendicular to each other

Explanation:
1. Acceleration is towards the centre and velocity is tangential.
2. They are always perpendicular.

Teacher's Note:
1. Uniform speed means no tangential component of acceleration.
2. Radial and tangential directions are perpendicular.

Answer: (a) The magnitude of acceleration does not remain constant.

Explanation:
1. The magnitude of the centripetal acceleration \( v^2/r \) is constant in uniform circular motion, so that statement is incorrect.
2. Direction of acceleration changes continuously (always towards the centre).

Teacher's Note:
1. Uniform means constant speed.
2. Acceleration vector rotates but its magnitude is fixed.

Answer: (c) 2 m

Explanation:
1. Tangential acceleration \( a = \alpha r \), so \( r = \frac{10}{5} = 2 \) m.

Teacher's Note:
1. a_t = α r.
2. Treat the linear acceleration as tangential.

Answer: (d) 8 N

Explanation:
1. \( T = \frac{mv^2}{r} = \frac{0.5\times16}{1} = 8 \) N.

Teacher's Note:
1. Tension supplies the centripetal force.
2. Horizontal circle ignoring gravity.

Answer: (b) 0.5%

Explanation:
1. \( T_{tension} = m\omega^2\ell = m\frac{4\pi^2}{T^2}\ell \), so \( T^2 \propto \ell \) for constant tension, i.e. \( T \propto \sqrt\ell \).
2. A 1% increase in ℓ gives a 0.5% increase in the period.

Teacher's Note:
1. Keep the tension constant and solve for period.
2. Use the small change approximation.

Answer: (a) constant speed and variable velocity

Explanation:
1. A force perpendicular to the velocity changes only the direction.
2. Speed stays constant while the velocity vector keeps rotating.

Teacher's Note:
1. Velocity is a vector; its direction changes.
2. Constant magnitude of centripetal force gives uniform circular motion.

Answer: (b) 40 N

Explanation:
1. \( \omega = 2\pi\times\frac{10}{\pi} = 20 \) rad/s.
2. \( T = m\omega^2r = 0.1\times400\times1 = 40 \) N.

Teacher's Note:
1. Convert revolutions per second to rad/s.
2. T = mω²r.

Answer: (a) \( 60^\circ \)

Explanation:
1. \( \omega = 2\pi\cdot\frac{\sqrt5}{\pi} = 2\sqrt5 \) rad/s, so \( \omega^2 = 20 \).
2. \( \cos\theta = \frac{g}{\omega^2l} = \frac{10}{20} = \frac12 \), so \( \theta = 60^\circ \).

Teacher's Note:
1. Conical pendulum: T cos θ = mg and T sin θ = m ω² l sin θ.
2. cos θ = g/(ω² l).

Answer: (c) 640 N

Explanation:
1. Quarter circle: \( \frac{\pi}{2}R = 628 \), so \( R = 400 \) m.
2. \( F = \frac{mv^2}{R} = \frac{1000\times256}{400} = 640 \) N.

Teacher's Note:
1. The arc length of 90° is πR/2.
2. Then use mv²/R.

Answer: (d) the car will not slip when moving with speed v

Explanation:
1. At the design speed the horizontal component of the normal reaction supplies exactly \( \frac{mv^2}{r} \): \( \tan\theta = \frac{v^2}{rg} \).
2. So no friction is needed and the car does not slip at that speed.

Teacher's Note:
1. Banking angle is set for one speed.
2. Friction extends the safe range.

Answer: (b) the cyclist has to counteract the centrifugal force, while in the case of the car only the passenger is thrown by this force

Explanation:
1. In the cyclist's frame the outward centrifugal tendency is balanced by leaning inward so that the resultant passes through the contact point.
2. A car passenger has no such compensation and tends to move outward relative to the car.

Teacher's Note:
1. Banking and leaning both reduce the outward tendency.
2. It is not a matter of mass or number of wheels.

Answer: (a) \( \sqrt{\mu rg} \)

Explanation:
1. Maximum friction \( \mu mg \ge \frac{mv^2}{r} \), so \( v_{max} = \sqrt{\mu rg} \).

Teacher's Note:
1. Friction is the only horizontal force.
2. Skidding starts when the demand exceeds μmg.

Answer: (b) \( mg \le \frac{mv^2}{R} \)

Explanation:
1. At the top the weight and the bucket's push provide the centripetal force: \( mg + N = \frac{mv^2}{R} \) with \( N \ge 0 \).
2. So \( mg \le \frac{mv^2}{R} \).

Teacher's Note:
1. The water stays in the bucket if the required centripetal force is at least mg.
2. Minimum speed gives equality: v² = gR.

Answer: (a) \( \frac{2g}{a} \)

Explanation:
1. Horizontal: \( N\sin\theta = m\omega^2(a\sin\theta) \), so \( N = m\omega^2a \). Vertical: \( N\cos\theta = mg \).
2. \( \omega^2 = \frac{g}{a\cos\theta} = \frac{g}{a\cos60^\circ} = \frac{2g}{a} \).

Teacher's Note:
1. Radius of the bead's circle is a sin θ.
2. Eliminate N.

Answer: (c) \( \sqrt{\frac{12g\ell}{5}} \)

Explanation:
1. Speed of the middle mass is \( v/2 \). Energy: \( \tfrac12mv^2 + \tfrac12m\frac{v^2}{4} = mg\ell + mg\frac\ell2 \).
2. \( \frac58v^2 = \frac32g\ell \), so \( v = \sqrt{\frac{12g\ell}{5}} \).

Teacher's Note:
1. Both masses are on the same rigid rod.
2. Speeds are proportional to distances from the pivot.

Answer: (a) \( 120^\circ \)

Explanation:
1. Let θ be the angle from the downward vertical. \( v^2 = \frac{7gL}{2} - 2gL(1-\cos\theta) \) and tension zero when \( \frac{mv^2}{L} + mg\cos\theta = 0 \).
2. \( \frac{3}{2}gL + 3gL\cos\theta = 0 \) gives \( \cos\theta = -\frac12 \), so \( \theta = 120^\circ \).

Teacher's Note:
1. Thread slackens when the tension becomes zero.
2. Radial equation: T + mg cos θ = mv²/L, measured from the downward vertical.

Answer: (d) lowest: mg − T₁; highest: mg + T₂

Explanation:
1. At the lowest point the string tension T₁ acts up and weight down, so the downward net force is mg − T₁.
2. At the highest point both tension and weight act down: mg + T₂.

Teacher's Note:
1. Take downward as positive.
2. Tension always points towards the centre of the circle.

Answer: (b) \( \sqrt{\frac{g\ell}{3}} \), \( \cos^{-1}\frac23 \)

Explanation:
1. At angle φ from the downward vertical: \( v^2 = g\ell - 2g\ell(1-\cos\phi) = g\ell(2\cos\phi-1) \).
2. \( T = \frac{mv^2}{\ell} + mg\cos\phi = mg(3\cos\phi-1) = mg \), so \( \cos\phi = \frac23 \) and \( v^2 = \frac{g\ell}{3} \).

Teacher's Note:
1. Write energy conservation and the radial equation.
2. Set T = mg.

Answer: (b) 10 m/s

Explanation:
1. \( T_{max} = \frac{mv_t^2}{\ell} + 5mg \) and \( T_{min} = \frac{mv_t^2}{\ell} - mg \). With ratio 4: \( X + 5mg = 4X - 4mg \), \( X = 3mg \).
2. \( v_t^2 = 3g\ell = 3\times10\times\frac{10}{3} = 100 \), so \( v_t = 10 \) m/s.

Teacher's Note:
1. Maximum tension is at the lowest point; minimum at the top.
2. Bottom speed² = top speed² + 4gℓ.

Answer: (c) \( \sqrt{gR} \)

Explanation:
1. At the top the particle leaves the surface if \( \frac{mv^2}{R} \ge mg \).
2. So the smallest speed is \( \sqrt{gR} \).

Teacher's Note:
1. Normal reaction becomes zero at the leaving point.
2. At the top the centripetal requirement alone decides.

Answer: (a) R/2

Explanation:
1. Drop in height from A to B: \( R(\cos37^\circ-\cos53^\circ) = 0.2R \), so \( v^2 = 0.4gR \).
2. At B the velocity is tangent to the track (37° with the horizontal); the component of g perpendicular to the velocity is \( g\cos37^\circ = 0.8g \). Radius of curvature \( = \frac{v^2}{0.8g} = \frac R2 \).

Teacher's Note:
1. Use energy conservation for the speed.
2. Radius of curvature uses the component of g perpendicular to the velocity.