JEE Physics Circular Motion MCQs Set 07

Answer: (d) 24 rad/s

Explanation:
1. \( \omega = \dfrac{d\theta}{dt} = 6t^2 \).
2. At t = 2 s, \( \omega = 6\times4 = 24 \) rad/s.

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

Answer: (d) \( \dfrac{\pi}{30} \) rad/s

Explanation:
1. The second hand completes one revolution in T = 60 s.
2. \( \omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{60} = \dfrac{\pi}{30} \) rad/s.

Teacher's Note:
1. Use \( \omega = 2\pi/T \). Similarly, the minute hand has T = 3600 s.

Answer: (a) \( -18\hat i - 13\hat j + 2\hat k \)

Explanation:
1. \( \vec v = \vec\omega\times\vec r \), expanded as a determinant with rows \( (3,-4,1) \) and \( (5,-6,6) \).
2. \( \hat i[(-4)(6)-(1)(-6)] - \hat j[(3)(6)-(1)(5)] + \hat k[(3)(-6)-(-4)(5)] \).
3. \( = -18\hat i - 13\hat j + 2\hat k \).

Teacher's Note:
1. The order matters in a cross product: \( \vec\omega\times\vec r \) and \( \vec r\times\vec\omega \) have opposite signs.

Answer: (a) \( 2R\sin\dfrac{\omega t}{2} \)

Explanation:
1. The angle turned is \( \theta = \omega t \). The displacement is the chord of this angle.
2. Chord \( = 2R\sin\dfrac{\theta}{2} = 2R\sin\dfrac{\omega t}{2} \).

Teacher's Note:
1. Displacement is the straight-line chord, not the arc. The arc length would be Rωt.

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

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. First find the speed from a = v²/r, then multiply by the mass.

Answer: (d) a, c, d

Explanation:
1. \( \vec\omega \) is along the axis, perpendicular to the plane of the circle, while \( \vec v \) and \( \vec a_c \) lie in that plane. So (a) and (d) are true.
2. \( \vec a_c \) points to the centre and \( \vec v \) is tangential, so (c) is true.
3. For circular motion in a plane, \( \vec\alpha \) is also along the axis, parallel or antiparallel to \( \vec\omega \), so (b) is false.

Teacher's Note:
1. Use the direction of each vector: the axis for ω and α, the tangent for v, and the radius for the centripetal acceleration.

Answer: (d) 2

Explanation:
1. Same force: \( \dfrac{2m v_1^2}{r} = \dfrac{m v_2^2}{2r} \).
2. \( v_2^2 = 4v_1^2 \Rightarrow v_2 = 2v_1 \), so n = 2.

Teacher's Note:
1. Equate the two expressions mv²/r and solve for the ratio of speeds.

Answer: (c) The magnitude of the acceleration is constant

Explanation:
1. The direction of velocity changes continuously, so the velocity is not constant.
2. With constant speed on a path of constant radius, the acceleration is centripetal with constant magnitude \( v^2/r \), but its direction keeps changing.

Teacher's Note:
1. Constant magnitude does not mean constant vector. A vector is constant only if both magnitude and direction are fixed.

Answer: (b) The tension T along the thread towards O, the weight W downwards and a centrifugal force F horizontally outward

Explanation:
1. In the rotating frame the mass is at rest, so the net force must be zero.
2. Gravity W acts downwards and the tension T acts along the thread towards O. Their resultant points inward, so a pseudo (centrifugal) force F acts horizontally outward to balance it.

Teacher's Note:
1. In an inertial frame there is no centrifugal force, only T and W. The pseudo force appears only when we describe the motion from the rotating frame.

Answer: (b) \( \sqrt{\left(\dfrac{v^2}{R}\right)^2 + a^2} \)

Explanation:
1. The radial (centripetal) acceleration is \( v^2/R \) and the tangential acceleration has magnitude a.
2. These two are perpendicular, so \( a_{net} = \sqrt{(v^2/R)^2 + a^2} \).

Teacher's Note:
1. Radial and tangential components are perpendicular, so add them as vectors, not as plain numbers.

Answer: (d) \( T \)

Explanation:
1. On the smooth table, the only horizontal force on the particle is the tension T, directed towards the centre.
2. So the net force towards the centre is T, and it equals \( mv^2/l \).

Teacher's Note:
1. Centripetal force is not a new force. It is just the name for the net inward force, here the tension.

Answer: (c) \( \theta = \tan^{-1}2 \)

Explanation:
1. 72 km/h = 20 m/s.
2. \( \tan\theta = \dfrac{v^2}{Rg} = \dfrac{400}{20\times10} = 2 \).
3. So \( \theta = \tan^{-1}2 \).

Teacher's Note:
1. Convert km/h to m/s before substituting (divide by 3.6).

Answer: (b) \( \sqrt{2g\ell} \)

Explanation:
1. To rise by ℓ, the kinetic energy must equal the gain in potential energy: \( \tfrac12mv^2 = mg\ell \).
2. \( v = \sqrt{2g\ell} \).

Teacher's Note:
1. Up to the horizontal level the string stays taut, so energy conservation is enough.

Answer: (c) \( \dfrac{ML\omega^2}{2} \)

Explanation:
1. The liquid acts as if its whole mass M were at its centre of mass, at distance L/2 from the axis.
2. Force \( = M\omega^2\dfrac{L}{2} = \dfrac{ML\omega^2}{2} \).

Teacher's Note:
1. For a uniform body in rotation, apply Mω²r with r measured to the centre of mass.

Answer: (d) 45°

Explanation:
1. The bob moves in a circle with the car: \( a = v^2/r = 100/10 = 10 \) m/s².
2. \( \tan\theta = a/g = 10/10 = 1 \Rightarrow \theta = 45^\circ \).

Teacher's Note:
1. A hanging bob in an accelerating frame tilts with tan θ = a/g. The rod length is irrelevant.

Answer: (b) \( g\sqrt{3\sin^2\theta+1} \)

Explanation:
1. By energy conservation, \( v^2 = 2g\ell\sin\theta \), so the centripetal acceleration is \( a_c = v^2/\ell = 2g\sin\theta \).
2. The tangential acceleration is \( a_t = g\cos\theta \).
3. \( a = \sqrt{4g^2\sin^2\theta + g^2\cos^2\theta} = g\sqrt{3\sin^2\theta+1} \).

Teacher's Note:
1. Treat the motion as non-uniform circular motion: find the radial part from energy and the tangential part from the component of gravity.

Answer: (d) \( h \geq \dfrac{5r}{2} \)

Explanation:
1. The minimum speed at the top of the loop is \( \sqrt{gr} \).
2. Energy: \( mgh = mg(2r) + \tfrac12m(gr) \), giving \( h = \tfrac{5r}{2} \) as the minimum.
3. So the block must start at \( h \geq \tfrac{5r}{2} \).

Teacher's Note:
1. The height is measured from the bottom of the loop. Including the equality case, the block just manages to complete the circle.

Answer: (a) \( \dfrac{v^2b}{Rg} \)

Explanation:
1. For safe passage without friction, \( \tan\theta = \dfrac{v^2}{Rg} \).
2. For a small banking angle, \( \tan\theta \approx \sin\theta = \dfrac{h}{b} \).
3. So \( h = \dfrac{v^2b}{Rg} \).

Teacher's Note:
1. Banked roads have small angles, so h/b is used for tan θ.

Answer: (d) \( \tan\theta = \dfrac{r\omega^2}{g} \)

Explanation:
1. The bob moves in a circle of radius r with angular speed ω, so it needs a centripetal acceleration \( \omega^2r \) towards the centre.
2. The horizontal component of tension supplies this: \( T\sin\theta = m\omega^2r \), and \( T\cos\theta = mg \).
3. Dividing, \( \tan\theta = \dfrac{r\omega^2}{g} \).

Teacher's Note:
1. Resolve the tension: the vertical part balances the weight, the horizontal part gives the centripetal force.

Answer: (d) x = 2L

Explanation:
1. The minimum speed at the top is \( v = \sqrt{gL} \), horizontal.
2. The body falls a height 2L to the level of the lowest point: \( 2L = \tfrac12gt^2 \Rightarrow t = 2\sqrt{L/g} \).
3. Range \( x = vt = \sqrt{gL}\times2\sqrt{L/g} = 2L \).

Teacher's Note:
1. After the string breaks, the body is a horizontal projectile. Find the speed from the circular condition, then use the fall time.

Answer: (c) 24.20 m

Explanation:
1. For a fixed banking angle, \( v^2 = Rg\tan\theta \), so \( R \propto v^2 \).
2. \( R' = 20\times(1.1)^2 = 20\times1.21 = 24.2 \) m.

Teacher's Note:
1. A 10% rise in speed needs a 21% rise in radius, because radius goes as the square of speed.

Answer: (d) 3 : 5 : 6

Explanation:
1. All three have the same ω. Take \( m\omega^2\ell = k \). The radii are ℓ, 2ℓ and 3ℓ for A, B and C.
2. Section BC: \( T_{BC} = m\omega^2(3\ell) = 3k \).
3. Section AB: \( T_{AB} - T_{BC} = m\omega^2(2\ell) \Rightarrow T_{AB} = 5k \).
4. Section OA: \( T_{OA} - T_{AB} = m\omega^2\ell \Rightarrow T_{OA} = 6k \).
5. So BC : AB : OA = 3 : 5 : 6.

Teacher's Note:
1. Write the equation of motion for each particle separately, starting from the outermost one.

Answer: (b) \( \sqrt{4g\ell} \)

Explanation:
1. A rod can push as well as pull, so the mass can just reach the top with zero speed.
2. Energy: \( \tfrac12mv^2 = mg(2\ell) \Rightarrow v = \sqrt{4g\ell} \).

Teacher's Note:
1. With a string the minimum bottom speed is √(5gℓ). With a rod it is √(4gℓ), because the top speed can be zero.

Answer: (a) \( \sqrt{2(u^2-g\ell)} \)

Explanation:
1. At the lowest point, the velocity is u horizontally. At the horizontal position the speed v satisfies \( v^2 = u^2 - 2g\ell \), and the velocity is vertical.
2. The two velocities are perpendicular: \( |\Delta\vec v| = \sqrt{u^2+v^2} = \sqrt{2u^2-2g\ell} \).
3. \( = \sqrt{2(u^2-g\ell)} \).

Teacher's Note:
1. Change in velocity is a vector difference. Here the two velocities are at right angles, so use Pythagoras.

Answer: (c) \( r \leq \dfrac{\mu g}{\omega^2} \)

Explanation:
1. The coin needs a centripetal force \( m\omega^2r \), which static friction must supply.
2. Condition: \( m\omega^2r \leq \mu mg \Rightarrow r \leq \dfrac{\mu g}{\omega^2} \).

Teacher's Note:
1. Friction can supply at most μmg. Coins nearer the centre need less force and stay put.

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

Explanation:
1. Friction provides the centripetal force: \( \dfrac{mv^2}{R} \leq \mu_smg \).
2. \( v_{max} = \sqrt{\mu_sRg} \).

Teacher's Note:
1. The maximum safe speed does not depend on the mass of the car.

Answer: (d) 0.1 m/s²

Explanation:
1. \( \tfrac12mv^2 = 8\times10^{-4} \) with m = 0.01 kg gives \( v^2 = 0.16 \) m²/s².
2. Distance in two revolutions: \( s = 2\times2\pi(0.064) \approx 0.804 \) m.
3. \( v^2 = 2a_ts \Rightarrow a_t = \dfrac{0.16}{2\times0.804} \approx 0.1 \) m/s².

Teacher's Note:
1. Use the equation v² = 2as along the arc. Convert grams and centimetres to SI units first.

Answer: (b) \( \sqrt{gR\dfrac{\tan\theta-\mu_s}{1+\mu_s\tan\theta}} \)

Explanation:
1. At the minimum speed the car tends to slide down, so friction acts up the incline.
2. \( N\cos\theta + \mu_sN\sin\theta = mg \) and \( N\sin\theta - \mu_sN\cos\theta = \dfrac{mv^2}{R} \).
3. Dividing: \( v_{min}^2 = Rg\dfrac{\tan\theta-\mu_s}{1+\mu_s\tan\theta} \).

Teacher's Note:
1. Maximum speed has friction down the slope and gives the plus-minus signs swapped. Keep the two cases separate.

Answer: (a) 5.7 m/s

Explanation:
1. The component of the acceleration towards the centre is \( a\cos30^\circ = 7.5\times\dfrac{\sqrt3}{2} = 6.5 \) m/s².
2. \( \dfrac{v^2}{R} = 6.5 \Rightarrow v^2 = 32.5 \).
3. \( v = 5.7 \) m/s.

Teacher's Note:
1. Only the radial component of the acceleration equals v²/R. The tangential component changes the speed.