JEE Physics Circular Motion MCQs Set 05

Answer: (b) 37°

Explanation:
1. For a speeding-up particle the tangential acceleration is along the velocity, which is tangent to the circle at P.
2. The radius OP is perpendicular to the direction \( (3,4) \), so OP is along \( (-4, 3) \).
3. The angle of OP with the negative x-axis satisfies \( \tan\theta = \dfrac{3}{4} \), so \( \theta = 37^\circ \).

Teacher's Note:
1. The radius is perpendicular to the tangent.
2. A 3-4-5 triangle gives the angles 37° and 53°.

Answer: (b) zero

Explanation:
1. Uniform speed on a fixed circle means constant \( \omega = v/r \).
2. So \( \alpha = d\omega/dt = 0 \).

Teacher's Note:
1. Angular acceleration measures change in angular speed.
2. The centripetal acceleration is a separate quantity.

Answer: (a) Centripetal force

Explanation:
1. A net force directed towards the centre is needed to keep changing the direction of velocity.
2. This is the centripetal force.

Teacher's Note:
1. Centrifugal force is a pseudo force seen only in the rotating frame.
2. The real force is always centripetal.

Answer: (d) away from the centre of the path

Explanation:
1. The bob must have the same centripetal acceleration as the car, so the net horizontal force on it points towards the centre.
2. This needs the string to tilt so that the bob is displaced away from the centre and the tension has a horizontal component towards the centre.

Teacher's Note:
1. Seen from the car, the bob swings outward due to the pseudo force.
2. It hangs at a tilt like a car leaning outward.

Answer: (c) \( a_r \ne 0 \) but \( a_t = 0 \)

Explanation:
1. Circular motion needs a radial (centripetal) acceleration, so \( a_r \ne 0 \).
2. Uniform speed means no tangential acceleration, so \( a_t = 0 \).

Teacher's Note:
1. Radial part bends the path and tangential part changes the speed.
2. Uniform circular motion needs only the first.

Answer: (a) perpendicular to each other

Explanation:
1. Velocity is along the tangent and acceleration is along the radius.
2. These two directions are perpendicular.

Teacher's Note:
1. This perpendicularity is why the speed does not change.
2. The acceleration only turns the velocity.

Answer: (c) 3F

Explanation:
1. \( F = m\omega^2 r \) with \( \omega = 2\pi f \) fixed.
2. So \( F \propto r \) and tripling r triples F.

Teacher's Note:
1. At constant frequency the force is proportional to the radius.
2. At constant speed it would be inversely proportional.

Answer: (d) 160 N

Explanation:
1. \( F = \dfrac{mv^2}{r} = \dfrac{0.5\times 64}{0.2} = 160 \) N.

Teacher's Note:
1. Square the speed before dividing by the radius.
2. A quick estimate: 32 divided by 0.2 gives 160.

Answer: (d) a, c

Explanation:
1. The centripetal force \( mv^2/r \) depends on the speed, which is unchanged, so its magnitude is unchanged.
2. It always points towards the centre, whichever way the body circulates, so its direction is unchanged.
3. So (a) and (c) are correct.

Teacher's Note:
1. Centripetal force depends on \( v^2 \), so the sign of velocity is irrelevant.
2. It always points to the centre.

Answer: (a) \( 36\times 10^{21} \)

Explanation:
1. \( F = M\omega^2 r = 6\times10^{24}\times(2\times10^{-7})^2\times 1.5\times10^{11} \).
2. \( = 6\times 4\times 1.5\times 10^{24-14+11} = 36\times 10^{21} \) N.

Teacher's Note:
1. Convert 1.5 \( \times 10^8 \) km to \( 1.5\times 10^{11} \) m.
2. Powers of ten are easiest to add separately.

Answer: (b) due to lack of proper centripetal force

Explanation:
1. On a turn, the friction between tyres and road must supply \( mv^2/r \).
2. If the speed is too high, the available friction is less than the required centripetal force and the car skids outward.

Teacher's Note:
1. Skidding means the required centripetal force exceeds the maximum friction.
2. Gravity and normal reaction act vertically.

Answer: (a) 2.5 m/s, 31.6 m/s\( ^2 \), 31.6 N

Explanation:
1. \( \omega = 2\pi\times 2 = 4\pi \) rad/s.
2. \( v = \omega r = 0.8\pi \approx 2.5 \) m/s.
3. \( a = \omega^2 r = 16\pi^2\times 0.2 \approx 31.6 \) m/s\( ^2 \), and \( T = ma \approx 31.6 \) N.

Teacher's Note:
1. Tension supplies the centripetal force when gravity is ignored.
2. Tension equals \( m\omega^2 r \).

Answer: (a) 0

Explanation:
1. The tension is perpendicular to the displacement at every point.
2. Gravity is perpendicular to the horizontal displacement.
3. So the net work done in a complete circle is zero.

Teacher's Note:
1. Work equals force times displacement along the force.
2. A perpendicular force does no work.

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

Explanation:
1. Vertically: \( T\cos\theta = mg \). Horizontally: \( T\sin\theta = m\omega^2 r \).
2. Dividing, \( \tan\theta = \dfrac{\omega^2 r}{g} \).

Teacher's Note:
1. Resolve the tension into a vertical part that balances weight and a horizontal part that gives centripetal force.
2. Divide the two equations to eliminate T and m.

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

Explanation:
1. Banking angle satisfies \( \tan\theta = \dfrac{v^2}{Rg} \).
2. For small angles \( \tan\theta \approx \sin\theta = \dfrac{h}{b} \).
3. So \( h = \dfrac{v^2 b}{Rg} \).

Teacher's Note:
1. The slope of the road is height divided by width.
2. Compare with \( \tan\theta = v^2/(Rg) \) for no friction.

Answer: (d) 57°

Explanation:
1. \( v = 180 \) km/h \( = 50 \) m/s.
2. \( \tan\theta = \dfrac{v^2}{rg} = \dfrac{2500}{1600} = 1.5625 \), so \( \theta \approx 57^\circ \).

Teacher's Note:
1. Convert km/h to m/s by dividing by 3.6.
2. \( \tan 57^\circ \approx 1.54 \).

Answer: (c) at the extreme positions

Explanation:
1. Along the string, \( T - mg\cos\theta = \dfrac{mv^2}{\ell} \).
2. So \( T = mg\cos\theta \) only when \( v = 0 \), which is at the extreme positions.

Teacher's Note:
1. At the extremes the bob is momentarily at rest.
2. At the mean position tension is greater than mg.

Answer: (c) the string is vertical with the stone at the lowest position

Explanation:
1. \( T = \dfrac{mv^2}{r} + mg\cos\theta \), where \( \theta \) is measured from the downward vertical.
2. At the bottom both the speed and \( \cos\theta \) are maximum, so T is maximum there.

Teacher's Note:
1. The stone is fastest at the lowest point and gravity also pulls outward from the centre there.
2. So that is where the string is most likely to break.

Answer: (a) zero

Explanation:
1. At the top, gravity is vertical (along the radius), perpendicular to the velocity.
2. So the tangential component of acceleration is zero.

Teacher's Note:
1. Gravity at the top is entirely radial.
2. Tangential acceleration appears only at positions away from the top and bottom.

Answer: (c) in a direction perpendicular to the string

Explanation:
1. At the instant of release the speed is zero, so the centripetal part \( v^2/\ell \) is zero.
2. Only the tangential acceleration \( g\sin30^\circ \) remains, which is perpendicular to the string.

Teacher's Note:
1. Zero speed means zero radial acceleration.
2. The tangential direction is perpendicular to the string.

Answer: (d) 3 m/s

Explanation:
1. Height gained \( h = \ell(1-\cos 60^\circ) = 0.35 \) m.
2. \( v^2 = 16 - 2gh = 16 - 7 = 9 \), so \( v = 3 \) m/s.

Teacher's Note:
1. Use energy conservation between the lowest point and the given position.
2. Height is measured from the lowest position.

Answer: (d) 18 N

Explanation:
1. At the top, \( T + mg = \dfrac{mv^2}{r} \).
2. \( T = \dfrac{3\times 64}{4} - 30 = 48 - 30 = 18 \) N.

Teacher's Note:
1. Both tension and weight point towards the centre at the top.
2. Subtract weight from the required centripetal force.

Answer: (a) \( \sqrt{5} \) rad/s

Explanation:
1. Tension is greatest at the bottom: \( T - mg = m\omega^2 r \).
2. \( 45 - 10 = 1\times\omega^2\times 7 \Rightarrow \omega^2 = 5 \).
3. \( \omega = \sqrt{5} \) rad/s.

Teacher's Note:
1. Use the lowest point, where the tension is greatest.
2. The string breaks first there.

Answer: (c) 3g

Explanation:
1. At the top, \( v_0^2 = gr \).
2. At the horizontal position the body has dropped through a height r, so \( v^2 = gr + 2gr = 3gr \).
3. \( a_c = \dfrac{v^2}{r} = 3g \).

Teacher's Note:
1. Energy conservation gives the speed after dropping through r.
2. The centripetal acceleration at the top is g, at the bottom 5g.

Answer: (b) the lowest point

Explanation:
1. Tension is \( T = \dfrac{mv^2}{r} + mg\cos\theta \) (θ from the downward vertical).
2. Both the speed and \( \cos\theta \) are largest at the lowest point, so the tension is maximum there.

Teacher's Note:
1. The string breaks where tension is maximum.
2. That is the bottom of the circle.