JEE Physics Circular Motion MCQs Set 17

Answer: (b) \( -\frac{K}{2r} \)

Explanation:
1. \( \frac{mv^2}{r} = \frac{K}{r^2} \Rightarrow \tfrac12mv^2 = \frac{K}{2r} \).
2. Potential energy of an inverse-square attraction is \( -\frac Kr \), so \( E = \frac K{2r}-\frac Kr = -\frac K{2r} \).

Teacher's Note:
1. Kinetic energy comes from the centripetal condition.
2. A bound orbit has negative total energy.

Answer: (d) (iii) and (iv)

Explanation:
1. Motion along a circle needs a non-zero radial (centripetal) acceleration to keep bending the path.
2. The tangential acceleration may be zero (uniform circular motion) or non-zero (speed changing).

Teacher's Note:
1. No radial acceleration means a straight path.
2. Tangential acceleration only changes speed.

Answer: (a) 1 : 2

Explanation:
1. About the centre, \( \omega_C = \frac va \).
2. By the inscribed-angle theorem the angle subtended at a point on the circle is half that at the centre, so \( \omega_A =rac{v}{2a} \). Ratio ω_A : ω_C = 1 : 2.

Teacher's Note:
1. Angle at the circumference is half the angle at the centre.
2. Differentiate the angle to get ω.

Answer: (c) 5 m/s²

Explanation:
1. Centripetal acceleration \( = \frac{v^2}{r} = \frac{3600}{1200} = 3 \) m/s².
2. Total \( = \sqrt{3^2+4^2} = 5 \) m/s².

Teacher's Note:
1. Radial and tangential components are perpendicular.
2. Add them as vectors.

Answer: (c) 36 m

Explanation:
1. Radial component \( = \sqrt{5^2-3^2} = 4 \) m/s².
2. \( r = \frac{v^2}{a_c} = \frac{144}{4} = 36 \) m.

Teacher's Note:
1. Resolve total acceleration into components.
2. a_c = v²/r.

Answer: (a) \( \frac a2\sqrt{4+\pi^2} \)

Explanation:
1. Tangential acceleration \( a_t = a \). Arc length \( s = \frac{at^2}{2} = \frac{2\pi r}{8} \Rightarrow r = \frac{2at^2}{\pi} \).
2. \( a_c = \frac{a^2t^2}{r} = \frac{\pi a}{2} \).
3. Total \( = \sqrt{a^2+\frac{\pi^2a^2}{4}} = \frac a2\sqrt{4+\pi^2} \).

Teacher's Note:
1. Find the radius from the arc traced.
2. Combine a_t and a_c.

Answer: (d) the necessary centripetal force is obtained from the horizontal component of the reaction of the track

Explanation:
1. Banking tilts the normal reaction so that its horizontal component supplies the centripetal force.
2. This reduces the sideways thrust on the rails; weight is unchanged.

Teacher's Note:
1. The normal force is perpendicular to the inclined surface.
2. Its horizontal component points to the centre.

Answer: (b) 0.675 rev/s

Explanation:
1. \( \omega^2 r = 90 \Rightarrow \omega^2 = 18 \Rightarrow \omega = 4.24 \) rad/s.
2. \( n = \frac{\omega}{2\pi} = 0.675 \) rev/s.

Teacher's Note:
1. Centripetal acceleration is ω²r.
2. Divide ω by 2π for rev/s.

Answer: (b) 30

Explanation:
1. \( v_{max} = \sqrt{\mu rg} = \sqrt{0.6\times150\times10} = 30 \) m/s.

Teacher's Note:
1. Friction supplies the centripetal force.
2. Take g = 10 m/s².

Answer: (c) at the bottom of the circle

Explanation:
1. At the bottom \( T = mg+\frac{mv^2}{r} = 19.6+32 = 51.6 \) N.
2. At the top \( T = \frac{mv^2}{r}-mg = 12.4 \) N, which does not match.

Teacher's Note:
1. Test the top and bottom points.
2. Weight adds to tension at the bottom.

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

Explanation:
1. The particle leaves at once if the normal reaction becomes zero at the top: \( mg = \frac{mv^2}{R} \).
2. \( v = \sqrt{gR} \).

Teacher's Note:
1. Condition for leaving the surface is N = 0.
2. At the top, gravity alone gives the centripetal force.

Answer: (d) maximum at B

Explanation:
1. At the bottom of a dip the centre of the circle is above the vehicle, so \( N-mg = \frac{mv^2}{r} \) and \( N = mg+\frac{mv^2}r \).
2. On a hill top \( N = mg-\frac{mv^2}{r} \) and on level ground \( N = mg \). So N is greatest at B.

Teacher's Note:
1. The sign of the centripetal term depends on where the centre lies.
2. Compare with N = mg at A.