## A solid sphere of mass 500 gm and radius 10 cm rolls down on an inclined plane with out slipping. The height of the centre of mass of the sphere from the ground is 0.8 m. The translational speed of the centre of mass of the sphere on reaching the bottom of inclined will be

Q: A solid sphere of mass 500 gm and radius 10 cm rolls down on an inclined plane with out slipping. The height of the centre of mass of the sphere from the ground is 0.8 m. The translational speed of the centre of mass of the sphere on reaching the bottom of inclined will be
(A) √5 m/s
(B)√6  m/s
(C) √10 m/s
(D) 20 m/s
Sol: mg (0.7) =½mv² + ½Icmω²
= ½mv² + ½(2mR²/5)(v/R)²
= 7mv²/10
v = √10 m/s

## When a body rolls without sliding up an inclined plane, the frictional force is

Q: When a body rolls without sliding up an inclined plane, the frictional force is
(A) directed up the plane
(B) directed down the plane
(C) zero
(D) dependent on its velocity
Solution : Since the body rolls without sliding,
ap = 0 & vp = 0
Hence, v – rω = 0  & a – rα = 0
v = rω &  a = rα

Since the body acceleration a is downwards parallel to the plane, therefore acceleration of the body must be in anticlockwise sense. That means the friction must act up the plane to produce an anticlockwise torque to produce anticlockwise angular acceleration.

## A particle m starts with zero velocity along a line y = 4d. The position of particle m varies as x = A sinωt. At ωt = π/2, its angular momentum with respect to the origin

Q: A particle of mass m starts with zero velocity along a line y = 4d. The position of particle m varies as x = A sinωt. At ωt = π/2 , its angular momentum with respect to the origin
(A) mAωd
(B) mωd/A
(D) zero
Sol:

At ωt = π/2
x = A sinωt
v = Aωcosωt = 0
So, angular momentum of particle m with respect to the origin should be zero.

## A particle of mass m is revolving in a horizontal circle of radius r with a constant angular speed ω . The areal velocity of the particle is

Q: A particle of mass m is revolving in a horizontal circle of radius r with a constant angular speed ω . The areal velocity of the particle is
(A) r2 ω
(B) r2 θ
(C) r2ω/2
(D) rω2/2
Sol: Solution :

Areal velocity = dA/dt  ;
where A = area of the sector =r2θ/2
dA/dt = d(r2θ/2)/dt
½(r2)dθ/dt = r2ω/2