A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is :
(3) 52
Pure rolling: v= R, and I= 25MR2 for a solid sphere.
[figure]
Krot= 12I 2= 12 25MR2 v2R2= 15Mv2, Ktrans= 12Mv2
KtransKrot= 1/21/5= 52
Answer: (3)
A solid sphere of mass ' m ' and radius ' r ' is allowed to roll without slipping from the highest point of an inclined plane of length ' L ' and makes an angle 30 with the horizontal. The speed of the particle at the bottom of the plane is v1. If the angle of inclination is increased to 45 while keeping L constant. Then the new speed of the sphere at the bottom of the plane is v2. The ratio v12:v22 is
Consider a circular disc of radius 20 cm with centre located at the origin. A circular hole of radius 5 cm is cut from this disc in such a way that the edge of the hole touches the edge of the disc. The distance of centre of mass of residual or remaining disc from the origin will be
A circular disk of radius R meter and mass M kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that (t)=5t2-8t, where (t) is the angular position of the rotating disc as a function of time t. How much power is delivered by the applied torque, when t=2 s ?
A uniform circular disc of radius ' R ' and mass ' M ' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius R/2 is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above. [figure]
The position vectors of two 1 kg particles, (A) and (B), are given by rA= ( 1t2 i+ 2t j+ 3t k )m and rB= ( 1t i+ 2t2 j+ 3t k )m, respectively; ( 1=1 m/s2, 2=3n m/s, 3=2 m/s, 1=2 m/s, 2=-1 m/s2, 3=4p m/s), where t is time, n and p are constants. At t=1 s, | VA |= | VB | and velocities VA and VB of the particles are orthogonal to each other. At t=1 s, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is L kg m2 s-1. The value of L is .
The torque due to the force (2 i+ j+2 k ) about the origin, acting on a particle whose position vector is ( i+ j+ k ), would be