A uniform sphere of mass 500 g rolls without slipping on a plane surface so that its centre moves at a speed of 0.02 m/s . The total kinetic energy of rolling sphere would be (in J)

 Consider v to be the velocity of the sphere’s centre. When rolling without sliding, the angular speed of the centre is equal to vr, where r is the sphere’s radius.

Total kinetic energy (E) of the sphere will be the sum of translational kinetic energy ( K.E1 ) and rotational kinetic energy ( K.E2 )

⇒E=1/2mv2+1/2Iω2

where m is the mass of the sphere

I is the moment of inertia of the sphere

We know that moment of inertia of a sphere about the diameter I=2/5MR2

where M is the mass of the sphere

R is its radius On substituting the values of the moment of inertia and angular velocity, E becomes

⇒ E = 1/2mv2+1/2(2MR2/5)ω2 =1/2mv2+1/5mv2 = 7/10mv2

Putting the value of mass and velocity in the above equation we get,

⇒ E = 7/10 × 1/2×0.022=1.4×10−4J