A 10 kg solid sphere of radius r = 0.8 m is rolling without slipping on a horizontal rough surface with 8 m/s; the force applied by the right half of the sphere on the left half is 30X newton. Find the value of X.
Physics · Mixed Chapter
JEE Mains

Question

A 10 kg solid sphere of radius r = 0.8 m is rolling without slipping on a horizontal rough surface with 8 m/s; the force applied by the right half of the sphere on the left half is 30X newton. Find the value of X.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: For pure rolling, the angular speed is ω=v/r=8/0.8=10 rad/s. Step 2: The centre of mass of a uniform solid hemisphere lies 3r/8 from the sphere centre. Thus the centre of mass of either half is at d=3×0.8/8=0.3 m from the whole sphere centre. Step 3: Its centre of mass has centripetal acceleration ω²d=10²×0.3=30 m/s². Step 4: Each half has mass 5 kg, so the required internal force is 5×30=150 N=30X N. Therefore X=5.