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Question

A uniform rod of mass M and length L lies radially on a disc of radius 2R rotating with angular speed ω in a horizontal plane about its axis. The rod does not slip on the disc and the centre of the rod is at a distance R from the centre of the disc. Then the kinetic energy of the rod is :

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Solution

The correct option is **A** 12mω2(R2+L212)

Rotational KE of the rod of mass m is: KErot=12Iω2

But MI about the centre of the disc by applying parallel axis theorem is:

I=mL212+mR2

Rotational KE of the rod of mass m is: KErot=12Iω2

But MI about the centre of the disc by applying parallel axis theorem is:

I=mL212+mR2

Hence, KErot=12(mL212+mR2)ω2=12mω2(L212+R2)

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