1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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 :

A
12mω2(R2+L212)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
12mω2R2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
124mω2L2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 12mω2(R2+L212)Rotational KE of the rod of mass m is: KErot=12Iω2But MI about the centre of the disc by applying parallel axis theorem is:I=mL212+mR2Hence, KErot=12(mL212+mR2)ω2=12mω2(L212+R2)

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Work Energy and Power
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program