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A uniform ring of mass m and radius r has a particle of mass 2m rigidly attached to it on the inside surface. The system rolls on a horizontal surface without slipping. In the position shown at a certain instant, it's centre c has a velocity v. Find the kinetic energy of the system

A uniform ring of mass m and radius r has a particle of mass 2m rigidly attached to it on the inside surface. The system rolls on a horizontal surface without slipping. In the position shown at a certain instant, it's centre c has a velocity v.  Find the kinetic energy of the system 

Grade:12

2 Answers

Arun
25750 Points
5 years ago
Dear Sonal
 
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Thiccsardar
13 Points
one year ago
NOTE- Instantaneous axis of rotation(the bottommost point) is stationary.
Moment of Inertia in frame of centre of mass= I of ring + I of particle= mR^2+ 2mr^2=3mR^2
By parallel axis theorem, Moment of inertia in frame of IAR = 3mR^2+ mR^2=4mR^2
Also, the ring performs rolling motion, So V0= RW
K.E = 1/2IW^2 = 1/2(4mR^2)(v0/R)^2 = 2mV0^2 ----ans.

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