a circular ring of mass M and radius R is roll on a horizontal surface with speed v. what is its kinetic energy ?
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Answered by
1
hey here's your answer__________
I(circular ring)=MR^2
w(omega-angular velocity)=v/R
---kinetic energy =1/2(I)(w^2)
---1/2Mv^2
I(circular ring)=MR^2
w(omega-angular velocity)=v/R
---kinetic energy =1/2(I)(w^2)
---1/2Mv^2
Answered by
0
Answer:Mv^2
K.E. 1/2Iw^2 + 1/2Mv^2
1/2MR^2v^2/R^2+1/2Mv^2
Mv^2
Explanation:
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