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Rotational motion
Asked by | 13 Mar, 2010, 01:44: PM

(c) Both will reach the wall simultaneously.

KE of a rolling body = KE translational + KE rotational

= (1/2)MVcm2 + (1/2)Iw2 =(1/2)[(MVcm2 + I(Vcm/R)2]

= (1/2)[(MVcm2 + I(Vcm/R)2] = (Vcm2 /2)(M + I/R2) = (MVcm2 /2)(1 + c)

... since I = cMR, for example, for ring I = MR2, hence c = 1.

Now since both have the same KE,

Vcm = (2KE/((1 + c)M))1/2

(Vcm)ring/(Vcm)cylinder = (1/((1+c)M)ring1/2/(1/((1+c)M)cylinder1/2

(Vcm)ring/(Vcm)cylinder = = ((1.5x0.4)/(0.3x2)1/2 =  1

Both are moving at the same speed, and will reach at the same time.

Regards,

Team,

TopperLearning.

Answered by | 13 Mar, 2010, 03:22: PM

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