A classical mechanics problem by Yogesh Ghadge

Classical Mechanics Level 3

A block of mass m m is supported by a massless string wound around a uniform hollow cylinder of mass m m and radius r r . If the string does not slip on the cylinder, with what acceleration a a will the block fall on release?

1 2 g m − 1 \frac12 gm^{-1} 5 6 g m − 1 \frac56 gm^{-1} g m − 1 gm^{-1} 2 3 g m − 1 \frac23 gm^{-1}

Yogesh Ghadge by Yogesh Ghadge , 22, India

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1 solution

Satvik Pandey
Oct 21, 2014

fig fig

From the f.b.d of block

m g − T = m a mg-T=ma .........(1)

Rotational eq. for pulley

T R = I α TR=I\alpha ......................(2) ( I = m R 2 I=mR^{2} )

As the string is not slipping so

a = r α a=r\alpha ..............................(3)

From eq(2) and (3)

T = m a T=ma

on putting this value in eq(1) we will get

m g − m a = m a mg-ma=ma or m g = 2 m a mg=2ma

So a = g 2 a=\frac{g}{2}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Isn't the value of solid cylinder's Inertia I = 1 2 m R 2 I = \frac{1}{2}mR^2 , is it ?

Andronikus Lumembang - 6 years, 5 months ago

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