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Question

A solid cylinder rolls down a fixed inclined plane of height 'h' and inclination θ. Calculate the time taken to reach bottom?

{Assume the mass be M and radius be R}


A

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B

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C

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D

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Solution

The correct option is D


The Free body drawing of the sphere is as shown

Assume the direction of 'a' and 'α' as shown,

Mgsinθfr=Ma ....(1)

Taking torque about centre O,

τ=fr×h=Iα

As we know ,I=MR22 {about centre 0}

R.fr=MR22×α

Also as the cyclinder is rolling purely,

a=Rα and v=Rω.

R.fr=MR22×aR

fr=Ma2 ............(ii)

from (i)and (ii),

Mg sinθMa2=Ma

a=23gsinθ .......(iii)

Now, assume the length of wedge be 'L'.

By simple trigonometry,

L sinθ=h

or L=h cosecθ . ....(iv)

Applying the equation of motion,

s=ut+12at2

From (i),(ii)and(iv)

h cosecθ=0+1/2/23gt2sinθ

3h cosec2θg=t2

t=3h cosec2θg

or {As taking only postive value}

t=cosecθ.3hg.

Hence (D)


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