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Question

A solid cylinder having a radius of 0.4 m, initially rotating (at t=0) with angular velocity ω0=54 rad/s is placed on a rough inclined plane with θ=37 having friction coefficient μ=0.5. The time taken by the cylinder to start pure rolling is
[Take g=10 m/s2]


A
5.4 s
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B
2.4 s
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C
4.4 s
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D
1.1 s
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Solution

The correct option is D 1.1 s
Given:
Initial angular speed, ω0=54 rad/s
Initial linear speed, v0=0

Bottommost point of the cylinder has a tendency to slip upwards.

Hence, direction of friction is downward along the inclined plane as we can see in the figure. Friction will act such that torque due to it will decrease the angular speed and increase the linear speed.


On drawing the free body diagram, we obtain.


So, mgsin37+f=ma

& N=mgcos37

mgsin37+μmgcos37=ma

[f=μN]

a=gsin37+μgcos37

Using, g=10 m/s2 and μ=0.5 we get,

a=10[35+0.5×45]=10 m/s2

Further we have the first equation of motion, v=u+at

v=0+10×t=10t ......(1)

Now, using the balance of torque, τ=fr=Iα [Anticlockwise]

μmgcos37×r=mr22×α

α=2μgcos37r
α=2×0.5×10×450.4=20 rad/s2

Further, ω=ω0+αt

ω=5420×t ......(2)
[As the acceleration is retarding]

When cylinder starts pure rolling, v=ωr

10t=(5420×t)×0.4

25t=5020t

t=1.1 s

Why this question?Concept - Direction of friction is dependent on whichmotion (translation or rotation) is more or less.If translation is less, it acts so as to increasetranslation and decrease rotation, so thatthe object starts pure rolling after some time.

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