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Question

A cylinder initially kept at rest on a horizontal rough surface is given an angular velocity 10 rad/s. What is the distance travelled by the centre of mass of the cylinder till it begins pure rolling ?
Given: Coefficient of kinetic and static frcition are equal (μ=0.5) and radius of cylinder R=3 m. Also, take g=10 m/s2.


A
5 m
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B
10 m
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C
1 m
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D
20 m
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Solution

The correct option is B 10 m

Initially, linear velocity vi=0
Friction force will oppose the rotational motion, hence kinetic friction will act towards right as shown in figure

Here a=aCM
fk=ma
a=fkm ...(i)
From equation of torque,
τcom=Icomα
fk.R=mR22.α ...(ii)
From eq. (i) & (ii)
ma.R=mR22.α
a=R2α ...(iii)
Conserving angular momentum about point of contact, since τfk=0
Li=Lf
0+(Iω)=mvR+(Iω)
mR22.ω=mvR+mR22(vR)
where v is the final velocity of COM.
[ for achieving pure rolling, v=ωR]
ωR22=3vR2
v=ωR3 ...(iii)

Since friction will be kinetic till pure rolling starts,
fk=μN=μmg
a=fkm=μg ...(iv)
From equation v=u+at
u=0v=at
Substitute from eq. (iii) & (iv),
ωR3=μgt
t=ωR3μg
Now, distance travelled till pure rolling is achieved:
S=12at2
S=12×(μg)×(ωR3μg)2=ω2R218μg
S=(10)2×(3)218×0.5×10=10 m

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