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Question

A cylinder having radius 0.4m, initially rotating (at t=0) with ωo=54rad.s-1 is placed on a rough inclined plane with θ=37o having friction coefficientμ=0.5. The time taken by the cylinder to start pure rolling is (g=10ms-2)


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Solution

Step 1: Given data

A cylinder having radius 0.4m, initially rotating (at t=0) with ωo=54rad.s-1 is placed on a rough inclined plane with θ=37o having friction coefficient μ=0.5.

Step 2: Formula used

τ=Iα[whereτ=torque,I=current,α=angularacceleration]τ=F.r[whereτ=torque,F=force,r=radius]F=ma[whereF=force,m=mass,a=acceleration]v=ωr[wherev=velocity,r=radius,ω=angularvelocity]

Step 3: Calculating time taken by the cylinder to start pure rolling

When the cylinder comes into contact with the plane, it rotates only and has no rectilinear motion. Because of the force of friction and gravitational force operating on the body, when the plane and the cylinder collide, the angular velocity decreases, and the linear velocity increases.
The friction force on the cylinder will be determined by Ff=μN=μmgcosθ

The force of gravity parallel to the surface of the plane will beFg=mgsinθ

The torque on the cylinder will be given by

τ=F.r=(μmgcosθ×R)+(mgsinθ×0)=μmgRcosθ

As the force of gravity will act on the center of the cylinder, there will be no torque due to it.
The angular acceleration due to this torque will be

τ=Iαα=τII=12Mr2α=μmgRcosθ12mr2=2μgcosθr

The linear acceleration on the cylinder due to the force will be

F=maa=Fma=μmgcosθ+mgsinθm=g(μcosθ+sinθ)

The angular velocity at time t will be given by ω(t)=ωo-αt

The linear velocity at time t will be given by v(t)=vo+αt=0+at=0

There will be pure rolling when v=ωr

This will happen at time t, for which the following equation is satisfied

at=(ωo-αt)rat=rωo-αrtat+αrt=rωot(a+αr)=rωot=rωoa+αrt=0.4×54g(μcosθ+sinθ)+2μgcosθrrt=0.4×54g(μcosθ+sinθ)+2μgcosθt=0.4×54g(3μcosθ+sinθ)t=0.4×54g(3μcos37o+sin37o)t=0.4×54103×0.5×45+35Ascos37o=45,sin37o=35t=21.6101.2+0.6t=21.618t=1.2s

Hence, the required time is 1.2 seconds.


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