Question

A uniform ring of radius im is performing rolling motion on the horizontal surface as shown in the figure. At t = 0 acceleration of COM of ring is 3 i m/s2 while angular acceleration & angular velocity are -3k rad/s2 and -2 k rad's where x-z plane is horizontal and y is vertical axis. A is in contact with surface and C is at 2m above the surface. B and D are 1m above the surface С ул B 0. D >X N N A Magnitude of acceleration (in m/s2) of point B with respect to Dis​

Answer 1

Explanation:

The motion of the two objects is caused by the frictional force acting on the objects=μ

k

mg

Thus acceleration due to frictional force=μ

k

g

Thus linear velocity attained in time t,v=u+at=0+μ

k

gt=μ

k

gt

Also a torque would act due to friction causing a rotation about the center.

Thus τ=Iα

⟹fr=Iα

Thus angular acceleration, α=

I

fr

=

I

μ

k

mgr

Thus angular velocity attained after time t, ω=ω

0

−αt

When rolling starts, v=rω

⟹μ

k

gt=r(ω

0

−

I

μ

k

mgr

t)

⟹t=

μ

k

g+

I

μ

k

mgr

2

rω

0

Since I for ring is greater than that for solid disc, ring takes longer time to achieve pure rolling. Thus the solid disc starts to roll earlier.

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