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A hollow sphere is released from rest on a rough inclined plane as shown in the figure. Find the velocity of point  of contact as a function of time. (m=mass of hollow sphere,R=radius of sphere)

Asked by everyteengamer | 29 Feb, 2020, 11:17: AM
Let us find acceleration due to slipping

As shown in figure, net force acting on hollow sphere = ( mg sin37 - μ mg cos37 )

Hence , acceleration = force / mass = g ( sin 37 - μ cos 37 ) = 9.8 ( 0.6 - 0.2 × 0.8 ) = 4.312 m/s2

Let us find acceleration due to rolling

friction force gives a torque τ to rotate the hollow sphere

we have , τ = I α ................(1)

where I is moment of inertia , I = (2/3)mR2 , where R is radius of hollow sphere and α is angular acceleration

Torque τ = force × perpendicular distance = μ mg cos37 × R  .............(2)

from (1) and (2), we have, μ mg cos37 × R = (2/3)mR2 × α ..............(3)

By substituting values for μ and g , we get after simplification, α = 2.352 / R

Hence linear acceleration  a = α × R = 2.352 m/s2

At point of contact acceleration due to slipping is downward , but acceleration due to rolling is in upward direction so that
it is opposite to the direction of acceleration due to slipping.

Hence net acceleration = ( 4.312 - 2.352 ) m/s2 = 1.96 m/s2

velocity of point of contact as a function of time = acceleration × time = 1.96 t m/s ≈ 2 t m/s
Answered by Thiyagarajan K | 29 Feb, 2020, 06:40: PM

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