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 | 29th Feb, 2020, 11:17: AM

Expert Answer:

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 | 29th Feb, 2020, 06:40: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.