A hollow sphere is released from rest on a rough inclined plane with friction of 0.2 find the velocity of a point of a contact as a function of time
Answers
The velocity of a point of a contact as a function of time is 2 t m/s .
Explanation:
Acceleration due to slipping
As shown in figure, net force acting on hollow sphere = ( mg sin37 - μ mg cos37 )
a = F / m = g ( sin 37 - μ cos 37 )
a = 9.8 ( 0.6 - 0.2 × 0.8 ) = 4.312 m/s^2
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)mR
Where
- R = radius of hollow sphere
- α = angular acceleration.
Torque τ = force × perpendicular distance = μ mg cos37 × R .............(2)
From (1) and (2), we have:
μ mg cos37 × R = (2/3) mR^2 × 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/s^2
Net acceleration = ( 4.312 - 2.352 ) m/s^2 = 1.96 m/s^2
Velocity of point of contact as a function of time = acceleration × time = 1.96 t m/s ≈ 2 t m/s
Thus the velocity of a point of a contact as a function of time is 2 t m/s .
Also learn more
A body has a mass of 50 kg. Its velocity is brought down from 20 m/s to 5 m/s by a resisting force in 5 s. The magnitude of resisting force is
(b) 150 N (a) 50 N (c) 750 N (d) 375 N
https://brainly.in/question/11132369