Pure rolling-1!

Classical Mechanics Level 3

A uniform disc of mass m m and radius R R is projected horizontally with velocity v 0 v_0 on a rough horizontal floor so that it starts off with a purely sliding motion at t = 0 t = 0 . After t 0 = v 0 n μ g \large t_0 = \frac{v_0}{nμg} second, it acquires a purely rolling motion. Find the value of n n .


The answer is 3.

Nishant Rai by Nishant Rai , 25, India

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1 solution

Tanishq Varshney
May 29, 2015

When the disc starts pure rolling v c m = R w v_{cm}=R w where w w is angular velocity

using conservation of angular momentum about any point on the floor

m v o R = m v c m R + m R 2 2 w mv_{o} R=mv_{cm}R+m\frac{R^2}{2}w

v c m = 2 v o 3 v_{cm}=\frac{2v_{o}}{3}

since only friction force f f acts

so f = − m a f=-ma

μ m g = − m a \mu mg=-ma

a = − μ g a=- \mu g

v c m = v 0 + a t v_{cm}=v_{0}+at

t = v o 3 μ g \huge{t=\frac{v_{o}}{3 \mu g}}

15 Helpful 3 Interesting 3 Brilliant 0 Confused

m v o R = m v c m R + m R 2 2 w mv_{o} R=mv_{cm}R+m\frac{R^2}{2}w

v c m = 2 v o 3 v_{cm}=\frac{2v_{o}}{3}

Please explain this portion

Md Zuhair - 4 years, 2 months ago

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This is just conservation of angular momentum from the point of contact considering condition of pure rolling i.e. v=wr

Vijay Kumar - 2 years, 8 months ago

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