Rotation 2

A solid cylinder of radius R R is set into rotation about its axis with angular speed ω o \omega_{o} then lowered with its lateral surface onto a horizontal plane and released. If the coefficient of friction is μ \mu , find after how much time (in seconds) will the cylinder start pure rolling.

If the time is given by t = R ω o n μ g t = \dfrac{ R \omega_{o}}{n \mu g} , find n n .


The answer is 3.

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1 solution

Tapas Mazumdar
Mar 25, 2017

We have the following equations

F ( ext. force ) = f k = μ m g = m a a = μ g F (\text{ext. force}) = f_k = \mu mg = ma \implies \color{#3D99F6}{a = \mu g}

Γ ( ext. torque ) = f k R = I α μ m g R = m R 2 2 α α = 2 μ g R \Gamma (\text{ext. torque}) = f_k R = I \alpha \implies\mu mg R = \dfrac{mR^2}{2} \alpha \implies \color{#20A900}{\alpha = \dfrac{2 \mu g}{R}}

Now using the equations

ω = ω 0 + α t \omega = \omega_0 + \alpha t

and

v = v 0 + a t v = v_0 + at

we get

ω = ω 0 2 μ g R t ( 1 ) ( negative sign because of opposite sense of α and ω 0 ) \omega = \omega_0 - \dfrac{2 \mu g}{R} t \qquad \cdots(1) \qquad \qquad \qquad (\small\text{negative sign because of opposite sense of } \alpha \text{ and } \omega_0)

v = μ g t ( 2 ) v = \mu g t \qquad \cdots(2)

At the instant when pure rolling starts, we will have v = R ω v = R \omega . Thus equation ( 2 ) (2) becomes

R ω = μ g t ( 3 ) R \omega = \mu g t \qquad \cdots(3)

Solving ( 1 ) (1) and ( 3 ) (3) , we get

t = R ω 0 3 μ g t = \boxed{\dfrac{R \omega_0}{3 \mu g}}

Well done.Nice son..

Spandan Senapati - 4 years ago

Or conserve angular momentum about any point on floor.

Priyanshu Mishra - 3 years ago

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