SATHEE: Rotation 5 Question 27

Rotation 5 Question 27

39. A uniform disc of mass $m$ and radius $R$ is projected horizontally with velocity $v_{0}$ on a rough horizontal floor, so that it starts off with a purely sliding motion at $t = 0$ . After $t_{0}$ seconds, it acquires a purely rolling motion as shown in figure.

(1997 C, 5M)

(a) Calculate the velocity of the centre of mass of the disc at $t_{0}$ .

(b) Assuming the coefficient of friction to be $μ$ , calculate $t_{0}$ . Also calculate the work done by the frictional force as a function of time and the total work done by it over a time $t$ much longer than $t_{0}$ .

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Answer:

Correct Answer: 39. (a) $\frac{2}{3} v_{0}$

(b) $\frac{v_{0}}{3 \propto g}$ , For $t \leq t_{0}, W_{f} = \frac{m \propto g t}{2} [3 \propto g t - 2 v_{0}], \frac{- m v_{0}^{2}}{6}$

Solution:

(a) Between the time $t = 0$ to $t = t_{0}$ . There is forward sliding, so friction $f$ is leftwards and maximum i.e. $μ m g$ . For time $t > t_{0}$ , friction $f$ will become zero, because now pure rolling has started i.e. there is no sliding (no relative motion) between the points of contact.

So, for time $t < t_{0}$

Linear retardation, $a = \frac{f}{m} = μ g$

and angular acceleration, $α = \frac{τ}{I} = \frac{f R}{\frac{1}{2} m R^{2}} = \frac{2 μ g}{R}$

Now, let $v$ be the linear velocity and $ω$ , the angular velocity of the disc at time $t = t_{0}$ , then

$v = v_{0} - a t_{0} = v_{0} - μ g t_{0}$

and

$ω = α t_{0} = \frac{2 μ g t_{0}}{R}$

For pure rolling to take place

$v = R ω$

i.e. $v_{0} - μ g t_{0} = 2 μ g t_{0} \Rightarrow t_{0} = \frac{v_{0}}{3 μ g}$

Substituting in Eq. (i), we have

$v = v_{0} - μ g \frac{v_{0}}{3 μ g} \Rightarrow v = \frac{2}{3} v_{0}$

(b) Work done by friction

For $t \leq t_{0}$ , linear velocity of disc at any time $t$ is $v = v_{0} - μ g t$ and angular velocity is $ω = α t = \frac{2 μ g t}{R}$ . From Work-energy theorem, work done by friction upto time $t =$ Kinetic energy of the disc at time $t$ - Kinetic energy of the disc at time $t = 0$

$∴ W = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} - \frac{1}{2} m v_{0}^{2}$

$= \frac{1}{2} m {[v_{0} - μ g t]}^{2} + \frac{1}{2} \frac{1}{2} m R^{2} \frac{2 μ g t^{2}}{R} - \frac{1}{2} m v_{0}^{2}$

$= \frac{1}{2} [m v_{0}^{2} + m μ^{2} g^{2} t^{2} - 2 m v_{0} μ g t + 2 m μ^{2} g^{2} t^{2} - m v_{0}^{2}]$

or $W = \frac{m μ g t}{2} [3 μ g t - 2 v_{0}]$

For $t > t_{0}$ , friction force is zero i.e. work done in friction is zero. Hence, the energy will be conserved.

Therefore, total work done by friction over a time $t$ much longer then $t_{0}$ is total work done upto time $t_{0}$ (because beyond this work done by friction is zero) which is equal to