Question 62: Centre of Mass and Rotational Motion

Question

A solid cylinder having radius $R$ and length $L$ is slipping on a rough horizontal plane. At time $t=0$ the cylinder has a translational velocity $v_{\mathrm{o}}=49 \mathrm{~m} / \mathrm{s}$, perpendicular to its axis and a rotational velocity $v_{\mathrm{o}} / 4 R$ about the centre. The time taken by the cylinder to start rolling is $\_\_\_\_$ seconds. (coefficient of kinetic friction $\mu_K=0.25$ and $g=9.8 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

The cylinder is slipping, which means its translational motion and rotational motion are not in order $(\mathrm{v}=\mathrm{r} \omega)$ for pure rolling.
Initial translational velocity is $\mathrm{v}_0=49 \mathrm{~m} / \mathrm{s}$.
Initial angular velocity is $\omega_0=\frac{v_0}{4 R}$, where $R$ is the radius of solid cylinder of length $L$.
For pure rolling to start, the condition $\mathrm{v}=\omega \mathrm{R}$ must be satisfied.
Initially, $\omega_0 \mathrm{R}=\frac{\mathrm{v}_0}{4}$, which is less than $\mathrm{v}_0$, so the cylinder is sliding forward.
Since the cylinder is sliding forward, kinetic friction $\left(\mathrm{f}_{\mathrm{k}}ight)$ acts backward at the point of contact.
$$ \mathrm{f}_{\mathrm{k}}=\mu_{\mathrm{k}} \mathrm{~N}=\mu_{\mathrm{k}} \mathrm{mg} . $$
Using Newton's second law of motion ( $\mathrm{F}=\mathrm{ma}$ ) linear retardation of cylinder is,
$$ \mathrm{F}=\mathrm{ma} \Rightarrow-\mu_{\mathrm{k}} \mathrm{mg}=\mathrm{ma} \Rightarrow \mathrm{a}=-\mu_{\mathrm{k}} \mathrm{~g} $$
Friction provides a torque $(	au)$ about the center of mass.
$$ 	au=\mathrm{f}_{\mathrm{k}} \cdot \mathrm{R}=\left(\mu_{\mathrm{k}} \mathrm{mg}ight) \mathrm{R} $$
We also know $	au=\mathrm{I} \alpha$.
For a solid cylinder, the moment of inertia $\mathrm{I}=\frac{1}{2} \mathrm{mR}^2$.
$$ \left(\mu_{\mathrm{k}} \mathrm{mg}ight) \mathrm{R}=\left(\frac{1}{2} \mathrm{mR}^2ight) \alpha \Rightarrow \alpha=\frac{2 \mu_{\mathrm{k}} \mathrm{~g}}{\mathrm{R}} $$
We need to find the time t when the final translational velocity v and final angular velocity $\omega$ satisfy $\mathrm{v}=\omega \mathrm{R}$.
Using equation of linear motion, final linear velocity is $\mathrm{v}=\mathrm{v}_0+\mathrm{at}=\mathrm{v}_0-\mu_{\mathrm{k}} \mathrm{gt}$
Using equation of rotational motion, final angular velocity is,
$$ \omega=\omega_0+\alpha \mathrm{t}=\frac{\mathrm{v}_0}{4 \mathrm{R}}+\frac{2 \mu_{\mathrm{k}} \mathrm{~g}}{\mathrm{R}} \mathrm{t} $$
Applying the rolling condition $\mathrm{v}=\omega \mathrm{R}$ :
$\mathrm{v}_0-\mu_{\mathrm{k}} \mathrm{gt}=\left(\frac{\mathrm{v}_0}{4 \mathrm{R}}+\frac{2 \mu_{\mathrm{k}} \mathrm{g}}{\mathrm{R}} \mathrm{t}ight) \mathrm{R}$
$$\Rightarrow $$ $\mathrm{v}_0-\mu_{\mathrm{k}} \mathrm{gt}=\frac{\mathrm{v}_0}{4}+2 \mu_{\mathrm{k}} \mathrm{gt}$
$$\Rightarrow $$ $\mathrm{v}_0-\frac{\mathrm{v}_0}{4}=2 \mu_{\mathrm{k}} \mathrm{gt}+\mu_{\mathrm{k}} \mathrm{gt}$
$$\Rightarrow $$ $\frac{3 \mathrm{v}_0}{4}=3 \mu_{\mathrm{k}} \mathrm{gt}$
$$\Rightarrow $$ $$ \mathrm{t}=\frac{\mathrm{v}_0}{4 \mu_{\mathrm{k}} \mathrm{~g}} $$
Substituting the given values, $\mathrm{v}_0=49 \mathrm{~m} / \mathrm{s}, \mu_{\mathrm{k}}=0.25$ and $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$
$$ \mathrm{t}=\frac{49}{4 	imes 0.25 	imes 9.8} $$
$$\Rightarrow $$ $$ t=\frac{49}{9.8} s=5 s $$
Therefore, the time taken for the cylinder to start rolling is 5 seconds.
Hence, the correct option is (B).

Answer

15

Answer

5

Answer

10

Answer

7.5

Centre of Mass and Rotational Motion

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