Question

A cylinder having radius $0.4m$, initially rotating (at $t=0$) with ${\omega }_o=54rad.s^{-1}$ is placed on a rough inclined plane with $\theta ={37}^o$ having friction coefficient$\mu =0.5$. The time taken by the cylinder to start pure rolling is ($g=10m{s}^{-2}$)

Answer 1

Step 1: Given data

A cylinder having radius $0.4m$, initially rotating (at $t=0$) with ${\omega }_o=54rad.s^{-1}$ is placed on a rough inclined plane with $\theta ={37}^o$ having friction coefficient $\mu =0.5$.

Step 2: Formula used

$$\begin{gathered} \tau =I\alpha [where\tau =torque,I=current,\alpha =angularacceleration] \\ \tau =F.r[where\tau =torque,F=force,r=radius] \\ F=ma[whereF=force,m=mass,a=acceleration] \\ v=\omega r[wherev=velocity,r=radius,\omega =angularvelocity] \end{gathered}$$

Step 3: Calculating time taken by the cylinder to start pure rolling

When the cylinder comes into contact with the plane, it rotates only and has no rectilinear motion. Because of the force of friction and gravitational force operating on the body, when the plane and the cylinder collide, the angular velocity decreases, and the linear velocity increases.
The friction force on the cylinder will be determined by $F_f=\mu N=\mu mg\cos \theta$

The force of gravity parallel to the surface of the plane will be$F_g=mg\sin \theta$

The torque on the cylinder will be given by

$\tau =F.r=(\mu mg\cos \theta \times R)+(mg\sin \theta \times 0)=\mu mgR\cos \theta$

As the force of gravity will act on the center of the cylinder, there will be no torque due to it.
The angular acceleration due to this torque will be

$$\begin{gathered} \tau =I\alpha \Rightarrow \alpha =\frac{\tau }{I} \\ I=\frac{1}{2}M{r}^2 \\ \alpha =\frac{\mu mgR\cos \theta }{\frac{1}{2}m{r}^2}=\frac{2\mu g\cos \theta }{r} \end{gathered}$$

The linear acceleration on the cylinder due to the force will be

$$\begin{gathered} F=ma\Rightarrow a=\frac{F}{m} \\ a=\frac{\mu mg\cos \theta +mg\sin \theta }{m}=g(\mu \cos \theta +\sin \theta ) \end{gathered}$$

The angular velocity at time t will be given by $\omega \left(t\right)={\omega }_o-\alpha t$

The linear velocity at time t will be given by $v\left(t\right)=v_o+\alpha t=0+at=0$

There will be pure rolling when $v=\omega r$

This will happen at time t, for which the following equation is satisfied

$$\begin{gathered} at=({\omega }_o-\alpha t)r \\ at=r{\omega }_o-\alpha rt \\ at+\alpha rt=r{\omega }_o \\ t(a+\alpha r)=r{\omega }_o \\ t=\frac{r{\omega }_o}{a+\alpha r} \\ t=\frac{0.4\times 54}{g(\mu \cos \theta +\sin \theta )+{\displaystyle \frac{2\mu g\cos \theta }{r}}r} \\ t=\frac{0.4\times 54}{g(\mu \cos \theta +\sin \theta )+2\mu g\cos \theta } \\ t=\frac{0.4\times 54}{g(3\mu \cos \theta +\sin \theta )} \\ t=\frac{0.4\times 54}{g(3\mu \cos {37}^o+\sin {37}^o)} \\ t=\frac{0.4\times 54}{10\left(3\times 0.5\times \frac{4}{5}+\frac{3}{5}\right)}\left[As\cos {37}^o=\frac{4}{5},\sin {37}^o=\frac{3}{5}\right] \\ t=\frac{21.6}{10\left(1.2+0.6\right)} \\ t=\frac{21.6}{18} \\ t=1.2s \end{gathered}$$

Hence, the required time is 1.2 seconds.