Question

A cylinder of mass $M$, radius $R$ is resting on a horizontal platform (which is parallel to $xy-$plane) with its axis fixed along the $y-$axis and only free to rotate about its axis. The platform is given a motion in $x-$direction given by $x=A\cos \left(\omega t\right)$. There is no slipping between the cylinder and platform. The maximum torque acting on the cylinder during its motion is

• A. $\frac{1}{2}MRA{\omega }^2$
• B. $MRA{\omega }^2$
• C. $2MRA{\omega }^2$
• D. $MRA{\omega }^2\times \cos \omega t$

Answer 1

The correct option is A $\frac{1}{2}MRA{\omega }^2$

The axis of solid cylinder is fixed hence it will only rotate about its axis of rotation ($y$-axis).


Motion of platform is given by,
$x=A\cos \left(\omega t\right)$
Acceleration of the platform is given by
$\Rightarrow a=\frac{d^{2}x}{d{t}^2}=-A{\omega }^2\cos \left(\omega t\right)$
Above equation represents the $SHM$ for the platform,
The maximum acceleration achieved by the platform.
$\Rightarrow \left|a_{\text{max}}\right|=A{\omega }^2$
For no slipping between the platform and cylinder,
angular acceleration of cylinder, $\alpha =\frac{a}{R}$
where $a=a_{\text{platform}}$
For maximum magnitude of torque acting on the cylinder, magniude of ${}^{\prime }{\alpha }^{\prime }$ and ${}^{\prime }{a}^{\prime }$ will be maximum.
i.e $\left|{\tau }_{max}\right|\propto \left|a_{max}\right|$
$\Rightarrow {\tau }_{max}=I{\alpha }_{max}$
Here, I $\to \text{MOI}$ of the solid cylinder about axis passing through its centre i.e $y-$axis.
$\Rightarrow {\tau }_{\text{max}}=\left(\frac{M{R}^2}{2}\right)\left(\frac{a_{\text{max}}}{R}\right)$
$\Rightarrow {\tau }_{\text{max}}=\frac{M{R}^2}{2}\times \frac{A{\omega }^2}{R}$
$\therefore {\tau }_{\text{max}}=\frac{1}{2}MRA{\omega }^2$