Question

A cylinder of mass(M), radius (R) is resting on a horizontal platform (which is parallel to (X-Y) plane) with its axis fixed along the (y) axis and free to rotate about its axis. The platform is given a motion in (X) direction given by: ($x=Acoswt$ ). 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 $\frac{1}{2}MRA{\omega }^2$


Motion of platform$\text{ABCD}$in $\text{X-}$direction is described by:

$x=Acoswt$

$v=\frac{dx}{dt}⟹v=\frac{d\left(Acoswt\right)}{dt}$

$v=-Awsinwt$

$a=\frac{dv}{dt}⟹a=\frac{d(-Awsinwt)}{dt}$

$a=-A{w}^{2}coswt$

$\therefore a_{max}=-A{w}^2$

As we know torque is given by,

${torque}_{max}=I{\alpha }_{max}$

${\alpha }_{max}=\frac{a_{max}}{R}$

Moment of inertia of rod is given by,$I=\frac{1}{2}M{R}^2$

$\therefore torqu{e}_{max}=\frac{1}{2}M{R}^2\times \frac{A{w}^2}{R}=\frac{1}{2}MRA{w}^2$

Hence, The maximum torque acting on the cylinder during its motion is: $\frac{1}{2}MRA{w}^2$