Question

A tangential force of 20 N is applied on a cylinder of mass 4 kg and moment of inertia about its own axis. The cylinder rolls without slipping then the linear acceleration of its center of mass will be,
A. \[6.7\,m/s^2\]
B. \[10\,m/s^2\]
C. \[3.3\,m/s^2\]
D. None of these

Answer 1

Hint:When force is applied to the cylinder tangentially then it produces a torque which makes the cylinder rotate. For pure rolling, the surface in contact must be instantaneously at rest.

Formula used:
\[\tau = FR\]
Here \[\tau \] is the torque due to applied force F at distance R from the axis of rotation.
\[\tau = I\alpha \]
Here \[\tau \] is the torque, I is the moment of inertia and \[\alpha \] is the angular acceleration.
\[a = \alpha R\]
Here, a is the linear acceleration of the center of mass and R is the radius of the cylinder.

Complete step by step solution:

Fig: The free diagram of cylinder

The net force acting on the cylinder is \[F + f\], where F is the applied force and f is the frictional force. Using Newton’s 2nd law of motion, the linear acceleration of the cylinder is,
\[a = \dfrac{{{F_{net}}}}{m}\]
\[\Rightarrow a = \dfrac{{F + f}}{m}\]
\[\Rightarrow a = \dfrac{{20 + f}}{4}\]
\[\Rightarrow a = 5 + \dfrac{f}{4} \ldots \left( i \right)\]

The net torque acting about the central axis of the cylinder is,
\[\tau = {f_r} - {F_r}\]
Using the rotational motion formula, \[\tau = I\alpha \]
The moment of inertia of the cylinder about its own axis is given as,
\[I = \dfrac{{m{R^2}}}{2}\]
\[\Rightarrow {f_r} - {F_r} = - \left( {\dfrac{{m{r^2}}}{2}} \right)\alpha \]
\[\Rightarrow \alpha = \dfrac{{2\left( {20 - f} \right)}}{{4r}}\]
\[\Rightarrow \alpha = \dfrac{1}{r}\left( {10 - \dfrac{f}{2}} \right) \ldots \left( {ii} \right)\]

Using the relation between the angular acceleration and the linear acceleration of the center of mass for the pure rolling motion,
\[a = \alpha r \\ \]
\[\Rightarrow 5 + \dfrac{f}{4} = \left( {\dfrac{1}{r}\left( {10 - \dfrac{f}{2}} \right)} \right)r \\ \]
\[\Rightarrow 5 + \dfrac{f}{4} = 10 - \dfrac{f}{2} \\ \]
\[\Rightarrow \dfrac{f}{2} + \dfrac{f}{4} = 10 - 5 \\ \]
\[\Rightarrow \dfrac{{3f}}{4} = 5N \\ \]
\[\Rightarrow \dfrac{f}{4} = \dfrac{5}{3}N \\ \]
So, the acceleration of the center of mass is,
\[a = \left( {5 + \dfrac{5}{3}} \right)m/{s^2}\]
\[\therefore a = 6.7\,m/{s^2}\]
Hence, on application of the tangential force of magnitude 20N the cylinder’s center of mass is having linear acceleration of \[6.7\,m/{s^2}\].

Therefore,the correct option is A.

Note: As the torque is proportional to the distance between the point of application of tangential force and the axis of rotation, we need to be careful about the point where force is applied. If the applied force is not tangential then we take the tangential component of the force to calculate the torque produced.