Question

A uniform cylinder of mass M and radius R is to be pulled over a step of height \[\left( {a < R} \right)\]by applying a force F at its center ‘O’ perpendicular to the plane through the axes of the cylinder on the edge of the step (see figure). Then find the minimum value of F.

A. \[Mg\sqrt {{{\left( {\dfrac{R}{{R - a}}} \right)}^2} - 1} \]
B. \[Mg\sqrt {1 - \dfrac{{{a^2}}}{{{R^2}}}} \]
C. \[Mg\dfrac{a}{R}\]
D. \[Mg\sqrt {1 - {{\left( {\dfrac{{R - a}}{R}} \right)}^2}} \]

Answer 1

Hint: Before we start addressing the problem, we need to know about the torque. It is a measure of the force that can cause an object to rotate about an axis and it is denoted by \[\tau \].

Formula Used:
To formula for torque is given by,
\[\tau = FR\]
Where, $F$ is force and $R$ is distance or radius.

Complete step by step solution:


Consider a uniform cylinder of mass M and radius R is to be pulled over a step of height by applying a force F at its center ‘O’ perpendicular to the plane passing through the axes of the cylinder on the edge of the step. Then we need to find the minimum value of F.

In order to lift the sphere, if you see a point A, there should be a net torque about point A in the clockwise direction. There are two forces because of which the torque is coming. The two forces are F and mg. The torque because of F is \[{\tau _F} = FR\] in clockwise direction and \[{\tau _{mg}} = mgR\cos \theta \] in anticlockwise direction. Since, we want the toque in the clockwise direction. Therefore,
\[{\tau _F} \ge {\tau _{mg}}\]
\[\Rightarrow {\tau _F} \ge MgR\cos \theta \]

From the diagram we have,
\[\cos \theta = \dfrac{{\sqrt {{R^2} - {{\left( {R - a} \right)}^2}} }}{R}\]
\[ \Rightarrow \cos \theta = \sqrt {\dfrac{{{R^2}}}{{{R^2}}} - {{\left( {\dfrac{{R - a}}{R}} \right)}^2}} \]

\[ \Rightarrow \cos \theta = \sqrt {1 - {{\left( {\dfrac{{R - a}}{R}} \right)}^2}} \]

We want the minimum force, that is,
\[{F_{\min }} = mg\cos \theta \]
\[ \Rightarrow {F_{\min }} = mg\sqrt {1 - {{\left( {\dfrac{{R - a}}{R}} \right)}^2}} \]
Therefore, the minimum value of force is, \[mg\sqrt {1 - {{\left( {\dfrac{{R - a}}{R}} \right)}^2}}\].

Hence, option D is the correct answer.

Note:Here, the torque has both clockwise and anti-clockwise direction. Since the sphere is moving in the forward direction, we need to take the torque which is acting only in the clockwise direction. Some of the daily life examples of the torque are, opening the cap of a bottle or turning a steering wheel etc.