Chapter 8, Problem 79GP

To determine

The minimum force F required on a wheel of mass M.

Explanation of Solution

Introduction:

The angular acceleration of the object/body is directly proportional to the torque applied and inversely proportional to the rotational inertia of the object. If two or more torques acting on the object and it is in equilibrium, the net torque will be zero.

A torque is a physical quantity that explains the effect of the force on rotation of an object. Considering the work done by the torque

  $W={\displaystyle \int \tau d\theta }$ ,

Step up and equate the work done with the negative change of potential energy to climb up ( $-W=\Delta U$ ) about the centre of mass.

The force $F=\frac{mgh(1-R)}{\sqrt{2hR-h^2}}$ (since h

Conclusion:  The horizontal force required to climb up is $F=\frac{mgh(1-R)}{\sqrt{2hR-h^2}}$ .