Question

Two forces of magnitude F are acting on a uniform disc kept on a horizontal rough surface as shown in the figure. The disc rolls on the horizontal surface. Friction force by the horizontal surface on the disc is nF. Find the value of n.

Answer 1

n = 2/3

Answer 2

We consider the case where one force F is applied through the center in one horizontal direction and the other force F is applied at the rim in the opposite horizontal direction so that the net applied force (without friction) is zero. However, because the rim force produces a torque, friction must act on the disc to provide the necessary net force that makes the disc roll without slipping.

Let the disc have mass m, radius R, and moment of inertia $I=\tfrac{1}{2}mR^2$.

Assume: • Force F at the center to the right. • Force F at the rim to the left. • Friction f acts to the right (providing the net force for translation). • The no–slip condition: $a=\alpha R$.

Step 1. Translational Equation

The net force on the disc (horizontally) is only due to the friction:

$$f = ma \quad\Longrightarrow\quad a=\frac{f}{m}.$$

Step 2. Rotational Equation about the Center

Only the force F at the rim produces torque (the force at the center produces no torque) and friction also produces a torque. Choosing the anticlockwise direction as positive:

• The rim force F (applied to the left at the top or bottom, as arranged appropriately) gives a torque $F R$ (say, anticlockwise).
• The friction force f, acting at the point of contact (a distance R vertically from the center), gives a torque $fR$ (but it tends to act in the opposite sense, i.e. clockwise).

Thus the net torque is:

$$\tau_{\text{net}} = FR - fR.$$

This must equal the angular acceleration times the moment of inertia:

$$FR - fR = I\alpha = \frac{1}{2}mR^2\alpha.$$

Divide through by R:

$$F - f = \frac{1}{2}mR\,\alpha.$$

Using the rolling condition, $\alpha = \frac{a}{R} = \frac{f}{mR}$, substitute:

$$F - f = \frac{1}{2}mR\,\left(\frac{f}{mR}\right)=\frac{1}{2} f.$$

This simplifies to:

$$F = f + \frac{1}{2}f=\frac{3}{2}f \quad\Longrightarrow\quad f = \frac{2}{3}F.$$

Thus the friction force is $\frac{2}{3}F$, which means $n=\frac{2}{3}$.