Question

A rigid circular disc of radius 0.2 m and mass 10 kg rolls without slip on the ground at A. The coefficient of static friction \( \mu \) between ground and disc is 0.7. A torque \( T \) of 9 Nm acts on the disc as shown. Given acceleration due to gravity \( g = 10 \, \text{m/s}^2 \). The friction force (in N) acting on the disc (in integer) is \(\underline{\hspace{2cm}}\). \includegraphics[width=0.5\linewidth]{image16.png}

Hint:

For rolling motion without slipping, the frictional force is given by the relationship \( T = F_f \times r \), where \( T \) is the applied torque and \( r \) is the radius.

Answer 1

Correct Answer: 30

For rolling without slipping, the frictional force \( F_f \) can be found using the following equation: \[ T = F_f \times r \] Where:
- \( T = 9 \, \text{Nm} \),
- \( r = 0.2 \, \text{m} \) (radius of the disc),
- \( F_f = \frac{T}{r} = \frac{9}{0.2} = 45 \, \text{N}. \) Thus, the friction force acting on the disc is \( 45 \, \text{N} \).