Question

A disc of radius 0.4 m and mass one kg rotates about an axis passing through its centre and perpendicular to its plane. The angular acceleration of the disc is $10\ \text{rad/s}^2$. The tangential force applied to the rim of the disc is

• A. 4 N
• B. 1 N
• C. 2 N
• D. 8 N

Hint:

Combining the expressions directly into one step saves time: $F = \frac{\left(\frac{1}{2}MR^2\right)\alpha}{R} = \frac{1}{2}MR\alpha$. Substituting the numbers directly gives $\frac{1}{2} \times 1 \times 0.4 \times 10 = 2\ \text{N}$ instantly!

Answer 1

The Correct Option is C

The moment of inertia ($I$) of a uniform solid disc rotating about its central axis perpendicular to its plane is: $$I = \frac{1}{2}MR^2$$ Substituting the given values ($M = 1\ \text{kg}$ and $R = 0.4\ \text{m}$): $$I = \frac{1}{2} \times 1 \times (0.4)^2 = \frac{0.16}{2} = 0.08\ \text{kg}\cdot\text{m}^2$$ Torque ($\tau$) can be expressed both in terms of angular acceleration ($\alpha$) and the applied tangential force ($F$) acting at the radius boundary: $$\tau = I\alpha \quad \text{and} \quad \tau = F \cdot R$$ Equating the two expressions to solve for the force $F$: $$F \cdot R = I\alpha \implies F = \frac{I\alpha}{R}$$ $$F = \frac{0.08 \times 10}{0.4} = \frac{0.8}{0.4} = 2\ \text{N}$$
Final Answer:
The tangential force applied to the rim of the disc is 2 N, which corresponds to option (C).