Question

A disc of radius 0.4 metre and mass 1 kg rotates about an axis passing through its centre and perpendicular to its plane. The angular acceleration is $10\ \text{rad}/\text{s}^2$. The tangential force applied to the rim of the disc is

• A. 2 N
• B. 3 N
• C. 4 N
• D. 5 N

Hint:

Combine the formulas directly to save time: $F \times R = (\frac{1}{2}MR^2)\alpha \implies F = \frac{1}{2}MR\alpha$. Simply plug in $0.5 \times 1 \times 0.4 \times 10 = 2\ \text{N}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the tangential force required to produce a specific angular acceleration on a rotating disc.

Step 2: Key Formula or Approach:
The moment of inertia of a solid disc is $I = \frac{1}{2}MR^2$.
Torque ($\tau$) is related to angular acceleration ($\alpha$) by $\tau = I\alpha$.
Torque is also the product of tangential force ($F$) and radius ($R$): $\tau = F \times R$.

Step 3: Detailed Explanation:
Given $M = 1\ \text{kg}$ and $R = 0.4\ \text{m}$.
Calculate the moment of inertia:
$$I = \frac{1}{2}(1)(0.4)^2 = \frac{1}{2}(0.16) = 0.08\ \text{kg}\cdot\text{m}^2$$ Calculate the required torque for $\alpha = 10\ \text{rad}/\text{s}^2$:
$$\tau = I\alpha = 0.08 \times 10 = 0.8\ \text{N}\cdot\text{m}$$ Since the force is applied tangentially at the rim, distance is $R$:
$$F = \frac{\tau}{R} = \frac{0.8}{0.4} = 2\ \text{N}$$

Step 4: Final Answer:
The tangential force is 2 N, matching option (A).