A mass of 10 kg connected at the end of a rod of negligible mass is rotating in a circle of radius 30 cm with an angular velocity of 10 rad/s. If this mass is brought to rest in 10 s by a brake, what is the magnitude of the torque applied ?

Options

A. 0.9 N-m
B. 1.2 N-m
C. 2.3 N-m
D. 0.5 N-m

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Identify the given values
- Mass (m) = 10 kg
- Radius (r) = 30 cm = 0.3 m (convert cm to m)
- Initial angular velocity (ω₁) = 10 rad/s
- Final angular velocity (ω₂) = 0 rad/s (since the mass is brought to rest)
- Time (t) = 10 s

### Step 2: Calculate the angular deceleration (α)
The angular deceleration can be calculated using the formula:
\[ \alpha = \frac{\omega_2 - \omega_1}{t} \]
Substituting the values:
\[ \alpha = \frac{0 - 10}{10} = \frac{-10}{10} = -1 \text{ rad/s}^2 \]
The negative sign indicates that it is a deceleration (retardation).

### Step 3: Calculate the moment of inertia (I)
The moment of inertia for a point mass rotating about an axis is given by:
\[ I = m r^2 \]
Substituting the values:
\[ I = 10 \times (0.3)^2 = 10 \times 0.09 = 0.9 \text{ kg m}^2 \]

### Step 4: Calculate the torque (τ)
The torque can be calculated using the formula:
\[ \tau = I \alpha \]
Substituting the values:
\[ \tau = 0.9 \times (-1) = -0.9 \text{ N m} \]
The magnitude of the torque is:
\[ |\tau| = 0.9 \text{ N m} \]

### Final Answer
The magnitude of the torque applied is **0.9 N m**.

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