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 outlined procedure:

### Step 1: Identify the given values
- Mass (m) = 10 kg
- Radius (R) = 30 cm = 0.3 m
- Initial angular velocity (ω₀) = 10 rad/s
- Final angular velocity (ω) = 0 rad/s (since the mass is brought to rest)
- Time (T) = 10 s

### Step 2: Calculate the angular acceleration (α)
We can use the formula for angular velocity:
\[ \omega = \omega_0 + \alpha T \]
Substituting the known values:
\[ 0 = 10 + \alpha \cdot 10 \]
Rearranging gives:
\[ \alpha \cdot 10 = -10 \implies \alpha = -1 \text{ rad/s}^2 \]

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

### Step 4: Calculate the torque (τ)
Using the relationship between torque, moment of inertia, and angular acceleration:
\[ \tau = I \alpha \]
Substituting the values we calculated:
\[ \tau = 0.9 \cdot (-1) = -0.9 \text{ N m} \]

### Step 5: Determine the magnitude of the torque
Since we are interested in the magnitude of the torque, we take the absolute value:
\[ |\tau| = 0.9 \text{ N m} \]

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

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