A sphere of mass 2 kg and radius 5 cm is rotating at the rate of 300 rpm. Calculate the torque required to stop it in 6.28 it in 6.28 revolution. Moment of inertia of the sphere about any diameter `=2/5 MR^(2)`

Answer

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

### Given Data:
- Mass of the sphere (m) = 2 kg
- Radius of the sphere (r) = 5 cm = 0.05 m
- Initial rotational speed (N) = 300 rpm
- Number of revolutions to stop (n) = 6.28 revolutions

### Step 1: Convert the rotational speed from rpm to rad/s
To convert revolutions per minute (rpm) to radians per second (rad/s), we use the conversion factor:
\[ \omega_i = N \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} \]
Substituting the values:
\[ \omega_i = 300 \times \frac{2\pi}{60} = 10\pi \text{ rad/s} \]

### Step 2: Calculate the angular displacement in radians
Since 6.28 revolutions is equivalent to:
\[ \theta = 6.28 \times 2\pi = 4\pi \text{ rad} \]

### Step 3: Calculate the moment of inertia (I) of the sphere
The moment of inertia (I) for a solid sphere about any diameter is given by:
\[ I = \frac{2}{5} m r^2 \]
Substituting the values:
\[ I = \frac{2}{5} \times 2 \times (0.05)^2 = \frac{2}{5} \times 2 \times 0.0025 = \frac{2 \times 0.005}{5} = \frac{0.01}{5} = 0.002 \text{ kg m}^2 \]

### Step 4: Use the kinematic equation to find angular acceleration (α)
Using the equation:
\[ \omega_f^2 = \omega_i^2 + 2\alpha\theta \]
Since the final angular velocity (\(\omega_f\)) is 0 (the sphere stops), we have:
\[ 0 = (10\pi)^2 + 2\alpha(4\pi) \]
Rearranging gives:
\[ \alpha = -\frac{(10\pi)^2}{2 \times 4\pi} = -\frac{100\pi^2}{8\pi} = -\frac{25\pi}{2} \text{ rad/s}^2 \]

### Step 5: Calculate the torque (τ) required to stop the sphere
Torque is related to moment of inertia and angular acceleration by:
\[ \tau = I \alpha \]
Substituting the values:
\[ \tau = 0.002 \times \left(-\frac{25\pi}{2}\right) = -\frac{0.05\pi}{2} \text{ N m} \]
Calculating the numerical value:
\[ \tau = -\frac{0.05 \times 3.14}{2} \approx -0.0785 \text{ N m} \]

### Final Answer:
The magnitude of the torque required to stop the sphere is approximately \(0.0785 \text{ N m}\).

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