A uniform rod of mass m and length 2a lies at rest on rotating with angular speed `omega_(0)=40rad//s` is placed between two smooth walls on a rough ground. Distance between the walls is slightly greater than the lenght of the rod . Cofficent of friction between the rod and the ground is `mu`=0.1. rod will stop rotating after time t=.......s.

Answer

To solve the problem, we need to determine the time it takes for the uniform rod to stop rotating after being placed on a rough ground. Here's the step-by-step solution:

### Step 1: Identify the given values
- Initial angular speed, \( \omega_0 = 40 \, \text{rad/s} \)
- Coefficient of friction, \( \mu = 0.1 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- Length of the rod, \( L = 2a \) (not directly needed for calculations)

### Step 2: Calculate the torque due to friction
The torque (\( \tau \)) acting on the rod due to friction can be calculated using the formula:
\[ \tau = \mu m g r \]
where \( r \) is the radius of the rod. For a uniform rod, we can consider \( r \) to be half the length of the rod, which is \( a \).

### Step 3: Calculate the moment of inertia of the rod
The moment of inertia (\( I \)) of a uniform rod about its end is given by:
\[ I = \frac{1}{3} m L^2 = \frac{1}{3} m (2a)^2 = \frac{4}{3} m a^2 \]

### Step 4: Calculate the angular retardation (\( \alpha \))
Using the relationship between torque, moment of inertia, and angular acceleration:
\[ \alpha = \frac{\tau}{I} \]
Substituting the values we have:
\[ \alpha = \frac{\mu m g r}{\frac{4}{3} m a^2} \]
Since \( r = a \):
\[ \alpha = \frac{\mu m g a}{\frac{4}{3} m a^2} = \frac{3 \mu g}{4a} \]
Substituting \( \mu = 0.1 \) and \( g = 10 \):
\[ \alpha = \frac{3 \times 0.1 \times 10}{4} = \frac{3}{4} = 0.75 \, \text{rad/s}^2 \]

### Step 5: Use the angular motion equation to find time (\( t \))
We use the equation for angular motion:
\[ \omega_f = \omega_i + \alpha t \]
Where:
- \( \omega_f = 0 \) (final angular velocity)
- \( \omega_i = 40 \, \text{rad/s} \)
- \( \alpha = -0.75 \, \text{rad/s}^2 \) (negative because it's retardation)

Setting up the equation:
\[ 0 = 40 - 0.75 t \]
Rearranging gives:
\[ 0.75 t = 40 \]
Thus,
\[ t = \frac{40}{0.75} = \frac{4000}{75} \approx 16 \, \text{s} \]

### Final Answer
The rod will stop rotating after approximately \( t = 16 \, \text{s} \).

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