A uniform solid cylinder of mass 5kg and radius 0.1m is resting on a horizontal platform (parallel to the x-y plane) and is free to rotate about its axis along the y-axis the platform is given a motion in the x direction given by x=0.2 cos (10t) m if there is no slipping then maximum torque acting on the cylinder during its motion is

Options

A. `0.2 N-m`
B. `2.0 N-m`
C. `5.0 N -m`
D. `10.0 N-m`

Answer

To find the maximum torque acting on the cylinder during its motion, we can follow these steps:

### Step 1: Understand the motion of the platform
The platform moves in the x-direction according to the equation:
\[ x(t) = 0.2 \cos(10t) \]

### Step 2: Find the velocity of the cylinder
To find the velocity of the cylinder, we differentiate the position with respect to time:
\[ v(t) = \frac{dx}{dt} = -0.2 \cdot 10 \sin(10t) = -2 \sin(10t) \, \text{m/s} \]

### Step 3: Find the acceleration of the cylinder
Next, we differentiate the velocity to find the acceleration:
\[ a(t) = \frac{dv}{dt} = -2 \cdot 10 \cos(10t) = -20 \cos(10t) \, \text{m/s}^2 \]

### Step 4: Determine the maximum acceleration
The maximum value of \( a(t) \) occurs when \( \cos(10t) = 1 \):
\[ a_{\text{max}} = 20 \, \text{m/s}^2 \]

### Step 5: Relate torque to angular acceleration
The torque \( \tau \) acting on the cylinder is given by:
\[ \tau = I \alpha \]
where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration. For a solid cylinder, the moment of inertia \( I \) is:
\[ I = \frac{1}{2} m r^2 \]
where \( m = 5 \, \text{kg} \) and \( r = 0.1 \, \text{m} \).

### Step 6: Calculate the moment of inertia
Substituting the values:
\[ I = \frac{1}{2} \times 5 \times (0.1)^2 = \frac{1}{2} \times 5 \times 0.01 = 0.025 \, \text{kg m}^2 \]

### Step 7: Relate linear acceleration to angular acceleration
Since there is no slipping, we have:
\[ a = \alpha r \]
Thus, the angular acceleration \( \alpha \) can be expressed as:
\[ \alpha = \frac{a}{r} = \frac{20}{0.1} = 200 \, \text{rad/s}^2 \]

### Step 8: Calculate the maximum torque
Now we can calculate the maximum torque:
\[ \tau_{\text{max}} = I \alpha = 0.025 \times 200 = 5 \, \text{N m} \]

### Final Answer
The maximum torque acting on the cylinder during its motion is:
\[ \tau_{\text{max}} = 5 \, \text{N m} \]

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