A homogeneous cylinder of mass Mand radius r is pulled on a horizontal plane by a horizontal force F acting through its centre of mass. Assuming rolling without slipping, find the angular acceleration of the cylinder,

Options

A. `(3F)/(2MR)`
B. `(2F)/(3MR)`
C. `(F)/(2MR)`
D. `(3F)/(4MR)`

Answer

To find the angular acceleration of a homogeneous cylinder being pulled by a horizontal force \( F \) while rolling without slipping, we can follow these steps:

### Step 1: Identify the forces and torques acting on the cylinder
The horizontal force \( F \) is applied at the center of mass of the cylinder. This force will cause the cylinder to both translate and rotate. The torque \( \tau \) about the center of mass due to the force \( F \) is given by:
\[ \tau = F \cdot r \]
where \( r \) is the radius of the cylinder.

### Step 2: Write the equation for torque
According to the rotational dynamics, the torque is also related to the moment of inertia \( I \) and the angular acceleration \( \alpha \) by the equation:
\[ \tau = I \cdot \alpha \]

### Step 3: Determine the moment of inertia of the cylinder
For a homogeneous cylinder, the moment of inertia \( I \) about an axis through its center is given by:
\[ I = \frac{1}{2} M r^2 \]

### Step 4: Set up the equations
From Step 1, we have:
\[ F \cdot r = I \cdot \alpha \]
Substituting the moment of inertia from Step 3 into this equation gives:
\[ F \cdot r = \left(\frac{1}{2} M r^2\right) \cdot \alpha \]

### Step 5: Solve for angular acceleration \( \alpha \)
Rearranging the equation to solve for \( \alpha \):
\[ \alpha = \frac{F \cdot r}{\frac{1}{2} M r^2} \]
This simplifies to:
\[ \alpha = \frac{2F}{M r} \]

### Final Answer
Thus, the angular acceleration \( \alpha \) of the cylinder is:
\[ \alpha = \frac{2F}{M r} \]

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