Calculate the acceleration of a hollow cylinder rolling down an inclined plane of inclination 30.

Answer

To calculate the acceleration of a hollow cylinder rolling down an inclined plane, we can follow these steps:

### Step 1: Identify the forces acting on the hollow cylinder
When the hollow cylinder is rolling down the inclined plane, two main forces act on it:
- The gravitational force (mg) acting downwards.
- The normal force (N) acting perpendicular to the surface of the incline.

### Step 2: Resolve the gravitational force
The gravitational force can be resolved into two components:
- The component parallel to the incline: \( mg \sin(\theta) \)
- The component perpendicular to the incline: \( mg \cos(\theta) \)

For an inclination angle of \( \theta = 30^\circ \):
- \( \sin(30^\circ) = \frac{1}{2} \)
- \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)

Thus, the parallel component is:
\[ F_{\text{parallel}} = mg \sin(30^\circ) = mg \cdot \frac{1}{2} = \frac{mg}{2} \]

### Step 3: Write the equation of motion
According to Newton's second law, the net force acting on the hollow cylinder is equal to the mass times its acceleration (a):
\[ F_{\text{net}} = ma \]

The net force acting down the incline is the component of gravitational force:
\[ \frac{mg}{2} = ma \]

### Step 4: Consider rotational motion
Since the cylinder is rolling without slipping, we also need to consider its rotational motion. The moment of inertia (I) for a hollow cylinder about its central axis is given by:
\[ I = mR^2 \]
where R is the radius of the cylinder.

The torque (\( \tau \)) caused by the gravitational force about the center of mass is:
\[ \tau = R \cdot F_{\text{parallel}} = R \cdot \frac{mg}{2} \]

Using the relation \( \tau = I \alpha \) (where \( \alpha \) is the angular acceleration), we can write:
\[ R \cdot \frac{mg}{2} = mR^2 \alpha \]

### Step 5: Relate linear and angular acceleration
For rolling motion without slipping, the linear acceleration (a) and angular acceleration (\( \alpha \)) are related by:
\[ a = R \alpha \]
Thus, we can substitute \( \alpha \) with \( \frac{a}{R} \):
\[ R \cdot \frac{mg}{2} = mR^2 \cdot \frac{a}{R} \]

### Step 6: Simplify the equation
This simplifies to:
\[ \frac{mg}{2} = ma \]

### Step 7: Solve for acceleration (a)
Now, we can solve for acceleration (a):
\[ \frac{mg}{2} = ma \]
Dividing both sides by m gives:
\[ \frac{g}{2} = a \]
Thus, the acceleration of the hollow cylinder is:
\[ a = \frac{g}{2} \]

### Step 8: Substitute the value of g
Using \( g \approx 9.8 \, \text{m/s}^2 \):
\[ a = \frac{9.8}{2} = 4.9 \, \text{m/s}^2 \]

### Final Answer:
The acceleration of the hollow cylinder rolling down the inclined plane is \( 4.9 \, \text{m/s}^2 \).

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