A body starts rolling down an inclined plane of length L and height h. This body reaches the bottom of the plane in time t. The relation between L and t is?

Options

A. `t prop L`
B. `t prop 1/L`
C. `L prop t^2`
D. `t prop 1/L^2`

Answer

To find the relation between the length \( L \) of the inclined plane and the time \( t \) taken by a body to roll down, we can follow these steps:

### Step 1: Understand the Problem
We have a body rolling down an inclined plane of length \( L \) and height \( h \). We need to derive a relationship between \( L \) and \( t \).

### Step 2: Identify Forces Acting on the Body
When the body rolls down, the forces acting on it include:
- The gravitational force \( mg \) acting downwards.
- The component of gravitational force along the incline, which is \( mg \sin \theta \).
- The frictional force that allows rolling without slipping.

### Step 3: Apply Newton's Second Law
Using Newton's second law, we can write the equation of motion for the body:
\[ F_{\text{net}} = mg \sin \theta - f = ma \]
Where \( f \) is the frictional force and \( a \) is the linear acceleration of the body.

### Step 4: Relate Linear Acceleration to Angular Acceleration
Since the body is rolling, we can relate the linear acceleration \( a \) to the angular acceleration \( \alpha \) using:
\[ a = r \alpha \]
Where \( r \) is the radius of the body.

### Step 5: Torque Due to Friction
The torque \( \tau \) due to friction is given by:
\[ \tau = r f = I \alpha \]
Where \( I \) is the moment of inertia of the body. For a general body, we can express \( I \) as \( K m r^2 \), where \( K \) is a constant depending on the shape of the body.

### Step 6: Substitute and Solve for Acceleration
From the torque equation, we can express the frictional force \( f \) as:
\[ f = \frac{I \alpha}{r} = \frac{K m r^2 \alpha}{r} = K m r \alpha \]
Substituting \( \alpha = \frac{a}{r} \) gives:
\[ f = K m a \]
Now substituting \( f \) back into the equation of motion:
\[ mg \sin \theta - K m a = ma \]
Rearranging gives:
\[ mg \sin \theta = (1 + K) m a \]
Thus, the acceleration \( a \) is:
\[ a = \frac{g \sin \theta}{1 + K} \]

### Step 7: Relate Distance, Acceleration, and Time
Using the second equation of motion for distance \( s \):
\[ s = ut + \frac{1}{2} a t^2 \]
Since the body starts from rest, \( u = 0 \):
\[ L = \frac{1}{2} a t^2 \]
Substituting \( a \):
\[ L = \frac{1}{2} \left(\frac{g \sin \theta}{1 + K}\right) t^2 \]

### Step 8: Establish the Relationship
From the equation \( L = \frac{g \sin \theta}{2(1 + K)} t^2 \), we can see that \( L \) is proportional to \( t^2 \):
\[ L \propto t^2 \]

### Conclusion
The relation between the length \( L \) of the inclined plane and the time \( t \) taken by the body to reach the bottom is:
\[ L \propto t^2 \]