A cylinder rolls down two different inclined planes of the same height but of different inclinations

Options

A. In both cases the speed and time of descent will be different
B. In both cases the speed and time of descent will be same
C. The speed will be different but time of descent will be same
D. The time of descent will be different but speed will be same

Answer

To solve the problem of a cylinder rolling down two different inclined planes of the same height but with different inclinations, we can follow these steps:

### Step 1: Identify Forces Acting on the Cylinder
When the cylinder rolls down the incline, 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 normal force \( N \) acting perpendicular to the incline, which is \( mg \cos \theta \).
- The frictional force \( f \) acting up the incline, which is necessary for rolling without slipping.

### Step 2: Write the Equations of Motion
For the motion along the incline, we can write the equation:
\[ mg \sin \theta - f = ma \quad \text{(1)} \]
where \( a \) is the linear acceleration of the center of mass of the cylinder.

For the rotational motion about the center of mass, we have:
\[ f \cdot R = I \alpha \quad \text{(2)} \]
where \( R \) is the radius of the cylinder, \( I \) is the moment of inertia, and \( \alpha \) is the angular acceleration.

### Step 3: Relate Linear and Angular Acceleration
Since the cylinder rolls without slipping, we have the relationship:
\[ a = \alpha R \quad \text{(3)} \]

### Step 4: Substitute Angular Acceleration
From equation (3), we can express \( \alpha \) in terms of \( a \):
\[ \alpha = \frac{a}{R} \]
Substituting this into equation (2) gives:
\[ f \cdot R = I \left(\frac{a}{R}\right) \]
Rearranging this, we find:
\[ f = \frac{I a}{R^2} \quad \text{(4)} \]

### Step 5: Substitute Friction in the Linear Motion Equation
Substituting equation (4) into equation (1):
\[ mg \sin \theta - \frac{I a}{R^2} = ma \]
Rearranging gives:
\[ mg \sin \theta = ma + \frac{I a}{R^2} \]
Factoring out \( a \):
\[ mg \sin \theta = a \left(m + \frac{I}{R^2}\right) \]
Thus, we can solve for \( a \):
\[ a = \frac{mg \sin \theta}{m + \frac{I}{R^2}} \quad \text{(5)} \]

### Step 6: Moment of Inertia for a Solid Cylinder
For a solid cylinder, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{2} m R^2 \]
Substituting this into equation (5):
\[ a = \frac{mg \sin \theta}{m + \frac{1}{2} m} = \frac{mg \sin \theta}{\frac{3}{2} m} = \frac{2g \sin \theta}{3} \]

### Step 7: Determine the Final Speed
Using energy conservation, the potential energy at the top converts to kinetic energy at the bottom:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
Substituting \( I \) and using \( \omega = \frac{v}{R} \):
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} m R^2\right) \left(\frac{v^2}{R^2}\right) \]
This simplifies to:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]
Solving for \( v \):
\[ v = \sqrt{\frac{4gh}{3}} \]

### Step 8: Time of Descent
Using the kinematic equation \( v = u + at \) (where \( u = 0 \)):
\[ t = \frac{v}{a} = \frac{\sqrt{\frac{4gh}{3}}}{\frac{2g \sin \theta}{3}} = \frac{3}{2g \sin \theta} \sqrt{\frac{4gh}{3}} = \frac{3h}{\sqrt{12g \sin^2 \theta}} \]

### Conclusion
The speed of the cylinder at the bottom of both inclines will be the same, but the time taken to reach the bottom will differ depending on the angle of inclination.

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