A solid sphere and a solid cylinder having the same mass and radius, rolls down the same incline. The ratio of their acceleration will be

Options

A. `15 : 14`
B. `14 : 15`
C. `5 : 3`
D. `3 : 5`

Explanation

To solve the problem of finding the ratio of acceleration of a solid sphere and a solid cylinder rolling down an incline, we can follow these steps:

### Step 1: Identify the forces acting on the objects
Both the solid sphere and the solid cylinder experience gravitational force acting down the incline, which can be expressed as:
\[ F_{\text{gravity}} = mg \sin \theta \]
where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of the incline.

### Step 2: Set up the equations of motion
For both objects, the net force acting down the incline can be expressed as:
\[ F_{\text{net}} = mg \sin \theta - f \]
where \( f \) is the frictional force. According to Newton's second law, this can be written as:
\[ mg \sin \theta - f = ma \]
where \( a \) is the linear acceleration of the object.

### Step 3: Relate linear acceleration to angular acceleration
Since both objects are rolling without slipping, we can relate the linear acceleration \( a \) to the angular acceleration \( \alpha \) using the relationship:
\[ a = \alpha r \]
where \( r \) is the radius of the object. This can be rearranged to find \( \alpha \):
\[ \alpha = \frac{a}{r} \]

### Step 4: Write the torque equation
The torque \( \tau \) due to the frictional force about the center of mass is given by:
\[ \tau = f \cdot r \]
This torque can also be expressed in terms of the moment of inertia \( I \) and angular acceleration \( \alpha \):
\[ \tau = I \alpha \]
Substituting for \( \alpha \):
\[ f \cdot r = I \left(\frac{a}{r}\right) \]
From this, we can express the frictional force:
\[ f = \frac{I a}{r^2} \]

### Step 5: Substitute into the net force equation
Now, substituting the expression for \( f \) into the net force equation:
\[ mg \sin \theta - \frac{I a}{r^2} = ma \]
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 \( a \):
\[ a = \frac{mg \sin \theta}{m + \frac{I}{r^2}} \]

### Step 6: Calculate the moment of inertia for both shapes
For a solid sphere:
\[ I_{\text{sphere}} = \frac{2}{5} m r^2 \]
For a solid cylinder:
\[ I_{\text{cylinder}} = \frac{1}{2} m r^2 \]

### Step 7: Substitute the moments of inertia into the acceleration formula
For the solid sphere:
\[ a_{\text{sphere}} = \frac{mg \sin \theta}{m + \frac{2/5 \cdot m r^2}{r^2}} = \frac{mg \sin \theta}{m + \frac{2}{5} m} = \frac{mg \sin \theta}{\frac{7}{5} m} = \frac{5g \sin \theta}{7} \]

For the solid cylinder:
\[ a_{\text{cylinder}} = \frac{mg \sin \theta}{m + \frac{1/2 \cdot m r^2}{r^2}} = \frac{mg \sin \theta}{m + \frac{1}{2} m} = \frac{mg \sin \theta}{\frac{3}{2} m} = \frac{2g \sin \theta}{3} \]

### Step 8: Find the ratio of their accelerations
Now, we can find the ratio of the acceleration of the solid cylinder to that of the solid sphere:
\[ \text{Ratio} = \frac{a_{\text{cylinder}}}{a_{\text{sphere}}} = \frac{\frac{2g \sin \theta}{3}}{\frac{5g \sin \theta}{7}} = \frac{2}{3} \cdot \frac{7}{5} = \frac{14}{15} \]

### Final Answer
The ratio of the accelerations of the solid cylinder to the solid sphere is:
\[ \frac{14}{15} \]