A solid sphere, disc and solid cylinder, all of the same mass, are allowed to roll down (from rest) on inclined plane, them

Options

A. solid sphere reaches the bottom first
B. solid sphere reaches the bottom last
C. disc will reach the bottom first
D. all reach the bottom at the same time

Answer

To solve the problem of which object (solid sphere, disc, or solid cylinder) reaches the bottom of an inclined plane first when rolling from rest, we can follow these steps:

### Step 1: Understand the Rolling Motion
When objects roll down an incline, they convert potential energy into both translational and rotational kinetic energy. The total mechanical energy at the top is equal to the total mechanical energy at the bottom.

### Step 2: Write the Energy Conservation Equation
The potential energy (PE) at the top is given by:
\[ PE = mgh \]
where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height of the incline.

At the bottom, the total kinetic energy (KE) is the sum of translational and rotational kinetic energy:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
where \( v \) is the linear velocity, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity. For rolling without slipping, \( \omega = \frac{v}{r} \).

### Step 3: Substitute Moment of Inertia
The moment of inertia \( I \) for each object is:
- Solid Sphere: \( I = \frac{2}{5} m r^2 \)
- Solid Cylinder: \( I = \frac{1}{2} m r^2 \)
- Disc: \( I = \frac{1}{2} m r^2 \)

### Step 4: Calculate the Effective Velocity
Using the relationship \( \omega = \frac{v}{r} \), we can rewrite the kinetic energy for each object:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left( \frac{I}{r^2} \right) v^2 \]

Substituting \( I \):
1. For the solid sphere:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left( \frac{2}{5} m r^2 \right) \frac{v^2}{r^2} = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2 \]

2. For the solid cylinder and disc:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left( \frac{1}{2} m r^2 \right) \frac{v^2}{r^2} = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]

### Step 5: Set Up the Equation for Each Object
From energy conservation:
\[ mgh = KE \]
This leads to:
1. For the solid sphere:
\[ mgh = \frac{7}{10} mv^2 \Rightarrow v^2 = \frac{10gh}{7} \]

2. For the solid cylinder and disc:
\[ mgh = \frac{3}{4} mv^2 \Rightarrow v^2 = \frac{4gh}{3} \]

### Step 6: Compare Velocities
To determine which object reaches the bottom first, we compare the velocities:
- Solid Sphere: \( v^2 = \frac{10gh}{7} \)
- Solid Cylinder/Disc: \( v^2 = \frac{4gh}{3} \)

Calculating the ratios:
- For the solid sphere: \( \frac{10}{7} \approx 1.43 \)
- For the solid cylinder/disc: \( \frac{4}{3} \approx 1.33 \)

Since \( \frac{10}{7} > \frac{4}{3} \), the solid sphere has the highest velocity.

### Conclusion
The solid sphere reaches the bottom of the incline first, followed by the solid cylinder and disc, which reach the bottom at the same time.