When a rigid sphere is projected horizontally with an initial linear velocity over a uniformly rough horizontal floor

Options

A. the kinetic energy of the sphere remains constant throughout the motion
B. the kinetic energy of translation is slowly shared by rotation, but the total kinetic energy decreases initially and finally attains a constant value
C. the total kinetic energy of the sphere decreases continuously on doing work against friction and finally reduces to zero
D. a constant frictional force, opposite to the translation motion acts on the sphere through out its motion

Explanation

To solve the problem of a rigid sphere projected horizontally with an initial linear velocity over a uniformly rough horizontal floor, we will analyze the motion step by step.

### Step 1: Understand the Initial Conditions
- A rigid sphere is projected horizontally with an initial velocity \( v \).
- The sphere is on a rough horizontal surface, which means there will be friction acting on it.

**Hint:** Identify the forces acting on the sphere and their directions.

### Step 2: Identify the Forces Acting on the Sphere
- The primary force acting on the sphere is the frictional force, which acts in the opposite direction to the motion (backward).
- This frictional force will cause a deceleration (retardation) in the linear motion of the sphere.

**Hint:** Remember that friction opposes the direction of motion.

### Step 3: Analyze the Motion
- As the sphere moves, the frictional force will reduce its linear velocity \( v \).
- Simultaneously, the friction will create a torque about the center of mass, causing the sphere to start rotating.

**Hint:** Consider the relationship between linear and angular motion.

### Step 4: Relate Linear and Angular Quantities
- The angular acceleration \( \alpha \) is related to the linear acceleration \( a \) by the equation:
\[ a = R \alpha \]
where \( R \) is the radius of the sphere.
- The linear velocity \( v \) and angular velocity \( \omega \) are related by:
\[ v = R \omega \]

**Hint:** Use the equations of motion for both linear and rotational dynamics.

### Step 5: Determine the Condition for Pure Rolling
- The sphere will eventually reach a state of pure rolling when the point of contact with the ground has zero velocity.
- This occurs when:
\[ v = R \omega \]
- At this point, the frictional force does no work, and the kinetic energy becomes constant.

**Hint:** Think about what happens to the velocities at the point of pure rolling.

### Step 6: Analyze Kinetic Energy Changes
- Initially, the kinetic energy of the sphere is:
\[ KE_{\text{initial}} = \frac{1}{2} mv^2 \]
- As the sphere slows down, the translational kinetic energy decreases while the rotational kinetic energy increases:
\[ KE_{\text{rotational}} = \frac{1}{2} I \omega^2 \]
- The total kinetic energy will initially decrease due to work done against friction, but after reaching pure rolling, it will remain constant.

**Hint:** Consider how energy is transferred between translational and rotational forms.

### Step 7: Conclusion
- The correct statements regarding the motion of the sphere are:
- The translational kinetic energy decreases while the rotational kinetic energy increases until pure rolling is achieved.
- The total kinetic energy decreases initially but becomes constant after pure rolling is established.

**Final Answer:** The correct option is that the kinetic energy of translation is shared by rotation, and the total kinetic energy decreases initially but becomes constant after pure rolling.