Question

A force of 49 N acts tangentially at the highest point of a sphere (solid) of mass 20 kg, kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is

• A. 3.5 \( m/s^2 \)
• B. 0.35 \( m/s^2 \)
• C. 2.5 \( m/s^2 \)
• D. 0.25 \( m/s^2 \)

Hint:

Apply the torque equation about the point of contact to find the angular acceleration. Use the relation between linear acceleration and angular acceleration for rolling without slipping.

Answer 1

The Correct Option is A

This problem requires determining the acceleration of the center of a solid sphere undergoing rolling without slipping on a rough horizontal surface. A tangential force is applied at the sphere's highest point.

Concepts Utilized:

The solution employs principles of linear and rotational motion for rigid bodies, coupled with the condition for rolling without slipping.

1. Newton's Second Law for Linear Motion: The sum of external forces on a body equals its mass multiplied by the acceleration of its center of mass.

\[ \Sigma F_{ext} = m a_{cm} \]

2. Newton's Second Law for Rotational Motion: The net external torque about the center of mass equals its moment of inertia about the center of mass multiplied by its angular acceleration.

\[ \Sigma \tau_{cm} = I_{cm} \alpha \]

3. Moment of Inertia of a Solid Sphere: For a solid sphere of mass m and radius r, its moment of inertia about its center is:

\[ I_{cm} = \frac{2}{5} m r^2 \]

4. Condition for Rolling without Slipping: The linear acceleration of the center of mass is directly proportional to the angular acceleration, with the radius as the constant of proportionality:

\[ a_{cm} = \alpha r \]

Solution Procedure:

Step 1: Force identification and equation of motion setup.

Let \( a \) denote the sphere's center acceleration, \( \alpha \) its angular acceleration, and \( f_s \) the static friction force at the point of contact. The applied force is \( F = 49 \, \text{N} \) and the mass is \( m = 20 \, \text{kg} \).

The applied force \( F \) drives the sphere's forward acceleration. It also generates a clockwise torque. To counteract this and prevent slipping, the static friction force \( f_s \) must act forward (to the right).

The linear motion equation for the horizontal direction is:

\[ F + f_s = ma \quad \cdots (1) \]

Step 2: Rotational motion equation setup.

Torques are considered about the center of mass. The applied force \( F \) produces a clockwise torque of magnitude \( F \cdot r \). The friction force \( f_s \) generates a counter-clockwise torque of magnitude \( f_s \cdot r \). Assuming clockwise angular acceleration is positive, the net torque is:

\[ \tau_{net} = F \cdot r - f_s \cdot r \]

Applying Newton's second law for rotation, \( \tau_{net} = I \alpha \):

\[ F \cdot r - f_s \cdot r = I \alpha \quad \cdots (2) \]

Step 3: Application of the rolling without slipping condition.

Substituting \( I = \frac{2}{5} m r^2 \) and \( \alpha = \frac{a}{r} \) into equation (2):

\[ r(F - f_s) = \left( \frac{2}{5} m r^2 \right) \left( \frac{a}{r} \right) \]

Simplifying by dividing both sides by r:

\[ F - f_s = \frac{2}{5} ma \quad \cdots (3) \]

Step 4: Resolution of the system of linear equations for acceleration \( a \).

The system of equations with unknowns \( a \) and \( f_s \) is:

Equation (1): \( F + f_s = ma \)

Equation (3): \( F - f_s = \frac{2}{5} ma \)

Adding these equations eliminates \( f_s \):

\[ (F + f_s) + (F - f_s) = ma + \frac{2}{5} ma \]\[ 2F = \left( 1 + \frac{2}{5} \right) ma = \frac{7}{5} ma \]

Solving for \( a \):

\[ a = \frac{2F}{\frac{7}{5}m} = \frac{10F}{7m} \]

Final Calculation & Outcome:

Inserting the given values \( F = 49 \, \text{N} \) and \( m = 20 \, \text{kg} \) into the acceleration formula:

\[ a = \frac{10 \times 49 \, \text{N}}{7 \times 20 \, \text{kg}} \]\[ a = \frac{490}{140} \, \text{m/s}^2 = \frac{49}{14} \, \text{m/s}^2 \]\[ a = \frac{7}{2} \, \text{m/s}^2 = 3.5 \, \text{m/s}^2 \]

The acceleration of the sphere's center is 3.5 m/s². This aligns with option (1).