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

The problem asks for the acceleration of the center of a solid sphere that is rolling without slipping on a rough horizontal plane. A tangential force is applied to the highest point of the sphere.

Concept Used:

To solve this problem, we will use the equations for linear and rotational motion of a rigid body, along with the condition for rolling without slipping.

1. Newton's Second Law for Linear Motion: The net external force on the body equals its mass times 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 the moment of inertia about the center of mass times the 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, the 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 related to the angular acceleration by:

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

Step-by-Step Solution:

Step 1: Identify the forces and set up the equations of motion.

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

The applied force \( F \) tends to cause the sphere to accelerate to the right. It also creates a clockwise torque that would cause the point of contact to slip backward. To prevent this, the static friction force \( f_s \) must act in the forward (right) direction.

The equation for linear motion in the horizontal direction is:

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

Step 2: Set up the equation for rotational motion.

We take torques about the center of mass. The applied force \( F \) creates a clockwise torque of magnitude \( F \cdot r \). The friction force \( f_s \) creates a counter-clockwise torque of magnitude \( f_s \cdot r \). Taking the direction of angular acceleration (clockwise) as positive, the net torque is:

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

Using the rotational form of Newton's second law, \( \tau_{net} = I \alpha \):

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

Step 3: Use the condition for rolling without slipping.

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

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

Dividing by r on both sides, we get:

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

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

We now have two equations with two unknowns (\( a \) and \( f_s \)):

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

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

Adding these two equations together 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 Computation & Result:

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

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

The acceleration of the center of the sphere is 3.5 m/s². This corresponds to option (1).

Answer 2

Torque about bottom point: \( F \times 2r = I\alpha \) 

\( 49 \times 2r = \frac{7}{5}mr^2\alpha \) \( 14 - 4r\alpha \) 

As sphere rolls without slipping \( a = r\alpha \) \( a = \frac{14}{4} = \frac{7}{2} = 3.5 m/s^2 \)