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 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
Show Hint
- 3.5 \( m/s^2 \)
- 0.35 \( m/s^2 \)
- 2.5 \( m/s^2 \)
- 0.25 \( m/s^2 \)
The Correct Option is A
Approach Solution - 1
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).
Approach Solution -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 \( a = r\alpha \) \( a = \frac{14}{4} = \frac{7}{2} = 3.5 m/s^2 \)
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