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 
3.5 m/s²
The problem involves a solid sphere rolling without slipping on a rough horizontal surface under the influence of an external force applied at its highest point. To find the acceleration of the center of mass ($a$), we apply the laws of translational and rotational dynamics along with the rolling condition.
Given Data
- Applied Force ($F$): $49\text{ N}$
- Mass of the sphere ($m$): $20\text{ kg}$
- Moment of inertia of a solid sphere ($I$): $\frac{2}{5}mR^2$
- Rolling condition: $a = \alpha R$ (where $\alpha$ is angular acceleration and $R$ is radius)
Equations of Motion
Let $f$ be the force of friction acting on the sphere at the point of contact. We assume $f$ acts in the same direction as $F$ to assist the rolling motion (we will verify the sign later).
- Translational Equation:
$F + f = ma$ ... (Equation 1) - Rotational Equation (about the Center of Mass):
The torque is provided by $F$ and $f$. $F$ creates a clockwise torque, while $f$ creates a counter-clockwise torque.
$F \cdot R - f \cdot R = I\alpha$
$F - f = \frac{I\alpha}{R}$
Substituting $I = \frac{2}{5}mR^2$ and $\alpha = \frac{a}{R}$:
$F - f = \frac{2}{5}ma$ ... (Equation 2)
Solving for Acceleration
Add Equation 1 and Equation 2 to eliminate friction ($f$):
$(F + f) + (F - f) = ma + \frac{2}{5}ma$
$2F = \frac{7}{5}ma$
$a = \frac{10F}{7m}$
Substitute the numerical values into the formula:
$a = \frac{10 \times 49}{7 \times 20}$
$a = \frac{490}{140} = \frac{49}{14}$
$a = 3.5\text{ m/s}^2$
Final Answer
The acceleration of the center of the sphere is $3.5\text{ m/s}^2$.
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