## Step 1: Identify the Torque Acting on the Sphere The tension \( T \) in the string pulls tangentially on the rod, which is wrapped at distance \( r = \frac{1}{2}R \) from the center of the sphere. The torque (\( \tau \)) due to this tension about the center of the sphere is: \[ \tau = r \cdot T = \left(\frac{1}{2}R\right)T \] **Explanation:** Torque is the product of force and the perpendicular distance from the axis of rotation. Here, force is the tension \( T \) and the distance is \( \frac{1}{2}R \). --- ## Step 2: Relate Torque to Angular Acceleration The sphere rotates about its center, and the moment of inertia for a solid sphere about its center is: \[ I = \frac{2}{5}MR^2 \] Using Newton's second law for rotation: \[ \tau = I\alpha \] Plug in the values: \[ \left(\frac{1}{2}R\right)T = \left(\frac{2}{5}MR^2\right) \alpha \] Solve for \( \alpha \): \[ \alpha = \frac{\left(\frac{1}{2}R\right)T}{\frac{2}{5}MR^2} = \frac{5T}{4MR} \] **Explanation:** We set the external torque equal to the product of the moment of inertia and the angular acceleration \( \alpha \), then solve for \( \alpha \). --- ## **Final Answer** \[ \boxed{\alpha = \frac{5T}{4MR}} \] This is the sphere's angular acceleration in terms of the given quantities.

# Problem Breakdown and Solution Steps ## Step 1: Identify the Torque Acting on the Sphere To find the angular acceleration of the sphere, we first need to calculate the torque (\( \tau \)) generated by the tension in the string. ### Calculation of Torque The tension \( T \) in the string pulls on the rod at a distance \( r \) from the center of the sphere. Given that \( r = \frac{1}{2}R \) (half the radius of the sphere), the torque is calculated as follows: \[ \tau = r \cdot T = \left(\frac{1}{2}R\right)T \] ### Explanation Torque is defined as the product of the force applied (in this case, the tension \( T \)) and the distance from the axis of rotation (the radius \( r \)). Here, the distance is \( \frac{1}{2}R \), which gives us the torque acting on the sphere. --- ## Step 2: Relate Torque to Angular Acceleration Next, we use the relationship between torque, moment of inertia, and angular acceleration. ### Moment of Inertia The moment of inertia \( I \) for a solid sphere about its center is given by: \[ I = \frac{2}{5}MR^2 \] ### Newton's Second Law for Rotation According to Newton's second law for rotational motion: \[ \tau = I\alpha \] Substituting the expressions we've derived: \[ \left(\frac{1}{2}R\right)T = \left(\frac{2}{5}MR^2\right) \alpha \] ### Solving for Angular Acceleration To find \( \alpha \), we rearrange the equation: \[ \alpha = \frac{\left(\frac{1}{2}R\right)T}{\frac{2}{5}MR^2} \] This simplifies to: \[ \alpha = \frac{5T}{4MR} \] ### Explanation In this step, we used the calculated torque and the moment of inertia to derive the sphere's angular acceleration. This shows how the applied tension affects the rotational motion of the sphere. --- ## Final Answer \[ \boxed{\alpha = \frac{5T}{4MR}} \] This expression gives the angular acceleration of the sphere in terms of the tension \( T \), mass \( M \), and radius \( R \).