Let’s break this classic **Atwood’s machine with rotating objects** problem into steps: Given data: - Sphere: Mass (\(m_s\)) = 5 kg Radius (\(R_s\)) = 8 cm = 0.08 m - Pulley: Mass (\(m_p\)) = 0.8 kg Radius (\(R_p\)) = 2 cm = 0.02 m - Hanging mass (\(m_b\)) = 3 kg - Height (\(h\)) = 4 m Assume the sphere rolls without slipping, the pulley rotates, and the string doesn't slip. --- ## a) Linear Acceleration of Sphere and Block Let: - \(T_1\) = tension in string above the pulley (holding the block) - \(T_2\) = tension in string between pulley and sphere - \(a\) = linear acceleration (same for sphere and block, since they're tied by a string) - \(g\) = 9.8 m/s² ### **Step 1: Write Equations of Motion** #### For the Hanging Mass (Block): \[ m_b a = m_b g - T_1 \] \[ T_1 = m_b g - m_b a \] #### For the Sphere (Rolling on Table): The sphere rolls without slipping, so: \[ a = \alpha_{sphere} R_s \] Translational motion: \[ m_s a = T_2 \] Rotational motion about the center of mass: \[ T_2 R_s = I_s \alpha_{sphere} \] The moment of inertia for a solid sphere: \[ I_s = \frac{2}{5} m_s R_s^2 \] So, \[ T_2 R_s = \left(\frac{2}{5} m_s R_s^2\right) \left(\frac{a}{R_s}\right) \] \[ T_2 = \frac{2}{5} m_s a \] But from above, \(m_s a = T_2\), so: \[ T_2 = m_s a \] But also \(T_2 = \frac{2}{5} m_s a\). This suggests we need to include both rotational and translational for the sphere: From translation: \[ m_s a = T_2 \] From rotation: \[ T_2 R_s = I_s \alpha_{sphere} \implies T_2 = I_s \frac{a}{R_s^2} \] \[ T_2 = \frac{2}{5} m_s a \] But this can't be. Let's clarify: Sum of forces on sphere: \[ T_2 = m_s a \] But it also provides the torque: \[ T_2 R_s = I_s \alpha_{sphere} \implies T_2 = \frac{I_s \alpha_{sphere}}{R_s} \] But with rolling without slipping, \(a = \alpha_{sphere} R_s\), so: \[ T_2 = \frac{I_s a}{R_s^2} \] Set equal to \(m_s a\): \[ m_s a = \frac{I_s a}{R_s^2} \implies m_s = \frac{I_s}{R_s^2} \implies m_s = \frac{\frac{2}{5} m_s R_s^2}{R_s^2} = \frac{2}{5} m_s \] No, that's inconsistent: actually, the net force on the sphere is \(T_2 = m_s a\), but the total force causing acceleration is \(T_2\), with friction providing the torque to roll, but not appearing in the equations since the sphere's not sliding. The best approach is to use energy or constraint equations, or treat the sphere as a rolling object: For a rolling object pulled by a tension \(T_2\): \[ T_2 = m_s a + F_{fr} \] But the friction provides the torque. Let's try an alternative method. #### For the Pulley: Net torque: \[ (T_1 - T_2) R_p = I_p \alpha_{pulley} \] Moment of inertia for a solid disk: \[ I_p = \frac{1}{2} m_p R_p^2 \] String doesn't slip, so: \[ \alpha_{pulley} = \frac{a}{R_p} \] So, \[ (T_1 - T_2) R_p = \frac{1}{2} m_p R_p^2 \cdot \frac{a}{R_p} \implies T_1 - T_2 = \frac{1}{2} m_p a \] --- ### **Step 2: Collect All Equations** From above: \[ (1)\quad m_b a = m_b g - T_1 \] \[ (2)\quad m_s a = T_2 \] \[ (3)\quad T_1 - T_2 = \frac{1}{2} m_p a \] --- ### **Step 3: Solve for \(a\)** Express all tensions in terms of \(a\): From (2): \(T_2 = m_s a\) From (3): \(T_1 = T_2 + \frac{1}{2} m_p a = m_s a + \frac{1}{2} m_p a\) Substitute into (1): \[ m_b a = m_b g - T_1 \implies T_1 = m_b g - m_b a \] But also \[ T_1 = m_s a + \frac{1}{2} m_p a \] Set equal: \[ m_b g - m_b a = m_s a + \frac{1}{2} m_p a \] \[ m_b g = m_b a + m_s a + \frac{1}{2} m_p a \] \[ m_b g = \left[ m_b + m_s + \frac{1}{2} m_p \right] a \] \[ a = \frac{m_b g}{m_b + m_s + \frac{1}{2} m_p} \] --- #### **But wait! This ignores the sphere's rotational inertia.** For rolling objects, the effective mass is increased by the rotational inertia: For a rolling sphere: \[ a = \frac{F}{m + I/R^2} \] \[ I_{sphere} = \frac{2}{5} m_s R_s^2 \] So, \[ a = \frac{T_2}{m_s + \frac{I_s}{R_s^2}} \] \[ a = \frac{T_2}{m_s + \frac{2}{5} m_s} = \frac{T_2}{m_s \left(1 + \frac{2}{5}\right)} = \frac{T_2}{\frac{7}{5} m_s} \] So, \[ T_2 = \frac{7}{5} m_s a \] Update equations: (2) \(T_2 = \frac{7}{5} m_s a\) (3) \(T_1 = T_2 + \frac{1}{2} m_p a = \frac{7}{5} m_s a + \frac{1}{2} m_p a\) Return to (1): \(T_1 = m_b g - m_b a\) So, \[ m_b g - m_b a = \frac{7}{5} m_s a + \frac{1}{2} m_p a \] \[ m_b g = m_b a + \frac{7}{5} m_s a + \frac{1}{2} m_p a \] \[ a = \frac{m_b g}{m_b + \frac{7}{5} m_s + \frac{1}{2} m_p} \] --- ### **Step 4: Plug in Numbers** - \(m_b = 3\) kg - \(m_s = 5\) kg - \(m_p = 0.8\) kg - \(g = 9.8\) m/s² \[ a = \frac{3 \times 9.8}{3 + \frac{7}{5} \times 5 + \frac{1}{2} \times 0.8} \] \[ \frac{7}{5} \times 5 = 7 \] \[ \frac{1}{2} \times 0.8 = 0.4 \] \[ \text{Denominator: } 3 + 7 + 0.4 = 10.4 \] \[ a = \frac{29.4}{10.4} = 2.827 \text{ m/s}^2 \] **Explanation:** We set up the equations accounting for the translational and rotational inertia of each component. The sphere’s rolling increases its effective inertia, and the pulley’s rotation also adds inertia. Solving the system gives the linear acceleration. --- ### **Step 5: Final Answers for (a)** \[ \boxed{a_{sphere} = a_{block} = 2.83 \ \text{m/s}^2} \] (rounded to two decimals) --- ## b) Angular Accelerations \[ \alpha_{pulley} = \frac{a}{R_p} \] \[ \alpha_{sphere} = \frac{a}{R_s} \] Plug in values: - \(a = 2.83\) m/s² - \(R_p = 0.02\) m - \(R_s = 0.08\) m \[ \alpha_{pulley} = \frac{2.83}{0.02} = 141.5 \text{ rad/s}^2 \] \[ \alpha_{sphere} = \frac{2.83}{0.08} = 35.4 \text{ rad/s}^2 \] **Explanation:** Angular acceleration relates to linear acceleration by dividing by the radius for rolling or rotating objects. --- ### **Final Answers for (b):** \[ \boxed{\alpha_{pulley} = 142 \ \text{rad/s}^2} \] \[ \boxed{\alpha_{sphere} = 35.4 \ \text{rad/s}^2} \] --- ## c) Tensions \(T_1\) and \(T_2\) Recall: - \(T_2 = \frac{7}{5} m_s a\) - \(T_1 = T_2 + \frac{1}{2} m_p a\) Plug in: - \(m_s = 5\) kg, \(a = 2.83\) m/s² \[ T_2 = \frac{7}{5} \times 5 \times 2.83 = 7 \times 2.83 = 19.8 \text{ N} \] \[ T_1 = 19.8 + 0.5 \times 0.8 \times 2.83 = 19.8 + 0.4 \times 2.83 = 19.8 + 1.13 = 20.93 \text{ N} \] **Explanation:** We use the earlier equations for tension, now with the correct acceleration value. \(T_2\) is the tension pulling the sphere, \(T_1\) is the tension pulling the block and supporting the pulley. --- ### **Final Answers for (c):** \[ \boxed{T_2 = 19.8 \ \text{N}} \] \[ \boxed{T_1 = 20.9 \ \text{N}} \] --- ## **Summary Table** | Quantity | Value (to 2 decimals) | Units | |-----------------|----------------------|-------------| | \(a_{sphere}\), \(a_{block}\) | 2.83 | m/s² | | \(\alpha_{pulley}\) | 142 | rad/s² | | \(\alpha_{sphere}\) | 35.4 | rad/s² | | \(T_2\) | 19.8 | N | | \(T_1\) | 20.9 | N | --- **Each step involved writing force and torque equations, solving for acceleration with all moments of inertia, then finding angular accelerations and tensions. The final answers are boxed above.**

The problem involves analyzing a system where a solid sphere rolls without slipping on a table connected via a pulley to a hanging mass. To determine the accelerations and tensions, we apply Newton's second law for translation to both the hanging mass and the rolling sphere, incorporating the rotational dynamics by relating the sphere’s angular acceleration to its linear acceleration through the no-slip condition. We account for the moments of inertia of both the sphere and the pulley to include their rotational effects. By setting up equations that connect tensions, accelerations, and moments, and solving the combined system, we find the linear acceleration, then derive the angular accelerations and tensions.