Let's break the problem into two logical steps: --- ## **Step 1: Relating Linear and Angular Variables** Since the string is inextensible, mass \( m \), and **does not slip** on the pulley (radius \( R \), mass \( M \)), the **unwinding string length** \( \ell(t) \) is related to the **angle** \( \theta(t) \) the pulley has rotated: - **Total length of string:** \( L = 20\pi R \) - **Initially hanging:** \( \ell_ = \pi R \) - **Unwrapped (hanging) length increases as the pulley rotates:** \[ \ell(t) = \ell_ + R\theta(t) \] where \( \theta(t) \) is the angle the pulley has rotated (in radians). ### **Explanation (Step 1):** As the string unwinds, the length hanging freely increases by exactly the length that leaves the pulley, which corresponds to the arc length \( R\theta \) due to rotation. --- ## **Step 2: Dynamics and Acceleration** **Forces and Accelerations:** - Let \( a \) be the downward acceleration of the hanging part. - Let \( T \) be the tension in the string. - The pulley rotates, so torque = tension × radius = \( TR \), and moment of inertia for a disk: \( I = \frac{1}{2}MR^2 \). - **No slipping:** acceleration of string \( a = R \alpha \) (where \( \alpha \) is angular acceleration). **Write equations:** 1. **For the mass \( m \):** \[ m a = m g - T \] 2. **For the pulley (rotational):** \[ TR = I \alpha = \frac{1}{2}MR^2 \alpha \] \[ \alpha = a/R \implies TR = \frac{1}{2}MR^2 \frac{a}{R} \implies T = \frac{1}{2} M a \] **Substitute \( T \) into (1):** \[ m a = m g - \frac{1}{2}Ma \] \[ a\left(m + \frac{1}{2}M\right) = mg \] \[ a = \frac{mg}{m + \frac{1}{2}M} \] **This is a constant acceleration.** Now, the hanging part grows as: \[ \ell(t) = \ell_ + \frac{1}{2} a t^2 \] (Since initial velocity is zero, and the end of the string accelerates downward with \( a \).) ### **Explanation (Step 2):** We have related the unwinding length to the angular rotation, and then used Newton’s laws and rotational dynamics to find that the end of the string accelerates downward at \( a = \frac{mg}{m + \frac{1}{2}M} \), so the length of the hanging part grows quadratically with time. --- ## **Final Answer** \[ \boxed{ \ell(t) = \pi R + \frac{1}{2} \left( \frac{mg}{m + \frac{1}{2}M} \right) t^2 } \] **This formula gives the length of the freely hanging portion of the string as a function of time, until the string leaves the pulley.**