Solution Outline

To solve this problem, we will break it into two parts as follows:

(a) Angular Velocity of the Stick and Putty System

1. Principle Used: Conservation of angular momentum.

   • Before collision: The system has angular momentum due to the putty moving linearly.
   • After collision: The angular momentum is transferred to the stick-putty system.

   Key equations:

   • Moment of inertia of the stick about its pivot: $I_{\text{stick}} = \frac{1}{3} M L^2$
   • Moment of inertia of the putty (at distance $x = 0.8 \, \text{m}$ from the pivot): $I_{\text{putty}} = m x^2$
   • Total moment of inertia: $I_{\text{total}} = I_{\text{stick}} + I_{\text{putty}}$
   • Conservation of angular momentum: $m v x = I_{\text{total}} \omega$

   Steps:

   • Compute $I_{\text{total}}$.
   • Solve for $\omega$ (angular velocity): $\omega = \frac{m v x}{I_{\text{total}}}$

(b) Initial Speed of the Putty

1. Principle Used: Conservation of energy after the collision.

   • Initial state: Kinetic energy is stored as rotational kinetic energy: $K_{\text{rot}} = \frac{1}{2} I_{\text{total}} \omega^2$
   • Final state: The stick and putty system has gravitational potential energy when horizontal: $U_{\text{grav}} = M g \frac{L}{2} + m g x$

   Energy conservation: $\frac{1}{2} I_{\text{total}} \omega^2 = M g \frac{L}{2} + m g x$

   Steps:

   • Solve for $\omega^2$ using energy conservation.
   • Use the value of $\omega$ to back-solve for $v$ using angular momentum conservation: $v = \frac{I_{\text{total}} \omega}{m x}$

Let me compute the exact values for you. M = 2.0 # mass of the stick (kg) L = 1.0 # length of the stick (m) m = 1.0 # mass of the putty (kg) x = 0.8 # distance of putty from the pivot (m) g = 9.8 # acceleration due to gravity (m/s^2)

Moments of inertia

I_stick = (1/3) * M * L2 I_putty = m * x2 I_total = I_stick + I_putty

Gravitational potential energy when stick is horizontal

U_grav = (M * g * L / 2) + (m * g * x)

Rotational kinetic energy equivalence to potential energy

omega_squared = (2 * U_grav) / I_total omega = omega_squared**0.5

Angular momentum conservation to find v

v = (I_total * omega) / (m * x)

I_stick, I_putty### Results:

1. Moments of Inertia:

   • Stick: $I_{\text{stick}} = 0.67 \, \text{kg·m}^2$
   • Putty: $I_{\text{putty}} = 0.64 \, \text{kg·m}^2$
   • Total: $I_{\text{total}} = 1.31 \, \text{kg·m}^2$

2. Angular Velocity ($\omega$):
   $\omega = 5.20 \, \text{rad/s}$

3. Speed of the Putty ($v$):
   $v = 8.49 \, \text{m/s}$

Would you like step-by-step derivations or further clarification on any part? Here are some follow-up questions to deepen your understanding:

1. How is angular momentum conserved during a collision?
2. Why is the center of mass important in calculating potential energy?
3. What is the significance of the pivot point in this problem?
4. How do the stick's mass and length influence the final outcome?
5. What happens if the putty strikes closer or farther from the pivot point?

Tip: Always verify whether energy conservation holds true when solving problems involving rotational dynamics.