Inelastic Collision Between a Sliding Block and a Pivoting Rod

Understanding the principles of rotational motion and angular momentum conservation is essential in physics. In this tutorial, we analyze a problem where a 50 g block slides down a frictionless incline from a height of 20 cm and undergoes a perfectly inelastic collision with a uniform rod of mass 100 g and length 40 cm, which pivots about a fixed point. We aim to determine the angle θ through which the rod swings before momentarily coming to rest.

Step-by-Step Solution:

1. Calculating the Block's Velocity Before Collision:

• • Utilizing energy conservation, the potential energy of the block is converted into kinetic energy as it slides down.
  • Formula: mgh = (1/2)mv²
  • Solving for v: v = √(2gh) = √(2 * 9.8 m/s² * 0.20 m) ≈ 1.98 m/s

1. Applying Angular Momentum Conservation During Collision:

• • At the moment of collision, angular momentum about the pivot point is conserved.
  • Initial angular momentum: L_initial = m * v * d, where d is the distance from the pivot to the point of collision.
  • Calculating L_initial: L_initial = (0.050 kg) * (1.98 m/s) * (0.40 m) ≈ 0.0396 kg·m²/s
  • Moment of inertia of the rod about the pivot: I_rod = (1/3) * M * d² = (1/3) * (0.10 kg) * (0.40 m)² ≈ 0.00533 kg·m²
  • Moment of inertia of the block (treated as a point mass at distance d): I_block = m * d² = (0.050 kg) * (0.40 m)² = 0.008 kg·m²
  • Total moment of inertia: I_total = I_rod + I_block = 0.00533 kg·m² + 0.008 kg·m² = 0.01333 kg·m²
  • Using L_initial = I_total * ω to find angular velocity ω: ω = L_initial / I_total = 0.0396 kg·m²/s / 0.01333 kg·m² ≈ 2.97 rad/s

1. Energy Transformation and Determining the Swing Angle θ:

• • After the collision, the system's rotational kinetic energy is converted into gravitational potential energy as the rod swings upward.
  • Initial rotational kinetic energy: KE_initial = (1/2) * I_total * ω² = 0.5 * 0.01333 kg·m² * (2.97 rad/s)² ≈ 0.059 J
  • Height gained by the block: H_block = d * (1 - cos θ)
  • Height gained by the rod's center of mass: H_rod = (d/2) * (1 - cos θ)
  • Total potential energy gained: PE = m * g * H_block + M * g * H_rod = 0.392 J * (1 - cos θ)
  • Setting KE_initial equal to PE to solve for θ: 0.059 J = 0.392 J * (1 - cos θ)
  • Solving for θ: θ ≈ 32°

Key Takeaways & Common Mistakes to Avoid:

• ✔ Conservation Laws: Recognize when to apply conservation of energy versus conservation of angular momentum. In this problem, energy conservation applies to the block's slide, while angular momentum conservation applies during the collision.
• ✔ Moment of Inertia Calculations: Accurately calculate the moments of inertia for both the rod and the point mass (block) about the pivot point. Miscalculations here can lead to significant errors.
• ✔ Energy Transformation Understanding: After the collision, understand that the system's kinetic energy transforms into potential energy as the rod swings upward. This insight is crucial for determining the maximum angle θ.
• ✔ Unit Consistency: Ensure all units are consistent throughout the calculations to avoid errors, particularly when dealing with mass (grams vs. kilograms) and length (centimeters vs. meters).

Summary:

In this problem, we analyzed the motion of a block sliding down a frictionless incline and colliding inelastically with a pivoting rod. By applying energy conservation principles to determine the block's velocity, using angular momentum conservation during the collision, and analyzing the subsequent energy transformation, we calculated that the rod swings to a maximum angle of approximately 32° before momentarily stopping.

This problem integrates key concepts in rotational dynamics, including energy conservation, angular momentum, and moment of inertia calculations, providing valuable insights for students studying advanced physics topics.