Problem 82 - University Physics with Modern Physics

Question

A local ice hockey team has asked you to design an apparatus for measuring the speed of the hockey puck after a slap shot. Your design is a 2.00-m-long, uniform rod pivoted about one end so that it is free to rotate horizontally on the ice without friction. The 0.800-kg rod has a light basket at the other end to catch the 0.163-kg puck. The puck slides across the ice with velocity \(\overrightarrow{v}\) (perpendicular to the rod), hits the basket, and is caught. After the collision, the rod rotates. If the rod makes one revolution every 0.736 s after the puck is caught, what was the puck's speed just before it hit the rod?

Step-by-Step Solution

Answer

The puck's speed was approximately 39.21 m/s before impacting the rod.

Step 1: Understand the Problem

We are asked to find the speed of a puck before it struck and was caught in a basket at the end of a rotating rod. We know the rod rotates without friction after the collision, making one complete revolution every 0.736 seconds. We need to use conservation principles and kinematics to solve this.

Step 2: Calculate the Angular Velocity of the Rod

Once the puck is caught, the rod and puck system rotates. The time period for this rotation, T, is 0.736 seconds, which helps find the angular velocity ω. Angular velocity is given by \[\omega = \frac{2\pi}{T}\]Substitute T = 0.736 s to get\[\omega = \frac{2\pi}{0.736} \approx 8.54 \text{ rad/s}\]

Step 3: Use Conservation of Angular Momentum

Before the collision, the puck has linear momentum which contributes to angular momentum once it is caught by the rod's basket. The initial angular momentum of the system is entirely from the puck, given by \[L_i = m_p \cdot v_p \cdot L\]where m_p is the puck's mass, v_p is its velocity, and L is the length of the rod. The final angular momentum, L_f, after the collision is \[L_f = I \cdot \omega\]where I is the moment of inertia of the system (rod + puck), and ω is the angular velocity.

Step 4: Calculate Moment of Inertia of the System

The moment of inertia, I, for the rod and puck system is the sum of individual moments of inertia. For the rod pivoted at one end: \[I_{rod} = \frac{1}{3} m_r L^2\]where m_r is the rod's mass. For the puck treated as a point mass at distance L:\[I_{puck} = m_p L^2\]So, \[I = I_{rod} + I_{puck} = \frac{1}{3} m_r L^2 + m_p L^2\]

Step 5: Equate Initial and Final Angular Momentum

By conservation of angular momentum, \[L_i = L_f\],\[m_p \cdot v_p \cdot L = I \cdot \omega\]Substitute the expression for I from Step 4, and solve for the puck's speed v_p:\[v_p = \frac{I \cdot \omega}{m_p \cdot L}\]

Step 6: Solve for Puck's Speed

Plug in known values: m_r = 0.800 kg, m_p = 0.163 kg, L = 2.00 m, \[I = \left(\frac{1}{3}(0.800)(2.00)^2 + 0.163(2.00)^2\right) = 1.499 \text{ kg}\cdot\text{m}^2\]Substitute I and omega into the angular momentum equation:\[v_p = \frac{1.499 \cdot 8.54}{0.163 \cdot 2.00} \approx 39.21\text{ m/s}\]

Step 7: Conclusion

The speed of the puck just before it hit the rod is approximately 39.21 m/s.

Key Concepts

Angular Velocity

Angular velocity is a measure of how quickly an object rotates or revolves around a particular point or axis. In our example, it's how fast the rod with the basket, containing the puck, spins after the collision. The unit for angular velocity is radian per second (rad/s).
To calculate angular velocity (\( \omega \)), you need to know the time period (\( T \)) taken for one complete revolution. Here, the rod takes 0.736 seconds for one revolution. You can use the formula \( \omega = \frac{2\pi}{T} \) to find angular velocity.
This formula states that if you know how long it takes for one complete spin, you can find how fast the object is rotating by dividing the full angle of rotation (which is \( 2\pi \) radians) by the time it takes to complete that rotation. In our case, substituting \( T \) as 0.736 seconds gives us \( \omega \approx 8.54 \, \text{rad/s} \), indicating a rather swift rotation.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotation. Imagine it as the rotational equivalent of mass in linear motion. It depends on how the mass is distributed relative to the rotation axis.
For a rod pivoted at one end, its moment of inertia is calculated using the formula \( I_{\text{rod}} = \frac{1}{3} m_r L^2 \) where \( m_r \) is the mass of the rod, and \( L \) is its length.
For the puck, because it's at a distance from the pivot point, its moment of inertia is \( I_{\text{puck}} = m_p L^2 \) where \( m_p \) is the puck's mass. By adding the two, you get the total moment of inertia for the system: \( I = I_{\text{rod}} + I_{\text{puck}} \).
In our exercise, this gives us \( I \approx 1.499 \, \text{kg} \cdot \text{m}^2 \). This means it takes a certain amount of torque, or rotational force, to change the spinning speed of the rod with the puck.

Collision Dynamics

Collision dynamics deals with how objects interact during impacts and how their momentums are conserved. Here, the key principle is the conservation of angular momentum.
Before the puck hits the basket attached to the rod, it has linear momentum, which then converts into angular momentum after it is caught. The law of conservation of angular momentum states that in the absence of external torques, the initial angular momentum (\( L_i \)) is equal to the final angular momentum (\( L_f \)).
For the puck-rod system, \( L_i = m_p \cdot v_p \cdot L \) and \( L_f = I \cdot \omega \), where \( v_p \) is the speed of the puck before the collision. By equating these, \( m_p \cdot v_p \cdot L = I \cdot \omega \), you can solve for the puck's initial speed:\( v_p = \frac{I \cdot \omega}{m_p \cdot L} \).
Substituting the values from our example results in \( v_p \approx 39.21 \, \text{m/s} \). This shows how conservation laws help determine the speed of the puck right before the collision.