Problem 67 - Fundamentals of Physics

Question

Is an over head view of a thin uniform rod of length \(0.600 \mathrm{~m}\) and mass \(M\) rotating horizontally at \(80.0 \mathrm{rad} / \mathrm{s}\) counterclockwise about an axis through its center. A particle of mass \(M / 3.00\) and traveling horizontally at speed \(40.0 \mathrm{~m} / \mathrm{s}\) hits the rod and sticks. The particle's path is perpendicular to the rod at the instant of the hit, at a distance \(d\) from the rod's center. (a) At what value of \(d\) are rod and particle stationary after the hit? (b) In which direction do rod and particle rotate if \(d\) is greater than this value?

Step-by-Step Solution

Answer

(a) \( d = 0.180 \, \text{m} \). (b) Clockwise rotation for \( d > 0.180 \, \text{m} \).

Step 1: Calculate Initial Angular Momentum of Rod

The moment of inertia of the rod rotating about its center is given by the equation \( I = \frac{1}{12}ML^2 \). For a thin rod with length \( L = 0.600 \, \text{m} \) and angular velocity \( \omega = 80.0 \, \text{rad/s} \), the initial angular momentum of the rod is\[ L_{\text{rod}} = I \cdot \omega = \left(\frac{1}{12} M (0.600)^2\right) \cdot 80.0. \]

Step 2: Calculate Initial Angular Momentum of Particle

The particle has angular momentum with respect to the axis of the rod. It is calculated as the perpendicular distance \( d \) times the linear momentum of the particle. Since before the hit, the particle has speed \( v = 40.0 \, \text{m/s} \), its mass is \( \frac{M}{3} \), so:\[ L_{\text{particle}} = d \cdot \left( \frac{M}{3} \right) \cdot 40.0. \]

Step 3: Use Conservation of Angular Momentum

Angular momentum is conserved during the collision:\[ L_{\text{initial}} = L_{\text{rod}} + L_{\text{particle}} = L_{\text{final}} = 0. \]Thus,\[ \left(\frac{1}{12} M (0.600)^2\right) \cdot 80.0 + \left( \frac{M}{3} \right) \cdot 40.0 \cdot d = 0. \]

Step 4: Solve for \( d \)

Substitute the known values into the equation and solve for \( d \):\[ \left(\frac{1}{12} \cdot 0.36 \cdot 80.0 \right) + \left( \frac{1}{3} \cdot 40.0 \cdot d \right) = 0. \]This yields:\[ d = -\frac{( \frac{1}{12} \times 0.36 \times 80 )}{ \frac{1}{3} \times 40 } = -0.180 \, \text{m}. \]

Step 5: Interpret the Value of \( d \)

Since \( d \) must be positive (as distance is always non-negative), the physical implication is that at \( d = 0.180 \, \text{m} \) (without the sign), the system will be stationary just after the collision.

Step 6: Determine Direction of Rotation for \( d > 0.180 \, \text{m} \)

For a greater distance \( d \), the particle will provide larger angular momentum, causing the combined system to rotate clockwise, opposite from the rod's initial counterclockwise spin, to conserve angular momentum.

Key Concepts

Rotational Motion

Rotational motion is a type of motion where an object spins around an internal axis. In this problem, the rod experiences rotational motion as it spins around its central axis. This movement is similar to how a merry-go-round spins, maintaining a constant path.The speed and position in rotational motion can be described using angular velocity, denoted by \( \omega \). For the rod, angular velocity is initially given as \(80.0 \, \text{rad/s}\). Angular velocity is important because it tells us how fast the object is spinning.Understanding rotational motion helps in visualizing how different forces affect spinning objects, like the collision described in the problem.

Moment of Inertia

Moment of inertia is the rotational analog to mass. It determines how much torque is needed for an object to achieve a certain angular acceleration. For a rigid body like a rod, the moment of inertia depends on the distribution of mass around the rotation axis.The moment of inertia of the rod about its center is calculated using the formula \( I = \frac{1}{12}ML^2 \). Here, \( M \) is the mass of the rod and \( L \) is its length. In this problem, with \( L = 0.600 \, \text{m} \), the formula helps determine the resistance of the rod to changes in its rotational motion.This concept is vital because it influences the calculation of angular momentum, which plays a key role in predicting the results of rotational collisions.

Angular Momentum Conservation

Angular momentum conservation states that if no external torque is acting on a system, the total angular momentum remains constant. This principle is analogous to linear momentum conservation and is fundamentally important in rotational dynamics.In the given exercise, the rod and particle system is initially rotating with certain angular momenta. Just before the collision, we calculate the individual angular momenta: one from the spinning rod and the other from the motion of the particle. When they collide, their combined angular momentum must equal the initial total angular momentum.The key equation is:

1. \( L_{\text{initial}} = L_{\text{final}} \)
2. Thus, \( L_{\text{rod}} + L_{\text{particle}} = 0 \).

This conservation helps us determine the distance \( d \) at which the system becomes stationary after the impact.

Collision Dynamics in Physics

Collisions can significantly alter the motion of objects, and understanding these dynamics is crucial in physics. During a collision, energy and momentum are transferred between objects. In the context of rotational motion, this involves angular momentum rather than linear momentum.When the particle collides with the rod, it sticks to it, affecting the overall angular motion. The collision dynamics here show how the momentum of the moving particle adds to or counteracts the momentum of the rotating rod.Given that angular momentum must be conserved, if the particle's path provides greater angular momentum than initially present in the rod, it can reverse the direction of spin. The analysis explains why for distances \( d \) greater than 0.180 m, the system rotates clockwise after collision. This showcases the intricate balance of forces in rotational collision dynamics.