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Chapter 10: Problem 79

A uniform rod of length \(L\) rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. (a) What is the final angular speed of the rod? (b) What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision?

Short Answer

Expert verified
Final angular speed is \(\omega_f = \frac{3v}{13L}\); energy ratio is about 0.073.

Step by step solution

01

Understand the System Setup

We have a rod of length \(L\) and mass \(M\) pivoted at one end, initially at rest, and a bullet of mass \(m\), where \(m = \frac{M}{4}\), striking the rod at its center. The speed of the bullet before impact is \(v\). After the bullet embeds in the rod, we need to find the system's final angular speed \(\omega_f\) and the ratio of kinetic energies.
02

Apply Conservation of Angular Momentum

The system is isolated, so angular momentum is conserved. The initial angular momentum just before the collision, \(L_i\), is due to the bullet: \[ L_i = m \cdot v \cdot \left(\frac{L}{2}\right) \]The moment of inertia of the rod about the pivot is \(I = \frac{1}{3}ML^2\), and the bullet adds \(m\left(\frac{L}{2}\right)^2\) to it, so the final moment of inertia \(I_f\) is:\[ I_f = \frac{1}{3}ML^2 + m\left(\frac{L}{2}\right)^2 \]Using the conservation principle:\[ L_i = I_f \cdot \omega_f \]Substitute expressions to solve for \(\omega_f\):\[ m \cdot v \cdot \left(\frac{L}{2}\right) = \left(\frac{1}{3}ML^2 + m\left(\frac{L}{2}\right)^2\right) \cdot \omega_f \]
03

Solve for Final Angular Speed

Substitute \(m = \frac{M}{4}\):\[ \left(\frac{M}{4}\right) \cdot v \cdot \left(\frac{L}{2}\right) = \left(\frac{1}{3}ML^2 + \frac{M}{4}\left(\frac{L}{2}\right)^2\right) \cdot \omega_f \]Solve for \(\omega_f\):\[ \omega_f = \frac{\left(\frac{M}{4}\right) \cdot v \cdot \left(\frac{L}{2}\right)}{\frac{1}{3}ML^2 + \frac{M}{4}\left(\frac{L^2}{4}\right)} \]Simplifying gives:\[ \omega_f = \frac{3v}{13L} \]
04

Determine Initial Kinetic Energy of the Bullet

The initial kinetic energy \(K_{i}\) of the bullet is:\[ K_{i} = \frac{1}{2}m v^2 = \frac{1}{2} \cdot \left(\frac{M}{4}\right) \cdot v^2 = \frac{Mv^2}{8} \]
05

Determine Final Kinetic Energy of the System

The final kinetic energy \(K_f\) is rotational:\[ K_f = \frac{1}{2} I_f \omega_f^2 = \frac{1}{2} \left(\frac{1}{3}ML^2 + \frac{M}{4} \left(\frac{L^2}{4}\right)\right) \left(\frac{3v}{13L}\right)^2 \]Plug in \(I_f = \frac{13}{48}ML^2\): \[ K_f = \frac{1}{2} \cdot \frac{13}{48}ML^2 \cdot \left(\frac{9v^2}{169L^2}\right) = \frac{13 \times 9Mv^2}{2 \times 48 \times 169} = \frac{13 \times 9Mv^2}{16272} \]
06

Calculate the Ratio of Final to Initial Kinetic Energy

Given the expressions for \(K_i\) and \(K_f\), we find the ratio \( \frac{K_f}{K_i} \):\[ \frac{K_f}{K_i} = \frac{\frac{117Mv^2}{16272}}{\frac{Mv^2}{8}} = \frac{117 \times 8}{16272} = \frac{936}{16272} = \frac{1}{13.68} \approx 0.073 \]
07

Final Step: Conclusions

The final angular speed of the rod after the collision is \(\frac{3v}{13L}\), and the ratio of the kinetic energy after the collision to the kinetic energy of the bullet before the collision is approximately 0.073.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotational Dynamics
Rotational dynamics is the branch of mechanics that deals with the motion of rotating systems and the forces and torques that produce this motion. It is analogous to linear dynamics but involves rotational quantities such as angular velocity, angular acceleration, and moments of inertia.
When dealing with rotational dynamics, it's important to understand a few key concepts:
  • **Angular Motion:** Just like linear motion, rotational dynamics is governed by angular analogs such as angular displacement, angular velocity, and angular acceleration.
  • **Rotational Motion:** This refers to an object revolving around an axis. In our exercise, the rod rotates about its pivot point.
  • **Torque:** In rotational motion, torque plays a role similar to force. It is the rotational equivalent of linear force and is calculated as the product of force and the distance from the point of rotation.
  • **Work-Energy Theorem:** In rotations, the work done on an object is related to its change in kinetic energy.
Understanding these concepts helps in solving problems involving rotating bodies by considering angular analogs of linear quantities and ensuring all forces and torques in the system are accounted for.
Angular Speed
Angular speed measures how quickly an object rotates or revolves around an axis and is typically expressed in radians per second (rad/s). In our scenario, after the collision, the rod and the embedded bullet have a common angular speed as they spin about the pivot.
To calculate angular speed, consider the conservation of angular momentum. Before the collision, the bullet has linear momentum that converts into angular momentum when it embeds into the rod:
  • **Initial Angular Momentum:** Calculated from the bullet's speed and distance to the pivot point.
  • **Final Angular Momentum:** The conserved quantity now spread across both the rod and the embedded bullet.
  • The formula is expressed as: \[ L_i = I_f \cdot \omega_f \] where \( L_i \) is the initial angular momentum and \( I_f \) is the final moment of inertia.
  • **Final Angular Speed (\( \omega_f \)):** Obtained by solving the equation above, reflecting how the angular momentum is dispersed through the system post-collision.
By using the appropriate formulas, we calculate the angular speed as \( \omega_f = \frac{3v}{13L} \), showing the correspondence between the velocity of the bullet and the resulting rotation of the system.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It depends on the mass distribution relative to the axis of rotation. In this exercise, understanding the combined moment of inertia of the rod and bullet is crucial for calculating the final rotational speed.
Here is what you need to know about the moment of inertia in this context:
  • **Rod's Moment of Inertia:** For a rod rotating about one end, it is calculated as \( I_{rod} = \frac{1}{3}ML^2 \).
  • **Bullet's Contribution:** On hitting the rod, the bullet adds its own moment of inertia, calculated using \( m(\frac{L}{2})^2 \), as it strikes at the midpoint.
  • **Combined Moment of Inertia:** The sum of individual moments gives us the final effective moment of inertia, \( I_f = \frac{1}{3}ML^2 + m(\frac{L}{2})^2 \).
  • **Role in Rotation:** The combined moment plays a crucial role in defining how fast the system will rotate post-collision, governing how effectively the system resists rotational changes.
Applying these calculations helps us determine how the bullet's impact alters the rotational characteristics of the rod, making the moment of inertia a fundamental concept in understanding rotational dynamics and angular momentum conservation.

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Most popular questions from this chapter

A large wooden turntable in the shape of a flat uniform disk has a radius of 2.00 m and a total mass of 120 kg. The turntable is initially rotating at 3.00 rad/s about a vertical axis through its center. Suddenly, a 70.0-kg parachutist makes a soft landing on the turntable at a point near the outer edge. (a) Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.) (b) Compute the kinetic energy of the system before and after the parachutist lands. Why are these kinetic energies not equal?

A uniform, 0.0300-kg rod of length 0.400 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.0200 kg, are mounted so that they can slide along the rod. They are initially held by catches at positions 0.0500 m on each side of the center of the rod, and the system is rotating at 48.0 rev/min. With no other changes in the system, the catches are released, and the rings slide outward along the rod and fly off at the ends. What is the angular speed (a) of the system at the instant when the rings reach the ends of the rod; (b) of the rod after the rings leave it?

A small block with mass 0.130 kg is attached to a string passing through a hole in a frictionless, horizontal surface (see Fig. E10.40). The block is originally revolving in a circle with a radius of 0.800 m about the hole with a tangential speed of 4.00 m/s. The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 N. What is the radius of the circle when the string breaks?

A uniform marble rolls down a symmetrical bowl, starting from rest at the top of the left side. The top of each side is a distance \(h\) above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. (a) How far up the smooth side will the marble go, measured vertically from the bottom? (b) How high would the marble go if both sides were as rough as the left side? (c) How do you account for the fact that the marble goes \(higher\) with friction on the right side than without friction?

A 42.0-cm-diameter wheel, consisting of a rim and six spokes, is constructed from a thin, rigid plastic material having a linear mass density of 25.0 g/cm. This wheel is released from rest at the top of a hill 58.0 m high. (a) How fast is it rolling when it reaches the bottom of the hill? (b) How would your answer change if the linear mass density and the diameter of the wheel were each doubled?
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