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Chapter 11: Problem 53

A uniform thin rod of length \(0.500 \mathrm{~m}\) and mass \(4.00 \mathrm{~kg}\) can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a \(3.00 \mathrm{~g}\) bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle \(\theta=60.0^{\circ}\) with the rod (Fig. 11-50). If the bullet lodges in the rod and the angular velocity of the rod is \(10 \mathrm{rad} / \mathrm{s}\) immediately after the collision, what is the bullet's speed just before impact?

Short Answer

Expert verified
The bullet's speed just before impact is approximately 1124 m/s.

Step by step solution

01

Understand the Problem

We need to find the speed of the bullet just before it hits the rod. The collision is an inelastic collision where the bullet gets embedded within the rod, causing it to rotate. We will use the conservation of angular momentum to find the solution.
02

Define the System and Knowns

The system consists of a bullet striking a rod. We know the mass of the rod (\(M = 4.00\, \mathrm{kg}\)), the mass of the bullet (\(m = 3.00\, \mathrm{g} = 0.003\, \mathrm{kg}\)), length of the rod (\(L = 0.500\, \mathrm{m}\)), and the final angular velocity (\(\omega = 10\, \mathrm{rad/s}\)). The angle \(\theta\) is given as \(60.0^{\circ}\).
03

Calculate Initial and Final Moment of Inertia

The moment of inertia of the rod about its center is given by \(I_{\text{rod}} = \frac{1}{12}ML^2\). With the bullet lodged at a distance \(\frac{L}{2}\) (end of rod), the final moment of inertia is \(I_{\text{final}} = I_{\text{rod}} + m(\frac{L}{2})^2\).
04

Apply Conservation of Angular Momentum

The initial angular momentum (before the collision) is given by \(L_i = m v \cdot \frac{L}{2} \cdot \cos(\theta)\). The final angular momentum (after the collision) is \(L_f = I_{\text{final}} \cdot \omega\). Set \(L_i = L_f\).
05

Solve for Bullet's Speed

Substitute all known values and solve for \(v\):- Initial angular momentum: \(m v \cdot \frac{L}{2} \cdot \cos(\theta)\)- Final angular momentum: \(\left(\frac{1}{12}ML^2 + m(\frac{L}{2})^2\right) \omega\)Solve \(m v \cdot \frac{L}{2} \cdot \cos(\theta) = \left(\frac{1}{12}ML^2 + m(\frac{L}{2})^2\right) \omega\) to get \(v = \frac{\left(\frac{1}{12}ML^2 + m(\frac{L}{2})^2\right) \omega}{m \cdot \frac{L}{2} \cdot \cos(\theta)}\).
06

Calculate Numerically

Input the known values and solve the equation derived in the previous step. After calculation, \(v \approx 1124\, \mathrm{m/s}\).

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Key Concepts

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

Inelastic Collision
An inelastic collision is when two objects collide and stick together, moving as one entity after the encounter. In this problem, a bullet collides with a rod and lodges into it. Unlike elastic collisions, energy is not conserved in inelastic collisions. However, momentum is still conserved. This is what we use to solve the problem.

For this scenario, angular momentum is particularly important. Before the bullet hits the rod, the system's angular momentum is only due to the bullet. After the collision, both the bullet and the rod rotate about the rod's center. This is a perfect example of an inelastic collision as the two bodies become one and rotate together.
  • Energy conservation: Not applicable. Energy is lost as heat and sound.
  • Angular momentum conservation: Remains constant for the system.
Understanding inelastic collisions helps physicists predict the resulting motion when two objects interact forcefully and join together.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to rotational motion. It's similar to how mass is related to linear motion. For the rod in this exercise, its moment of inertia determines how difficult it is to get the rod rotating.

The moment of inertia depends on how the mass is distributed relative to the axis of rotation. In this problem, the rod's axis is through its center. The formula used is \( I_{\text{rod}} = \frac{1}{12}ML^2 \), which represents this unique distribution.
  • The closer the mass is to the axis, the smaller the moment of inertia.
  • An object with a larger moment of inertia will be harder to spin.
  • Adding the bullet increases the moment of inertia since it is away from the axis.
The increased moment of inertia due to the bullet's mass can be calculated and added to find the new resistance to rotational motion of the combined system.
Rotational Dynamics
Rotational dynamics explores the effect of forces on rotational motion. In this exercise, we see how the bullet's impact affects the rod鈥檚 rotation.

The concept of torque, angular velocity, and angular momentum are essential. Angular momentum is particularly crucial here since it is conserved during the collision. This allows us to calculate the bullet's speed before impact. The formula\(L_i = L_f\), where \(L_i\) is the initial angular momentum and \(L_f\) is the final angular momentum, helps to solve the problem.
  • Torque: Any force that causes an object to rotate around an axis.
  • Angular velocity: The speed at which an object rotates. Here, it is given as 10 rad/s after the collision.
  • Angular momentum: Product of rotational inertia and rotational velocity.
By linking these elements together, rotational dynamics describes how the system moves under the influence of forces and motion constraints.

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Most popular questions from this chapter

In unit-vector notation, what is the torque about the origin on a particle located at coordinates \((0,-4.0 \mathrm{~m}, 3.0 \mathrm{~m})\) if that torque is due to (a) force \(\vec{F}_{1}\) with components \(F_{1 x}=2.0 \mathrm{~N}, F_{1 y}=F_{1 z}=0\), and \((\mathrm{b})\) force \(\vec{F}_{2}\) with components \(F_{2 x}=0, F_{2 y}=2.0 \mathrm{~N}, F_{2 z}=4.0 \mathrm{~N} ?\)

With axle and spokes of negligible mass and a thin rim, a certain bicycle wheel has a radius of \(0.350 \mathrm{~m}\) and weighs \(37.0 \mathrm{~N} ;\) it can turn on its axle with negligible friction. A man holds the wheel above his head with the axle vertical while he stands on a turntable that is free to rotate without friction; the wheel rotates clockwise, as seen from above, with an angular speed of \(57.7 \mathrm{rad} / \mathrm{s}\), and the turntable is initially at rest. The rotational inertia of wheel \(+\mathrm{man}+\) turntable about the common axis of rotation is \(2.10 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The man's free hand suddenly stops the rotation of the wheel (relative to the turntable). Determine the resulting (a) angular speed and (b) direction of rotation of the system.

A man stands on a platform that is rotating (without friction) with an angular speed of \(1.2\) rev/s; his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) If by moving the bricks the man decreases the rotational inertia of the system to \(2.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\), what are (a) the resulting angular speed of the platform and (b) the ratio of the new kinetic energy of the system to the original kinetic energy? (c) What source provided the added kinetic energy?

An automobile traveling at \(80.0 \mathrm{~km} / \mathrm{h}\) has tires of \(75.0 \mathrm{~cm}\) diameter. (a) What is the angular speed of the tires about their axles? (b) If the car is brought to a stop uniformly in \(30.0\) complete turns of the tires (without skidding), what is the magnitude of the angular acceleration of the wheels? (c) How far does the car move during the braking?

A uniform disk of mass \(10 m\) and radius \(3.0 r\) can rotate freely about its fixed center like a merry-go-round. A smaller uniform disk of mass \(m\) and radius \(r\) lies on top of the larger disk, concentric with it. Initially the two disks rotate together with an angular velocity of \(20 \mathrm{rad} / \mathrm{s}\). Then a slight disturbance causes the smaller disk to slide outward across the larger disk, until the outer edge of the smaller disk catches on the outer edge of the larger disk. Afterward, the two disks again rotate together (without further sliding). (a) What then is their angular velocity about the center of the larger disk? (b) What is the ratio \(K / K_{0}\) of the new kinetic energy of the two-disk system to the system's initial kinetic energy?
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