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Chapter 6: Problem 178

The wad of clay of mass \(m\) is initially moving with a horizontal velocity \(v_{1}\) when it strikes and sticks to the initially stationary uniform slender bar of mass \(M\) and length \(L\). Determine the final angular velocity of the combined body and the \(x\) -component of the linear impulse applied to the body by the pivot \(O\) during the impact.

Short Answer

Expert verified
The final angular velocity is \( \omega_f = \frac{m v_1 \frac{L}{2}}{\frac{1}{3}ML^2 + m \frac{L^2}{4}} \). The impulse is given by \( I_x = m v_1 - (m + M)V_x \).

Step by step solution

01

Understand the Conservation of Angular Momentum

Before the collision, only the clay has angular momentum as the bar is stationary. The clay's position is \( \frac{L}{2} \) (midpoint of the bar) from the pivot. Use the conservation of angular momentum about point \( O \): \[ m \cdot v_1 \cdot \frac{L}{2} + 0 = (I_\text{total}) \cdot \omega_f + 0 \] where \(I_\text{total}\) is the moment of inertia of the system after the collision.
02

Calculate the Moment of Inertia of the Bar

The moment of inertia of a slender bar about its end is \( \frac{1}{3}ML^2 \). The clay can be considered a point mass at a distance \( \frac{L}{2} \), so its moment of inertia is \( m \left(\frac{L}{2}\right)^2 \). The total moment of inertia is: \[ I_\text{total} = \frac{1}{3}ML^2 + m \frac{L^2}{4} \]
03

Solve for the Final Angular Velocity

Substitute the expression for the total moment of inertia into the angular momentum equation and solve for the final angular velocity \( \omega_f \). \[ m v_1 \frac{L}{2} = \left( \frac{1}{3}ML^2 + m \frac{L^2}{4} \right) \omega_f \] Solve this equation to find \( \omega_f \): \[ \omega_f = \frac{m v_1 \frac{L}{2}}{\frac{1}{3}ML^2 + m \frac{L^2}{4}} \]
04

Understand Linear Impulse and Conservation of Linear Momentum

The linear impulse applied by the pivot is needed due to the change in linear momentum of the system. The total linear momentum along the x-axis before the impact is \( m v_1 \) since the bar was at rest. After impact, the x-component of linear momentum must equal this because momentum is conserved when considering only x-components, except the impulse from \( O \).
05

Calculate the x-component of the Linear Impulse

Apply the conservation of linear momentum: \( I_x = m v_1 - (m + M)V_x \) where \( V_x \) is the velocity of the center of mass of the system. To determine \( V_x \), suppose the center of mass velocity is a result of the angular velocity: \( V_x = \omega_f \frac{L}{2} \) at the bar's midpoint. Substitute \( \omega_f \) found earlier to solve for impulse \( I_x \).

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Key Concepts

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

Moment of Inertia
In physics, the moment of inertia is a crucial concept that represents an object's resistance to changes in its state of rotation. It is analogous to mass in linear motion. In the context of our problem, the moment of inertia (I_ ext{total}) describes how difficult it is to rotate the combined system of the bar and the clay after the collision.

  • For the slender bar, which is uniform, its moment of inertia about one end is given by the formula: \( \frac{1}{3}ML^2 \).
  • The clay, acting as a point mass located at the midpoint of the bar, contributes its own moment of inertia: \( m \left(\frac{L}{2}\right)^2 = m \frac{L^2}{4} \).
  • The total moment of inertia of the system is the sum of these two contributions: \( I_ ext{total} = \frac{1}{3}ML^2 + m \frac{L^2}{4} \).
Understanding this sum is essential because it affects the entire motion of the system post-collision.
Linear Impulse
Linear impulse in a physical context is related to the change in momentum resulting from a force applied over a period of time. In our scenario, the linear impulse comes into play at the pivot point, impacting the overall motion of the system.

  • The linear momentum of the system prior to collision is entirely due to the moving clay, calculated by the product of its mass and velocity, \( m \cdot v_1 \).
  • Post-collision, the total linear momentum must remain equal to the initial momentum due to the conservation of momentum principle, but here we account for the impulse that the pivot applies to the system.
To comprehend the linear impulse, we analyze the change in motion induced by this interaction at the pivot, understanding how the impulse helps maintain the system's momentum balance.
Conservation of Momentum
The conservation of momentum is a fundamental law stating that the total linear and angular momentum of a system remains constant in the absence of external forces. In this exercise, we apply this principle to both linear and angular aspects.

  • For linear momentum: Prior to the impact, the momentum is entirely with the clay. After, for the momentum to be conserved horizontally, the linear impulse applied by the pivot must balance any changes.
  • Angular momentum: Before impact, the clay's motion contributes angular momentum about the pivot only. This must equal the angular momentum of the entire system after the collision.
This law allows us to solve for unknowns, like the linear impulse or final angular velocity, ensuring the momentum distributions pre- and post-collision are consistent with natural laws.
Angular Velocity
Angular velocity (\omega_f) represents how fast an object rotates around a point. In this problem, determining the angular velocity of the bar-clay system post-collision is crucial for understanding the rotational dynamics.

  • Initially, the clay's movement translates into angular momentum at the bar's pivot, contributing to the system's post-collision rotation.
  • We derived \omega_f using the formula from conservation of angular momentum: \[\omega_f = \frac{m v_1 \frac{L}{2}}{\frac{1}{3}ML^2 + m \frac{L^2}{4}}\].
  • This equation results from balancing the initial angular momentum (from the clay) with the rotational momentum of the combined system.
Understanding this concept gives insight into how colliding bodies affect each other's rotation, giving depth to real-life applications like machinery or planetary motion.

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Most popular questions from this chapter

The 50 -kg flywheel has a radius of gyration \(\bar{k}=0.4 \mathrm{m}\) about its shaft axis and is subjected to the torque \(M=2\left(1-e^{-0.1 \theta}\right) \mathrm{N} \cdot \mathrm{m},\) where \(\theta\) is in radians. If the flywheel is at rest when \(\theta=0,\) determine its angular velocity after 5 revolutions.

The cargo box of the food-delivery truck for aircraft servicing has a loaded mass \(m\) and is elevated by the application of a couple \(M\) on the lower end of the link which is hinged to the truck frame. The horizontal slots allow the linkage to unfold as the cargo box is elevated. Determine the upward acceleration of the box in terms of \(h\) for a given value of \(M .\) Neglect the mass of the links.

The nose-wheel assembly is raised by the application of a torque \(M\) to link \(B C\) through the shaft at \(B\) The arm and wheel \(A O\) have a combined weight of 100 lb with center of mass at \(G\), and a centroidal radius of gyration of 14 in. If the angle \(\theta=30^{\circ},\) determine the torque \(M\) necessary to rotate link \(A O\) with a counterclockwise angular velocity of 10 deg/sec that is increasing at the rate of 5 deg/sec every second. Additionally, determine the total force supported by the pin at \(A\). The mass of links \(B C\) and \(C D\) may be neglected for this analysis.

The gear train shown starts from rest and reaches an output speed of \(\omega_{C}=240\) rev/min in 2.25 s. Rotation of the train is resisted by a constant \(150 \mathrm{N} \cdot \mathrm{m}\) moment at the output gear \(C .\) Determine the required input power to the \(86 \%\) efficient motor at \(A\) just before the final speed is reached. The gears have masses \(m_{A}=6 \mathrm{kg}, m_{B}=10 \mathrm{kg},\) and \(m_{C}=24 \mathrm{kg}\) pitch diameters \(d_{A}=120 \mathrm{mm}, d_{B}=160 \mathrm{mm},\) and \(d_{C}=240 \mathrm{mm},\) and centroidal radii of gyration \(k_{A}=\) \(48 \mathrm{mm}, k_{B}=64 \mathrm{mm},\) and \(k_{C}=96 \mathrm{mm}\)

A space telescope is shown in the figure. One of the reaction wheels of its attitude-control system is spinning as shown at \(10 \mathrm{rad} / \mathrm{s}\), and at this speed the friction in the wheel bearing causes an internal moment of \(10^{-6} \mathrm{N} \cdot \mathrm{m}\). Both the wheel speed and the friction moment may be considered constant over a time span of several hours. If the mass moment of inertia of the entire spacecraft about the \(x\) -axis is \(150\left(10^{3}\right) \mathrm{kg} \cdot \mathrm{m}^{2},\) determine how much time passes before the line of sight of the initially stationary spacecraft drifts by 1 arcsecond, which is \(1 / 3600\) degree. All other elements are fixed relative to the spacecraft, and no torquing of the reaction wheel shown is performed to correct the attitude drift. Neglect external torques.
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