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Ch 12: Rotation of a Rigid Body
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 12: Rotation of a Rigid BodyProblem 82
Chapter 12, Problem 82

FIGURE P12.82 shows a cube of mass m sliding without friction at speed v0. It undergoes a perfectly elastic collision with the bottom tip of a rod of length d and mass M = 2m. The rod is pivoted about a frictionless axle through its center, and initially it hangs straight down and is at rest. What is the cube's velocity—both speed and direction—after the collision?

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Analyze the system: The collision is perfectly elastic, so both momentum and kinetic energy are conserved. The system consists of a cube of mass m moving with velocity v₀ and a rod of mass M = 2m, which is initially at rest and pivoted at its center. The rod can rotate about its center, so angular momentum conservation will also be important.
Set up the conservation of linear momentum: Before the collision, the cube has momentum p₁ = m * v₀, and the rod has no linear momentum. After the collision, the cube will have a new velocity v₁, and the rod will gain angular momentum. Write the equation for linear momentum conservation: m * v₀ = m * v₁ + (M * v_cm), where v_cm is the velocity of the rod's center of mass.
Set up the conservation of angular momentum: The rod is pivoted at its center, so the angular momentum of the system about the pivot point must be conserved. Before the collision, the cube has angular momentum L₁ = m * v₀ * d/2 (where d/2 is the perpendicular distance from the pivot to the cube's line of motion). After the collision, the rod will have angular momentum L₂ = I * ω (where I = (1/12) * M * d² is the moment of inertia of the rod about its center, and ω is its angular velocity), and the cube will have angular momentum L₃ = m * v₁ * d/2. Write the equation for angular momentum conservation: m * v₀ * d/2 = (1/12) * M * d² * ω + m * v₁ * d/2.
Set up the conservation of kinetic energy: Since the collision is perfectly elastic, the total kinetic energy before and after the collision is conserved. Before the collision, the kinetic energy is KE₁ = (1/2) * m * v₀². After the collision, the kinetic energy is KE₂ = (1/2) * m * v₁² + (1/2) * I * ω². Write the equation for kinetic energy conservation: (1/2) * m * v₀² = (1/2) * m * v₁² + (1/2) * (1/12) * M * d² * ω².
Solve the system of equations: You now have three equations: one for linear momentum conservation, one for angular momentum conservation, and one for kinetic energy conservation. Solve these equations simultaneously to find the cube's final velocity v₁ (both magnitude and direction) and the angular velocity ω of the rod. Substitute M = 2m and simplify the equations as needed to express v₁ in terms of v₀, m, and d.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Conservation of Momentum

In a closed system, the total momentum before a collision is equal to the total momentum after the collision. This principle is crucial for analyzing collisions, as it allows us to relate the velocities of the objects involved before and after the event. In this scenario, we can apply conservation of momentum to determine the final velocity of the cube and the rod after the elastic collision.
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Elastic Collision

An elastic collision is one in which both kinetic energy and momentum are conserved. This means that not only do the objects bounce off each other without losing energy, but their total kinetic energy before and after the collision remains the same. Understanding this concept is essential for solving the problem, as it dictates how the velocities of the cube and the rod will change after the collision.
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Rotational Dynamics

Rotational dynamics involves the study of the motion of objects that rotate about an axis. In this problem, the rod pivots around a frictionless axle, which means we need to consider the effects of torque and angular momentum. The interaction between the cube and the rod during the collision will impart angular momentum to the rod, affecting its rotational motion and the final velocities of both objects.
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Related Practice
Textbook Question

The two blocks in FIGURE CP12.86 are connected by a massless rope that passes over a pulley. The pulley is 12 cm in diameter and has a mass of 2.0 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.50 N m. If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?

Textbook Question

A 10 g bullet traveling at 400 m/s strikes a 10 kg, 1.0-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open. What is the angular velocity of the door just after impact?

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Textbook Question

A merry-go-round is a common piece of playground equipment. A 3.0-m-diameter merry-go-round with a mass of 250 kg is spinning at 20 rpm. John runs tangent to the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 30 kg. What is the merry-go-round's angular velocity, in rpm, after John jumps on?

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Textbook Question

A satellite follows the elliptical orbit shown in FIGURE P12.77. The only force on the satellite is the gravitational attraction of the planet. The satellite's speed at point 1 is 8000 m/s. What is the satellite's speed at point 2?

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Textbook Question

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Textbook Question

Objects that rotate in air or water experience a torque due to drag. With quadratic drag, a drag torque that's negligible at low rpm quickly becomes significant as the rpm increases. Consider a square bar with cross section a x a and length L. It is rotating on an axle through its center at angular velocity ω in a fluid of density ρ. Assume that the drag coefficient C𝒹 is constant along the length of the bar. Find an expression for the magnitude of the drag torque on the bar. Hint: Begin by considering the drag force on a small piece of the bar of length dr at distance r from the axle.

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