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Ch 10: Dynamics of Rotational Motion
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 10, Problem 50a

A uniform, 4.5-kg, square, solid wooden gate 1.5 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.1-kg raven flying horizontally at 5.0 m/s flies into this door at its center and bounces back at 2.0 m/s in the opposite direction. What is the angular speed of the gate just after it is struck by the unfortunate raven?

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First, identify the conservation of angular momentum principle. The angular momentum before the collision must equal the angular momentum after the collision, as no external torques are acting on the system.
Calculate the initial angular momentum of the raven. Use the formula for linear momentum, \( p = mv \), where \( m \) is the mass of the raven and \( v \) is its velocity. The raven's initial angular momentum relative to the pivot is \( L_{initial} = m_{raven} \cdot v_{raven} \cdot r \), where \( r \) is the distance from the pivot to the point of impact (0.75 m, half the side of the gate).
Calculate the final angular momentum of the raven after it bounces back. Use the same formula, but with the final velocity of the raven. The direction of the velocity is opposite, so the angular momentum will be negative: \( L_{final} = m_{raven} \cdot v_{final} \cdot r \).
Determine the change in angular momentum of the raven, \( \Delta L = L_{initial} - L_{final} \). This change in angular momentum is transferred to the gate.
Use the formula for the moment of inertia of a square gate about the pivot point, \( I = \frac{1}{3} m_{gate} \cdot L^2 \), where \( m_{gate} \) is the mass of the gate and \( L \) is the length of the side of the gate. Then, use the relationship \( \Delta L = I \cdot \omega \) to solve for the angular speed \( \omega \) of the gate: \( \omega = \frac{\Delta L}{I} \).

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Key Concepts

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

Conservation of Angular Momentum

The conservation of angular momentum states that if no external torque acts on a system, the total angular momentum remains constant. In this scenario, the raven's impact on the gate is an internal interaction, so the angular momentum before and after the collision must be equal, allowing us to calculate the gate's angular speed post-collision.
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Moment of Inertia

Moment of inertia is a measure of an object's resistance to changes in its rotation. For a square gate pivoted at its upper edge, the moment of inertia can be calculated using the formula for a solid square plate. This value is crucial for determining the gate's angular speed after the collision, as it relates angular momentum to angular velocity.
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Elastic Collision

An elastic collision is one where kinetic energy is conserved. In this problem, the raven bounces back after hitting the gate, indicating an elastic collision. Understanding this concept helps in analyzing the energy transfer during the collision, which affects the gate's motion and allows us to apply conservation laws effectively.
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Related Practice
Textbook Question

The rotor (flywheel) of a toy gyroscope has mass 0.140 kg. Its moment of inertia about its axis is 1.20 × 10-4 kg m2. The mass of the frame is 0.0250 kg. The gyroscope is supported on a single pivot (Fig. E10.51) with its center of mass a horizontal distance of 4.00 cm from the pivot. The gyroscope is precessing in a horizontal plane at the rate of one revolution in 2.20 s. Find the upward force exerted by the pivot.

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Textbook Question

A uniform, 4.5-kg, square, solid wooden gate 1.5 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.1-kg raven flying horizontally at 5.0 m/s flies into this door at its center and bounces back at 2.0 m/s in the opposite direction. During the collision, why is the angular momentum conserved but not the linear momentum?

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Textbook Question

A thin uniform rod has a length of 0.500 m0.500\(\text{ m}\) and is rotating in a circle on a frictionless table. The axis of rotation is perpendicular to the length of the rod at one end and is stationary. The rod has an angular velocity of 0.400 rad/s0.400\(\text{ rad/s}\) and a moment of inertia about the axis of 3.00×103kg/m23.00\(\times\)10^{-3}\(\text{kg/m}\)^2. A bug initially standing on the rod at the axis of rotation decides to crawl out to the other end of the rod. When the bug has reached the end of the rod and sits there, its tangential speed is 0.160 m/s0.160\(\text{ m/s}\). The bug can be treated as a point mass. What is the mass of the rod.

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Textbook Question

A certain gyroscope precesses at a rate of 0.50 rad/s when used on earth. If it were taken to a lunar base, where the acceleration due to gravity is 0.165g, what would be its precession rate?

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Textbook Question

A small 10.0-g bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 50.0 g and is 100 cm in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 20.0 cm/s relative to the table. What is the angular speed of the bar just after the frisky insect leaps?

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Textbook Question

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth's mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

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