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Ch 10: Dynamics of Rotational Motion
Chapter 10, Problem 10

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•m^2. 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. Diagram illustrating the gyroscope's rotation axes and components for angular momentum conservation.
(a) Find the upward force exerted by the pivot.

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Calculate the total mass of the gyroscope by adding the mass of the rotor and the mass of the frame: m_total = m_rotor + m_frame.
Determine the gravitational force acting on the gyroscope using F_gravity = m_total * g, where g is the acceleration due to gravity (9.8 m/s^2).
Since the gyroscope is supported on a single pivot, the upward force exerted by the pivot must balance the gravitational force. Therefore, the upward force F_pivot is equal to F_gravity.
Express the upward force exerted by the pivot as F_pivot = m_total * g.
Substitute the values of m_total and g into the equation to find the upward force exerted by the pivot.

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Key Concepts

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

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion about an axis. It depends on the mass distribution relative to the axis of rotation. For a rigid body, it is calculated by summing the products of each mass element and the square of its distance from the axis. In this case, the gyroscope's moment of inertia is crucial for understanding its rotational dynamics.
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Intro to Moment of Inertia

Precession

Precession is the phenomenon where the axis of a spinning object, like a gyroscope, moves in a circular path due to an external torque. This occurs when the gravitational force acting on the center of mass creates a torque about the pivot point. The rate of precession can be calculated using the angular momentum of the gyroscope and the torque applied, which is essential for analyzing the gyroscope's motion in this problem.

Torque and Equilibrium

Torque is the rotational equivalent of linear force and is calculated as the product of the force and the distance from the pivot point. In the context of the gyroscope, the upward force exerted by the pivot must balance the torque due to the weight of the gyroscope to maintain equilibrium. Understanding how these forces interact is key to solving for the upward force in this scenario.
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Related Practice
Textbook Question
A cord is wrapped around the rim of a solid uniform wheel 0.250 m in radius and of mass 9.20 kg. A steady horizontal pull of 40.0 N to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (b) Find the magnitude and direction of the force that the axle exerts on the wheel.

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Textbook Question
A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m>s when it is at point P in Fig. E10.35.

(b) If the only force acting on the rock is its weight, what is the rate of change (magnitude and direction) of its angular momentum at this instant?
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Textbook Question
A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m>s when it is at point P in Fig. E10.35.

(a) At this instant, what are the magnitude and direction of its angular momentum relative to point O?
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Textbook Question

A machine part has the shape of a solid uniform sphere of mass 225 g and diameter 3.00 cm. It is spinning about a frictionless axle through its center, but at one point on its equator it is scraping against metal, resulting in a friction force of 0.0200 N at that point. (b) How long will it take to decrease its rotational speed by 22.5 rad/s?

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Textbook Question

A solid ball is released from rest and slides down a hillside that slopes downward at 65.0° from the horizontal. (c) In part (a), why did we use the coefficient of static friction and not the coefficient of kinetic friction?

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Textbook Question

A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg•m^2 about a vertical axle through its center, and it turns with negligible friction. (a) A child applies an 18.0-N force tangentially to the edge of the merry-go-round for 15.0 s. If the merry-go-round is initially at rest, what is its angular speed after this 15.0-s interval?

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