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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
Chapter 12, Problem 81

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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Determine the initial angular velocity of the merry-go-round in radians per second. Convert the given angular velocity from revolutions per minute (rpm) to radians per second using the formula: \( \omega = \frac{2\pi \times \text{rpm}}{60} \).
Calculate the initial moment of inertia of the merry-go-round. For a solid disk, the moment of inertia is given by \( I = \frac{1}{2} M R^2 \), where \( M \) is the mass of the disk and \( R \) is its radius. The radius is half the diameter.
Determine John's linear momentum before he jumps onto the merry-go-round. His linear momentum is given by \( p = m v \), where \( m \) is John's mass and \( v \) is his velocity. Convert this linear momentum into angular momentum relative to the center of the merry-go-round using \( L = p R \), where \( R \) is the radius of the merry-go-round.
Calculate the total moment of inertia after John jumps onto the merry-go-round. Add John's contribution to the moment of inertia, which is \( I_J = m R^2 \), to the initial moment of inertia of the merry-go-round.
Apply the principle of conservation of angular momentum. The total angular momentum before John jumps (merry-go-round's angular momentum + John's angular momentum) equals the total angular momentum after he jumps. Solve for the final angular velocity \( \omega_f \) using \( L_{initial} = I_{total} \omega_f \). Convert \( \omega_f \) back to rpm using \( \text{rpm} = \frac{\omega_f \times 60}{2\pi} \).

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Key Concepts

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

Angular Momentum

Angular momentum is a measure of the rotational motion of an object and is defined as the product of an object's moment of inertia and its angular velocity. In a closed system, the total angular momentum before an event must equal the total angular momentum after the event, assuming no external torques act on the system. This principle is crucial for solving problems involving rotating bodies, such as the merry-go-round in this scenario.
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Moment of Inertia

Moment of inertia is a property of a body that quantifies its resistance to changes in its rotational motion. It depends on the mass distribution relative to the axis of rotation. For a solid disk, like the merry-go-round, the moment of inertia can be calculated using the formula I = (1/2) m r², where m is the mass and r is the radius. Understanding moment of inertia is essential for calculating the effects of added mass, such as John jumping onto the merry-go-round.
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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 problem, when John jumps onto the merry-go-round, the system's angular momentum before he jumps must equal the angular momentum after he jumps. This principle allows us to set up an equation to find the new angular velocity of the merry-go-round after John's addition to the system.
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Related Practice
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 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

During most of its lifetime, a star maintains an equilibrium size in which the inward force of gravity on each atom is balanced by an outward pressure force due to the heat of the nuclear reactions in the core. But after all the hydrogen 'fuel' is consumed by nuclear fusion, the pressure force drops and the star undergoes a gravitational collapse until it becomes a neutron star. In a neutron star, the electrons and protons of the atoms are squeezed together by gravity until they fuse into neutrons. Neutron stars spin very rapidly and emit intense pulses of radio and light waves, one pulse per rotation. These 'pulsing stars' were discovered in the 1960s and are called pulsars. a. A star with the mass (M = 2.0 X 1030 kg) and size (R = 7.0 x 108 m) of our sun rotates once every 30 days. After undergoing gravitational collapse, the star forms a pulsar that is observed by astronomers to emit radio pulses every 0.10 s. By treating the neutron star as a solid sphere, deduce its radius.

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Textbook Question

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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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. Does the satellite experience any torque about the center of the planet? Explain.

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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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