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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 78

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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Step 1: Identify the conservation principle involved. This problem involves the conservation of angular momentum because the bullet embeds itself in the door, creating a rotational motion. Angular momentum before the collision equals angular momentum after the collision.
Step 2: Calculate the initial angular momentum of the bullet. The bullet has linear momentum, which can be converted to angular momentum relative to the hinge. Use the formula: L=mvr, where m is the mass of the bullet, v is its velocity, and r is the distance from the hinge to the point of impact (1.0 m).
Step 3: Determine the moment of inertia of the door after the bullet embeds itself. The door can be treated as a rectangular object rotating about one edge (hinge). Use the formula for the moment of inertia of a rectangular object: I=13ML2, where M is the mass of the door and L is its width. Add the contribution of the bullet's mass to the moment of inertia using I=mr2, where m is the bullet's mass and r is the distance from the hinge.
Step 4: Apply the conservation of angular momentum. The total angular momentum before the collision (from the bullet) equals the total angular momentum after the collision (rotational motion of the door and embedded bullet). Use the equation: L=Iω, where ω is the angular velocity of the door after impact.
Step 5: Solve for the angular velocity ω. Rearrange the conservation of angular momentum equation to isolate ω: ω=LI. Substitute the values for L (angular momentum of the bullet) and I (moment of inertia of the door and bullet system) to find the angular velocity.

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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 principle of conservation of angular momentum states that if no external torque acts on a system, the total angular momentum of that system remains constant. In this scenario, the bullet's linear momentum is converted into angular momentum when it strikes the door, allowing us to analyze the system's behavior post-impact.
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Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion, depending on the mass distribution relative to the axis of rotation. For the door, the moment of inertia can be calculated using its mass and the distance from the hinge to where the bullet strikes, which is crucial for determining the resulting angular velocity.
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Angular Velocity

Angular velocity is a vector quantity that represents the rate of rotation of an object around an axis, typically measured in radians per second. After the bullet embeds itself in the door, the angular velocity can be calculated using the conservation of angular momentum, allowing us to understand how fast the door will swing open immediately after the impact.
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Related Practice
Textbook Question

The sphere of mass M and radius R in FIGURE P12.75 is rigidly attached to a thin rod of radius r that passes through the sphere at distance (1/2)R from the center. A string wrapped around the rod pulls with tension T. Find an expression for the sphere's angular acceleration. The rod's moment of inertia is negligible.

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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

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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