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

Knight Calc 5th Edition

Ch 12: Rotation of a Rigid Body

Problem 73a

Chapter 12, Problem 73a

A long, thin rod of mass M and length L is standing straight up on a table. Its lower end rotates on a frictionless pivot. A very slight push causes the rod to fall over. As it hits the table, what are the angular velocity

Verified step by step guidance

Step 1: Recognize that the problem involves rotational motion and energy conservation. The rod starts in a vertical position and falls due to gravity, converting potential energy into rotational kinetic energy.

Step 2: Write the expression for the initial gravitational potential energy of the rod. Since the rod is uniform, its center of mass is located at a height of L/2. The potential energy is given by \( U = M g \frac{L}{2} \), where \( g \) is the acceleration due to gravity.

Step 3: Write the expression for the rotational kinetic energy of the rod as it hits the table. The rotational kinetic energy is given by \( K = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia of the rod about the pivot point, and \( \omega \) is the angular velocity.

Step 4: Determine the moment of inertia \( I \) of the rod about the pivot point. For a thin rod rotating about one end, \( I = \frac{1}{3} M L^2 \). Substitute this into the rotational kinetic energy expression.

Step 5: Apply the principle of conservation of energy. The initial potential energy \( U \) is completely converted into rotational kinetic energy \( K \) as the rod hits the table. Set \( M g \frac{L}{2} = \frac{1}{2} \left( \frac{1}{3} M L^2 \right) \omega^2 \) and solve for \( \omega \), the angular velocity.

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Key Concepts

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

Angular Velocity

Angular velocity is a measure of how quickly an object rotates around a pivot point, expressed in radians per second. In the context of the falling rod, it describes the rate at which the angle of the rod changes as it falls. Understanding angular velocity is crucial for analyzing the motion of rotating bodies and can be calculated using the formula ω = θ/t, where θ is the angle in radians and t is the time taken.

Moment of Inertia

Moment of inertia is a property of a body that quantifies its resistance to angular acceleration about a pivot point. For a rod rotating about one end, the moment of inertia is given by I = (1/3)ML², where M is the mass and L is the length of the rod. This concept is essential for determining how the mass distribution affects the rod's rotational motion as it falls.

Conservation of Energy

The principle of conservation of energy states that the total energy in a closed system remains constant. In the case of the falling rod, potential energy is converted into kinetic energy as it falls. This concept helps in calculating the angular velocity at the moment of impact by equating the initial potential energy at the upright position to the kinetic energy just before it hits the table.

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Related Practice

Textbook Question

A long, thin rod of mass M and length L is standing straight up on a table. Its lower end rotates on a frictionless pivot. A very slight push causes the rod to fall over. As it hits the table, what are the speed of the tip of the rod?

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Textbook Question

A solid spherical marble shot up a frictionless 15° slope rolls 2.50 m to its highest point. If the marble is shot with the same speed up a slightly rough 15° slope, it rolls only 2.30 m. What is the coefficient of rolling friction on the second slope?

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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

The 5.0 kg, 60-cm-diameter disk in FIGURE P12.71 rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released. What is the cylinder's angular velocity when it is directly below the axle?

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Textbook Question

A 750 g disk and a 760 g ring, both 15 cm in diameter, are rolling along a horizontal surface at 1.5 m/s when they encounter a 15° slope. How far up the slope does each travel before rolling back down?

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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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