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Chapter 10: Problem 52

A bowling ball has mass \(M,\) radius \(R,\) and a moment of inertia of \(\frac{2}{5} M R^{2} .\) If it starts from rest, how much work must be done on it to set it rolling without slipping at a linear speed \(v ?\) Express the work in terms of \(M\) and \(v\)

Short Answer

Expert verified
The work needed to set the bowling ball rolling without slipping at a linear speed \(v\) is \(\frac{7}{10} M v^2\).

Step by step solution

01

Analyze the problem and define the types of kinetic energy

When a bowling ball starts rolling from rest, two types of kinetic energy come into play: translational kinetic energy and rotational kinetic energy. The translational kinetic energy is due to the movement of the ball along the linear path and the rotational kinetic energy is due to the spinning of the ball around its axis.
02

Express the translational and rotational kinetic energy

The translational kinetic energy (\(KE_{trans}\)) can be represented as \(\frac{1}{2} m v^2\), where \(m\) represents the mass and \(v\) the velocity. The rotational kinetic energy (\(KE_{rot}\)) can be represented as \(\frac{1}{2} I \omega^2\), where \(I\) is the moment of inertia and \(\omega\) the angular velocity. Since the ball is rolling without slipping, the linear velocity \(v\) and the angular velocity \(\omega\) are related by \(v = R \omega\). Hence, \(KE_{rot} = \frac{1}{2} I \left(\frac{v}{R}\right)^2\).
03

Express the work done in terms of the final kinetic energy

The work done to set the ball rolling from rest will be equal to the final kinetic energy of the ball. Thus, the work \(W\) can be written as \(W = KE_{trans} + KE_{rot} \). Substituting the expressions for the kinetic energies from step 2 into this equation, we get \(W = \frac{1}{2} M v^2 + \frac{1}{2} \left(\frac{2}{5} M R^2\right) \left(\frac{v}{R}\right)^2 = \frac{7}{10} M v^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Translational Kinetic Energy
When an object moves along a linear path, it possesses translational kinetic energy. This type of energy is due to the mass and velocity of the object, given by the formula \( KE_{trans} = \frac{1}{2} m v^2 \). Here, \( m \) is the mass of the object, and \( v \) is its velocity.
  • Consider a bowling ball with mass \( M \) moving at speed \( v \).
  • The translational kinetic energy is calculated by substituting these values into the formula.
  • For our bowling ball, this becomes \( \frac{1}{2} M v^2 \).
Translational kinetic energy describes the straight-line movement energy of an object. It signifies the capacity of an object to do work due to its motion along a path. It is crucial to understand this concept when analyzing moving objects like a rolling bowling ball.
Rotational Kinetic Energy
Objects that rotate or spin also possess rotational kinetic energy. This form of energy relates to an object's moment of inertia and its angular velocity. It is given by \( KE_{rot} = \frac{1}{2} I \omega^2 \). Here, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity.
  • For a rolling object, angular velocity \( \omega \) and linear velocity \( v \) are related by \( \omega = \frac{v}{R} \), where \( R \) is the radius.
  • This means that for a bowling ball, \( KE_{rot} = \frac{1}{2} I \left(\frac{v}{R}\right)^2 \).
  • As per the exercise, the moment of inertia \( I = \frac{2}{5} M R^2 \).
  • Substituting values, \( KE_{rot} = \frac{1}{2} \left(\frac{2}{5} M R^2\right) \left(\frac{v}{R}\right)^2 \), simplifying further.
For our rolling ball, both translational and rotational kinetic energies must be considered to fully understand its motion. Rotational kinetic energy represents the energy from spinning around an axis.
Moment of Inertia
The moment of inertia, \( I \), essentially governs how hard it is to change an object's rotational state. It's comparable to mass in linear motion but applies to rotational motion. It depends on both the object's mass and the distribution of that mass relative to the axis of rotation.
  • For a spherical object like a bowling ball, the moment of inertia is \( I = \frac{2}{5} M R^2 \).
  • Moment of inertia factors in both the mass and the geometry of the object.
  • This value explains how much torque is needed for a given angular acceleration.
  • A larger \( I \) means it's tougher to change the object's rotation.
In understanding kinetic energies, knowing the moment of inertia helps to break down the contributions of both translational and rotational components. It acts as a bridge in rotation-related problems, allowing you to translate angular motion into work and energy calculations.

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Most popular questions from this chapter

A large, cylindrical roll of tissue paper of initial radius \(R\) lies on a long, horizontal surface with the outside end of the paper nailed to the surface. The roll is given a slight shove \(\left(v_{i} \approx 0\right)\) and commences to unroll. Assume the roll has a uniform density and that mechanical energy is conserved in the process. (a) Determine the speed of the center of mass of the roll when its radius has diminished to \(r\) (b) Calculate a numerical value for this speed at \(r=1.00 \mathrm{mm},\) assuming \(R=6.00 \mathrm{m} .\) (c) What If? What happens to the energy of the system when the paper is completely unrolled?

The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for \(8.00 \mathrm{s}\), at which time it is turning at 5.00 rev/s. At this point the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub smoothly slows to rest in \(12.0 \mathrm{s}\) Through how many revolutions does the tub turn while it is in motion?

(a) Without the wheels, a bicycle frame has a mass of \(8.44 \mathrm{kg} .\) Each of the wheels can be roughly modeled as a uniform solid disk with a mass of \(0.820 \mathrm{kg}\) and a radius of \(0.343 \mathrm{m} .\) Find the kinetic energy of the whole bicycle when it is moving forward at \(3.35 \mathrm{m} / \mathrm{s}\). (b) Before the invention of a wheel turning on an axle, ancient people moved heavy loads by placing rollers under them. (Modern people use rollers too. Any hardware store will sell you a roller bearing for a lazy susan.) A stone block of mass 844 kg moves forward at \(0.335 \mathrm{m} / \mathrm{s}\), supported by two uniform cylindrical tree trunks, each of mass \(82.0 \mathrm{kg}\) and radius \(0.343 \mathrm{m}\) No slipping occurs between the block and the rollers or between the rollers and the ground. Find the total kinetic energy of the moving objects.

An airliner arrives at the terminal, and the engines are shut off. The rotor of one of the engines has an initial clockwise angular speed of 2000 rad/s. The engine's rotation slows with an angular acceleration of magnitude \(80.0 \mathrm{rad} / \mathrm{s}^{2} .\) (a) Determine the angular speed after \(10.0 \mathrm{s}\) (b) How long does it take the rotor to come to rest?

A horizontal \(800-\mathrm{N}\) merry-go-round is a solid disk of radius \(1.50 \mathrm{m},\) started from rest by a constant horizontal force of \(50.0 \mathrm{N}\) applied tangentially to the edge of the disk. Find the kinetic energy of the disk after \(3.00 \mathrm{s}\)
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