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What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface? (a) A uniform solid cylinder; (b) a uniform sphere; (c) a thin-walled, hollow sphere; (d) a hollow cylinder with outer radius \(R\) and inner radius \(R / 2\).

Short Answer

Expert verified
The Fraction of total kinetic energy that is rotational for (a) a solid cylinder is \(1/3\), (b) a solid sphere is \(2/5\), (c) a thin-walled hollow sphere is \(2/5\) and (d) a hollow cylinder with outer radius \(R\) and inner radius \(R / 2\) is \(5 / 13\).

Step by step solution

01

Identify the Moment of Inertia

First, recall the appropriate formula for the moment of inertia for each object, as follows: for a solid cylinder, \(I=0.5MR^2\); for a solid sphere, \(I=0.4MR^2\); for a thin-walled hollow sphere, \(I=0.67MR^2\); and for a hollow cylinder with outer radius \(R\) and inner radius \(R/2\), \(I=0.375MR^2\). Here, \(M\) represents the mass and \(R\) the radius of our objects.
02

Find the Rotational and Translational Kinetic Energy

The kinetic energy for an object rolling without slipping is divided into rotational kinetic energy (\( KE_{rot} = 0.5Iω^2 \)) and translational kinetic energy (\( KE_{trans} = 0.5Mv^2 \)). Since the objects are rolling without slipping \( v = Rω \), hence \( KE_{rot} = 0.5Iω^2 = 0.25MR^2ω^2 \) and \( KE_{trans} = 0.5Mv^2 = 0.5MR^2ω^2 \). So, the total kinetic energy is \( KE_{total} = KE_{rot} + KE_{trans} \)
03

Calculate the Fraction of Total Kinetic Energy

The fraction of the total kinetic energy that is rotational is given by \(Fraction = KE_{rot} / KE_{total} \) Substitute the values of KE_{rot} and KE_{total} which is equivalent to \( Fraction = (0.5*I*ω^2) / ((0.5*I*ω^2) + 0.5Mv^2) \). After simplifying, the fraction will be \( Fraction = (0.5*I) / (0.5*I + 0.5M*R^2) = I / (I + M*R^2) \). Substituting moment of inertia of each object we have (a) For solid cylinder \( I = 0.5MR^2 \), Fraction = 1 / 3 (b) For solid sphere \( I = 0.4MR^2 \), Fraction = 2 / 5 (c) For thin-walled hollow sphere \( I = 0.67MR^2 \), Fraction = 2 / 5 (d) For hollow cylinder with outer radius \( R \) and inner radius \( R / 2 \), \( I = 0.375MR^2 \), Fraction = 5 / 13

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Key Concepts

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

Moment of Inertia
The moment of inertia is a fundamental concept in the physics of rotational motion, analogous to mass in linear motion. It refers to the distribution of mass in an object and its resistance to angular acceleration when a torque is applied. In essence, it's a measure of how difficult it is to change an object's rotational speed.

Each object has a unique moment of inertia, which depends not only on its mass but also on the distribution of that mass relative to the axis of rotation. For example, two objects with the same mass but different shapes will generally have different moments of inertia. The calculations provided in the textbook solutions highlight this by using different formulas for each type of object:
  • A uniform solid cylinder has a moment of inertia described by the formula: \(I = 0.5MR^2\).
  • A solid sphere's moment of inertia is given as: \(I = 0.4MR^2\).
  • A thin-walled, hollow sphere has: \(I = 0.67MR^2\).
  • A hollow cylinder with an outer radius \(R\) and inner radius \(R/2\) has: \(I = 0.375MR^2\).

To comprehend the moment of inertia's impact on rotational kinetic energy, it helps to picture each object rolling. The energy required to keep a solid cylinder spinning is different from that needed to spin a hollow one because their mass distributions are different. These differences in moment of inertia significantly determine how each object behaves when rolling, which directly affects their kinetic energy aspects.
Rolling Without Slipping
Rolling without slipping is an important special case of motion that applies to objects like wheels, balls, or cylinders when they move across a surface. It implies that the point of the object in contact with the surface is momentarily at rest relative to the surface. This condition ensures there is no sliding or skidding, and thus no energy is lost to friction as heat.

When an object rolls without slipping, there's a direct relationship between its translational motion (the movement of its center of mass) and its rotation. Specifically, the linear velocity \(v\) of the center of mass and the angular velocity \(ω\) of rotation are related by the expression \(v = Rω\), where \(R\) is the radius of the object.
  • This condition serves to link linear and rotational motion, allowing us to analyze the kinetic energy of rolling objects with a unified approach.
  • Understanding this concept is essential for determining the portions of kinetic energy attributed to both the translational and rotational movements of an object.
Translational Kinetic Energy
Translational kinetic energy is the energy an object possesses due to its motion through space. When an object moves or translates from one location to another, it has translational kinetic energy, expressed as \( KE_{trans} = 0.5Mv^2 \), where \(M\) is the mass and \(v\) is the linear velocity.

For objects rolling without slipping, such as those in the exercise, translational kinetic energy is only part of the story. As these objects roll, they also spin, which means they have rotational kinetic energy as well. Here's how they differ:
  • Translational kinetic energy relates to the object's overall motion across a surface.
  • Rotational kinetic energy is concerned with the energy due to an object's rotation around its axis.

This dual nature of kinetic energy in rolling objects clarifies why we must calculate both types to understand the full kinetic energy present. By combining translational and rotational kinetic energy, we get a complete picture of the object's total kinetic energy during rolling motion, allowing us to determine how much of the energy is because of its rotation, as seen in the exercise solutions.

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Most popular questions from this chapter

Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of a neutron star is roughly \(10^{14}\) times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was \(7.0 \times 10^{5} \mathrm{~km}\) (comparable to our sun); its final radius is \(16 \mathrm{~km}\). If the original star rotated once in 30 days, find the angular speed of the neutron star.

When an object is rolling without slipping, the rolling friction force is much less than the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3 ). When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that \(a_{x}\) and \(\alpha_{z}\) are approximately zero and \(v_{x}\) and \(\omega_{z}\) are approximately constant. Rolling without slipping means \(v_{x}=r \omega_{z}\) and \(a_{x}=r \alpha_{z}\). If an object is set in motion on a surface without these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established. A solid cylinder with mass \(M\) and radius \(R\), rotating with angular speed \(\omega_{0}\) about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is \(\mu_{\mathrm{k}}\). (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force on the cylinder. Calculate the accelerations \(a_{x}\) of the center of mass and \(\alpha_{z}\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \(\omega_{z}=\omega_{0}\) but \(v_{x}=0\) Rolling without slipping sets in when \(v_{x}=r \omega_{z} .\) Calculate the distance the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

A thin-walled, hollow spherical shell of mass \(m\) and radius \(r\) starts from rest and rolls without slipping down a track (Fig. \(\mathbf{P 1 0 . 7 2}\) ). Points \(A\) and \(B\) are on a circular part of the track having radius \(R\). The diameter of the shell is very small compared to \(h_{0}\) and \(R,\) and the work done by rolling friction is negligible. (a) What is the minimum height \(h_{0}\) for which this shell will make a complete loop-the-loop on the circular part of the track? (b) How hard does the track push on the shell at point \(B,\) which is at the same level as the center of the circle? (c) Suppose that the track had no friction and the shell was released from the same height \(h_{0}\) you found in part (a). Would it make a complete loop-theloop? How do you know? (d) In part (c), how hard does the track push on the shell at point \(A,\) the top of the circle? How hard did it push on the shell in part (a)?

A hollow, thin-walled sphere of mass \(12.0 \mathrm{~kg}\) and diameter \(48.0 \mathrm{~cm}\) is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by \(\theta(t)=A t^{2}+B t^{4},\) where \(A\) has numerical value 1.50 and \(B\) has numerical value \(1.10 .\) (a) What are the units of the constants \(A\) and \(B ?\) (b) At the time \(3.00 \mathrm{~s}\), find (i) the angular momentum of the sphere and (ii) the net torque on the sphere.

(a) Calculate the magnitude of the angular momentum of the earth in a circular orbit around the sun. Is it reasonable to model it as a particle? (b) Calculate the magnitude of the angular momentum of the earth due to its rotation around an axis through the north and south poles, modeling it as a uniform sphere. Consult Appendix \(\mathrm{E}\) and the astronomical data in Appendix F.

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