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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

A uniform rod of length \(L\) rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed \(v\) strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. (a) What is the final angular speed of the rod? (b) What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision?

A metal bar is in the \(x y\) -plane with one end of the bar at the origin. A force \(\overrightarrow{\boldsymbol{F}}=(7.00 \mathrm{~N}) \hat{\imath}+(-3.00 \mathrm{~N}) \hat{\jmath}\) is applied to the bar at the point \(x=3.00 \mathrm{~m}, y=4.00 \mathrm{~m}\). (a) In terms of unit vectors \(\hat{\imath}\) and \(\hat{\jmath},\) what is the position vector \(\vec{r}\) for the point where the force is applied? (b) What are the magnitude and direction of the torque with respect to the origin produced by \(\overrightarrow{\boldsymbol{F}} ?\)

(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.

While the turntable is being accelerated, the person suddenly extends her legs. What happens to the turntable? (a) It suddenly speeds up; (b) it rotates with constant speed; (c) its acceleration decreases; (d) it suddenly stops rotating.

A \(50.0 \mathrm{~kg}\) grindstone is a solid disk \(0.520 \mathrm{~m}\) in diameter. You press an ax down on the rim with a normal force of \(160 \mathrm{~N}\) (Fig. \(\mathbf{P} 10.58\) ). The coefficient of kinetic friction between the blade and the stone is \(0.60,\) and there is a constant friction torque of \(6.50 \mathrm{~N} \cdot \mathrm{m}\) between the axle of the stone and its bearings. (a) How much force must be applied tangentially at the end of a crank handle \(0.500 \mathrm{~m}\) long to bring the stone from rest to 120 rev \(/ \min\) in \(9.00 \mathrm{~s} ?\) (b) After the grindstone attains an angular speed of 120 rev \(/ \mathrm{min}\), what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 rev \(/ \min ?\) (c) How much time does it take the grindstone to come from \(120 \mathrm{rev} / \mathrm{min}\) to rest if it is acted on by the axle friction alone?

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