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Chapter 8: Problem 56

A uniform sphere and a uniform cylinder with the same mass and radius roll at the same velocity side by side on a level surface without slipping. If the sphere and the cylinder approach an inclined plane and roll up it without slipping, will they be at the same height on the plane when they come to a stop? If not, what will be the percentage difference of the heights?

Short Answer

Expert verified
No, the sphere and cylinder won't reach the same height; the cylinder will reach 6.25% less height than the sphere.

Step by step solution

01

Identifying Kinetic Energy Components

Both the sphere and the cylinder possess translational kinetic energy due to their linear motion and rotational kinetic energy due to their rotation. The translational kinetic energy (TKE) for each object is given by \( \frac{1}{2} m v^2 \), where \( m \) is mass and \( v \) is velocity. The rotational kinetic energy (RKE) for each is \( \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is angular velocity.
02

Moment of Inertia and Angular Velocity

The moment of inertia for a solid sphere is \( I = \frac{2}{5}mr^2 \), and for a solid cylinder, \( I = \frac{1}{2}mr^2 \). The relation between angular velocity \( \omega \) and linear velocity \( v \) is given as \( v = r\omega \), so \( \omega = \frac{v}{r} \) for both objects.
03

Calculating Total Energy for Each Object

For each object, the total kinetic energy (TKE + RKE) is calculated. For the sphere: \( \text{Total Energy} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{7}{10}mv^2 \). For the cylinder, \( \text{Total Energy} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2 = \frac{3}{4}mv^2 \).
04

Conservation of Energy as They Roll Up the Incline

As both roll up the incline without slipping, their kinetic energy converts entirely into potential energy (PE) at the maximum height. Potential energy at height \( h \) is \( mgh \), where \( g \) is the acceleration due to gravity. Set \( mgh \) equal to the total kinetic energy from Step 3 to find height \( h \).
05

Calculating Maximum Heights

For the sphere, \( mgh_s = \frac{7}{10}mv^2 \) gives \( h_s = \frac{7v^2}{10g} \). For the cylinder, \( mgh_c = \frac{3}{4}mv^2 \) gives \( h_c = \frac{3v^2}{4g} \).
06

Determine Percentage Difference in Heights

Calculate the percentage difference using the formula: \( \text{Percentage Difference} = \left( \frac{h_c - h_s}{h_s} \right) \times 100 \% \). Substitute and simplify: \( \text{Percentage Difference} = \left( \frac{\frac{3}{4} - \frac{7}{10}}{\frac{7}{10}} \right) \times 100 \% = 6.25 \% \).

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion, which can be both translational and rotational. For objects like a sphere or a cylinder moving along a surface, we need to account for both these forms.
  • Translational kinetic energy is given by the formula: \( \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the velocity of the object. This formula represents the energy due to the object moving along a path.
  • Rotational kinetic energy, on the other hand, depends on how the object spins. It is calculated as \( \frac{1}{2} I \omega^2 \), where \( I \) represents the moment of inertia, and \( \omega \) is the angular velocity.
By understanding both forms, we see that a rolling object combines these energies. This combination must be considered to understand how objects behave when they encounter frictionless inclined planes.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotation. It depends on the object's mass distribution relative to the axis of rotation.
  • A solid sphere, which is more compact, has a moment of inertia given by \( I = \frac{2}{5}mr^2 \).
  • A solid cylinder, which spreads its mass further from the axis, has a moment of inertia \( I = \frac{1}{2}mr^2 \).
The lower the moment of inertia, the easier it is to change the object's rotational state. Thus, understanding the moment of inertia helps in predicting how fast a body will rotate for a given amount of applied energy.
Conservation of Energy
The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In our scenario with the sphere and cylinder, kinetic energy is converted to potential energy as they roll up the incline.
  • When an object rolls up an incline, its kinetic energy decreases while its potential energy increases.
  • The total energy remains constant throughout the motion as long as no external forces (like friction or air resistance) do significant work on the objects.
By setting kinetic energy equal to potential energy at the maximum height, we determine how high each object can climb on the incline.
Rotational Motion
Rotational motion describes the movement of an object around a central axis. Understanding the connection between linear and rotational motion is crucial in physics.
  • The relationship between linear velocity \( v \) and angular velocity \( \omega \) is expressed as \( v = r \omega \). For rolling without slipping, this holds true for both spherical and cylindrical objects.
  • This equation allows us to express everything in terms of linear velocity, which makes the analysis of kinetic energy cleaner when considering objects that roll.
Understanding these principles helps in modeling systems that have parts moving along curved paths or rotating about pivots.
Percentage Difference Calculation
Percentage difference calculations help compare differences in quantities relative to their sizes, a useful tool in physics.
To find the percentage difference between two heights, we use the formula:\[ \text{Percentage Difference} = \left( \frac{h_c - h_s}{h_s} \right) \times 100 \% \]Here:
  • \( h_c \) is the height for the cylinder, and
  • \( h_s \) is the height for the sphere.
Substituting the values, the percentage difference is calculated as 6.25%. This tells us how much more, in percentage terms, one object climbs in comparison to another.

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Most popular questions from this chapter

A constant torque of \(10 \mathrm{~m} \cdot \mathrm{N}\) is applied to the rim of a 10-kg uniform disk of radius \(0.20 \mathrm{~m}\). What is the angular speed of the disk about an axis through its center after it rotates 2.0 revolutions from rest?

(a) How many uniform, identical textbooks of width \(25.0 \mathrm{~cm}\) can be stacked on top of each other on a level surface without the stack falling over if each successive book is displaced \(3.00 \mathrm{~cm}\) in width relative to the book below it? (b) If the books are \(5.00 \mathrm{~cm}\) thick, what will be the height of the center of mass of the stack above the level surface?

While repairing his bicycle, a student turns it upside down and sets the front wheel spinning at 2.00 rev \(/ \mathrm{s}\). Assume the wheel has a mass of \(3.25 \mathrm{~kg}\) and all of the mass is located on the rim, which has a radius of \(41.0 \mathrm{~cm}\). To slow the wheel, he places his hand on the tire, thereby exerting a tangential force of friction on the wheel. It takes \(3.50 \mathrm{~s}\) to come to rest. Use the change in angular momentum to determine the force he exerts on the wheel. Assume the frictional force of the axle is negligible.

IE A small heavy object of mass \(m\) is attached to a thin string to make a simple pendulum whose length is \(L\) When the object is pulled aside by a horizontal force \(F\) it is in static equilibrium and the string makes a constant angle \(\theta\) from the vertical. (a) The tension in the string should be (1) the same as, (2) greater than, or (3) less than the object's weight, \(m g .\) (b) Use the force condition for static equilibrium (along with a free-body diagram of the object) to prove that the string tension is \(T=\frac{m g}{\cos \theta}>m g\). Use the same procedure to show that \(F=m g \tan \theta\). (c) Prove the same result for \(F\) as in part (b) using the torque condition, summing the torques about the string's tied end. Explain why you cannot use this method to determine the string tension.

A flat cylindrical grinding wheel is spinning at 2000 rpm (clockwise when viewed head-on) when its power is suddenly turned off. Normally, if left alone, it takes 45.0 s to coast to rest. Assume the grinder has a moment of inertia of \(2.43 \mathrm{~kg} \cdot \mathrm{m}^{2}\). (a) Determine its angular acceleration during this process. (b) Determine the tangential acceleration of a point on the grinding wheel if the wheel is \(7.5 \mathrm{~cm}\) in diameter. (c) The slowing down is caused by a frictional torque on the axle of the wheel. The axle is \(1.00 \mathrm{~cm}\) in diameter. Determine the frictional force on the axle. (d) How much work was done by friction on the system?
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