/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 78 Two identical wheels are moving ... [FREE SOLUTION] | 91Ó°ÊÓ

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Two identical wheels are moving on horizontal surfaces. The center of mass of each has the same linear speed. However, one wheel is rolling. while the other is sliding on a frictionless surface without rolling. Each wheel then encounters an incline plane. One continues to roll up the incline, while the other continues to slide up. Eventually they come to a momentary halt, because the gravitational force slows them down. Each wheel is a disk of mass \(2.0 \mathrm{kg} .\) On the horizontal surfaces the center of mass of each wheel moves with a linear speed of \(6.0 \mathrm{m} / \mathrm{s}\). (a) What is the total kinctic energy of each wheel? (b) Determine the maximum height reached by each wheel as it moves up the incline.

Short Answer

Expert verified
(a) Sliding KE: 36 J, Rolling KE: 54 J. (b) Sliding height: 1.83 m, Rolling height: 2.75 m.

Step by step solution

01

Calculate the Kinetic Energy of the Sliding Wheel

The sliding wheel only has translational kinetic energy because it slides without rolling. The translational kinetic energy \( KE_t \) is given by:\[KE_t = \frac{1}{2} m v^2\]Substitute \( m = 2.0 \, \text{kg} \) and \( v = 6.0 \, \text{m/s} \) into the formula:\[KE_t = \frac{1}{2} \times 2.0 \, \text{kg} \times (6.0 \, \text{m/s})^2 = 36 \, \text{J}\]
02

Calculate the Kinetic Energy of the Rolling Wheel

The rolling wheel has both translational and rotational kinetic energy. The total kinetic energy \( KE_{total} \) is given by:\[KE_{total} = KE_t + KE_r = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2\]For a disk, the moment of inertia \( I \) is \( \frac{1}{2} m r^2 \) and \( \omega = \frac{v}{r} \). Substitute these into the formula:\[KE_r = \frac{1}{2} \left( \frac{1}{2} m r^2 \right) \left( \frac{v}{r} \right)^2 = \frac{1}{4} m v^2\]Therefore, the total kinetic energy is:\[KE_{total} = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2\]Substitute the values \( m = 2.0 \, \text{kg} \) and \( v = 6.0 \, \text{m/s} \):\[KE_{total} = \frac{3}{4} \times 2.0 \, \text{kg} \times (6.0 \, \text{m/s})^2 = 54 \, \text{J}\]
03

Calculate Maximum Height for Sliding Wheel

The maximum height \( h \) can be found using the conservation of energy:\[KE = mgh\]For the sliding wheel:\[36 \, \text{J} = 2.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times h\]Solve for \( h \):\[h = \frac{36}{19.62} \approx 1.83 \, \text{m}\]
04

Calculate Maximum Height for Rolling Wheel

Use the conservation of energy again, this time considering the total initial kinetic energy:\[54 \, \text{J} = 2.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times h\]Solve for \( h \):\[h = \frac{54}{19.62} \approx 2.75 \, \text{m}\]

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Key Concepts

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

Rolling Motion
Rolling motion is a fascinating combination of both translational and rotational motion. When an object like a wheel rolls, every point on its surface traces a circular path. This means the wheel not only moves forward but also spins.
This dual movement involves two types of kinetic energy:
  • Translational Kinetic Energy: Comes from the motion of the center of mass of the wheel moving linearly forward.
  • Rotational Kinetic Energy: Comes from the wheel spinning around its axis.
Together, these energies explain why a rolling object behaves differently from one that slides without rolling.
Linear Speed
Linear speed refers to how fast a point on an object's surface moves in a straight line. For both rolling and sliding wheels, this refers to the speed at which the center of mass is moving. In our example:
  • The initial linear speed of each wheel's center of mass is given as 6.0 m/s.
  • It’s crucial for determining the translational kinetic energy of the wheel.
While both wheels have the same linear speed, their energies differ due to one having rotational energy as well.
Moment of Inertia
Moment of inertia is like the rotational equivalent of mass. It measures how much torque is needed for an object to rotate. A disk, such as our wheel, has a specific moment of inertia given by \( I = \frac{1}{2} m r^2 \).
This formula shows that:
  • Mass (m): More mass means more resistance to rotation.
  • Radius (r): Larger radius increases the moment of inertia.
For the rolling wheel, its moment of inertia is crucial in calculating rotational kinetic energy, which adds to its total kinetic energy.
Conservation of Energy
The principle of conservation of energy tells us that energy can neither be created nor destroyed, only transformed from one type to another. In our scenario:
  • Initial Kinetic Energy: Each wheel starts with a certain amount of kinetic energy (36 J for sliding and 54 J for rolling).
  • Potential Energy at Height: As the wheels rise on the incline, their kinetic energy transforms into potential energy.
The maximum height reached by each wheel is determined by equating their initial kinetic energy to their gravitational potential energy at the peak of their climb: \( mgh \), where \( g \approx 9.81 \text{ m/s}^2 \). This energy transformation illustrates how different motions result in different heights.

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Most popular questions from this chapter

A block (mass \(=2.0 \mathrm{kg}\) ) is hanging from a massless cord that is wrapped around a pulley (moment of inertia \(=1.1 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2}\) ), as the drawing shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of \(0.040 \mathrm{m}\) during the block's descent. Find the angular acceleration of the pulley and the tension in the cord.

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has a length of \(0.80 \mathrm{m}\) and is uniform. It is hanging vertically straight downward. The end of the rod nearest the floor is given a linear speed \(v_{0,}\) so that the rod begins to rotate upward about the pivot. What must be the value of \(v_{0}\) such that the rod comes to a momentary halt in a straight-up orientation, exactly opposite to its initial orientation?

As seen from above, a playground carousel is rotating counterclockwise about its center on frictionless bearings. A person standing still on the ground grabs onto one of the bars on the carousel very close to its outer edge and climbs aboard. Thus, this person begins with an angular speed of zero and ends up with a nonzero angular speed, which means that he underwent a counterclockwise angular acceleration. The carousel has a radius of \(1.50 \mathrm{m},\) an initial angular speed of \(3.14 \mathrm{rad} / \mathrm{s},\) and a moment of inertia of \(125 \mathrm{kg} \cdot \mathrm{m}^{2} .\) The mass of the person is \(40.0 \mathrm{kg} .\) Find the final angular speed of the carousel after the person climbs aboard.

The wheel of a car has a radius of 0.350 m. The engine of the car applies a torque of 295 N ? m to this wheel, which does not slip against the road surface. Since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a counter- torque. Moreover, the car has a constant velocity, so this counter torque balances the applied torque. What is the magnitude of the static frictional force?

A bowling ball encounters a \(0.760-\mathrm{m}\) vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is \(3.50 \mathrm{m} / \mathrm{s}\) at the bottom of the rise. Find the translational speed at the top.

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