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A solid disk is rolling without slipping on a level surface at a constant speed of 2.50 \(\mathrm{m} / \mathrm{s}\) . (a) If the disk rolls up a \(30.0^{\circ}\) ramp, how far along the ramp will it move before it stops? (b) Explain why your answer in part (a) does not depend on either the mass or the radius of the disk.

Short Answer

Expert verified
The disk travels 0.956 m up the ramp; mass and radius don't affect the result.

Step by step solution

01

Identify Known Values and Concepts

The disk is rolling on a level surface before moving up a 30-degree ramp. We have the initial speed of 2.50 m/s. We need to find out how far it will travel up the ramp before stopping. The main concepts involved here are the conservation of energy and the equations of motion with gravitational potential energy change.
02

Use Conservation of Energy Principle

The disk's initial kinetic energy is converted into gravitational potential energy as it moves up the ramp. The equation illustrating this conversion is \( \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh \), where \( I \) is the moment of inertia for a solid disk \( I = \frac{1}{2}mr^2 \) and \( \omega = \frac{v}{r} \). Thus, the energy equation becomes \( m \left(\frac{3}{4}v^2\right) = mgh \).
03

Calculate Potential Height

Simplify the energy equation by cancelling \( m \) and solving for \( h \):\[ h = \frac{3}{4}\frac{v^2}{g} \] with \( v = 2.50 \) m/s and \( g = 9.81 \) m/s². Substitute these values to find \( h \).\[ h = \frac{3}{4}\frac{(2.50)^2}{9.81} \approx 0.478 \text{ m} \].
04

Relate Height to Ramp Distance

Since the disk moves along a ramp inclined at 30 degrees, use the relationship \( h = d \sin(30^{\circ}) \) to find the distance \( d \):\[ d = \frac{h}{\sin(30^{\circ})} \]With \( h = 0.478 \) m, \( d \approx 0.956 \text{ m} \).
05

Analyze Mass and Radius Dependence

Notice that in the energy conservation equation, both mass \( m \) and radius \( r \) canceled out when deriving \( h \). Thus, neither variable affects the final distance \( d \) traveled up the ramp, demonstrating it is independent of both mass and radius.

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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 a fundamental concept in physics that describes the energy an object possesses due to its motion. For an object like a solid disk rolling along a level surface, this energy is distributed between translational motion and rotational motion.

Translational kinetic energy is given by the formula \( \frac{1}{2}mv^2 \), where \( m \) stands for mass and \( v \) represents velocity. For rotational motion, the kinetic energy depends on the moment of inertia \( I \) and angular velocity \( \omega \), expressed as \( \frac{1}{2}I\omega^2 \).

In the exercise, as the disk transitions from rolling on a flat surface to climbing a ramp, its kinetic energy is gradually converted into gravitational potential energy. Understanding this transformation is key to analyzing the distance it will travel up the incline until all kinetic energy has been transferred, causing the disk to come to a stop.
Gravitational Potential Energy
Gravitational potential energy is the energy stored in an object as a result of its position in a gravitational field. It is a crucial concept when investigating how objects move in vertical and inclined situations. The expression for this type of energy is \( mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) on Earth), and \( h \) is the height above a reference point.

In our example of a disk rolling up a ramp, as the disk rises, its gravitational potential energy increases while its kinetic energy decreases proportionately. This exchange follows the conservation of energy principle, which states that energy cannot be created or destroyed, only transformed. Therefore, by calculating the height \( h \), we can determine how far the disk will move up the ramp before completely stopping.
Moment of Inertia
The moment of inertia describes how mass is distributed in a rotating object and determines how easily the object can change its rotational motion. It is similar to mass in linear motion and is fundamental in equations involving rotational dynamics. The moment of inertia \( I \) for a solid disk is given by \( \frac{1}{2}mr^2 \), where \( m \) is the mass of the disk and \( r \) is its radius.

In this problem, the moment of inertia is crucial in determining the rotational kinetic energy portion of the disk's total kinetic energy as it rolls. The rolling motion combines both translational and rotational kinetic energy. Interestingly, when calculating the final potential height or distance traveled up a ramp, mass and radius cancel each other out. This highlights a key benefit of the conservation of energy method, showing how complex factors can simplify under certain conditions.
Equations of Motion
Equations of motion serve as foundational tools in solving problems related to an object's movement. They describe the mathematical relationship between an object's speed, acceleration, position, and time. Specifically, concerning this problem, the relationships tie into the disk's transition from movement on a flat surface to an inclined one.

For the exercise at hand, we use the conservation of energy principle to connect the object's kinetic and gravitational potential energies. The energy conversion equation \( \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh \) encapsulates the transition. Most notably, the factor of mass \( m \) cancels out during calculations, emphasising that neither the mass nor the radius significantly affects the final solution.

These insights underline that equations of motion and energy conservation offer a pathway to simplify and solve seemingly complex motion problems, especially in scenarios where various physical properties might initially appear influential.

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Most popular questions from this chapter

Two metal disks, one with radius \(R_{1}=2.50 \mathrm{cm}\) and mass \(M_{1}=0.80 \mathrm{kg}\) and the other with radius \(R_{2}=5.00 \mathrm{cm}\) and mass \(M_{2}=1.60 \mathrm{kg}\) , are welded together and mounted on a frictionless axis through their common center, as in Problem \(9.89 .\) (a) A light string is wrapped around the edge of the smaller disk, and a 1.50 \(\mathrm{kg}\) block is suspended from the free end of the string. What is the magnitude of the downward acceleration of the block after it is released? (b) Repeat the calculation of part (a), this time with the string wrapped around the edge of the larger disk. In which case is the acceleration of the block greater? Does your answer make sense?

A thin, horizontal rod with length \(l\) and mass \(M\) pivots about a vertical axis at one end. A force with constant magnitude \(F\) is applied to the other end, causing the rod to rotate in a horizontal plane. The force is maintained perpendicular to the rod and to the axis of rotation. Calculate the magnitude of the angular acceleration of the rod.

A uniform hollow disk has two pieces of thin light wire wrapped around its outer rim and is supported from the ceiling (Fig. 10.51 ). Suddenly one of the wires breaks, and the remaining wire does not slip as the disk rolls down. Use energy conservation to find the speed of the center of this disk after it has fallen a distance of 1.20 \(\mathrm{m} .\)

Neutron Star Glitches. Occasionally, a rotating neutron star (see Exercise 10.39 ) undergoes a sudden and unexpected speedup called a glitch. One explanation is that a ghitch occurs when the crust of the neutron star settles slightly, decreasing the moment of inertia about the rotation axis. A neutron star with angular speed \(\omega_{0}=70.4\) rad/s underwent such a glitch in October 1975 that increased its angular speed to \(\omega=\omega_{0}+\Delta \omega,\) where \(\Delta \omega / \omega_{0}=2.01 \times 10^{-6} .\) If the radius of the neutron star before the glitch was \(11 \mathrm{km},\) by how much did its radius decrease in the star-quake? Assume that the neutron star is a uniform sphere.

A cord is wrapped around the rim of a solid uniform wheel 0.250 \(\mathrm{m}\) in radius and of mass 9.20 \(\mathrm{kg}\) . A steady horizontal pull of 40.0 \(\mathrm{N}\) to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) Compute the angular acceleration of the wheel and the acceleration of the part of the cord that has already been pulled off the wheel. (b) Find the magnitude and direction of the force that the axe exerts on the wheel. (c) Which of the answers in parts (a) and (b) would change if the pull were upward instead of horizontal?

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