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A solid disk is rolling without slipping on a level surface at a constant speed of 3.60 \(\mathrm{m} / \mathrm{s}\) (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 moves 1.986 m up the ramp; mass and radius do not affect the result due to energy conservation.

Step by step solution

01

Understand the Problem

The problem involves a solid disk rolling without slipping on a level surface and then moving up a ramp. We need to determine how far it travels up a 30-degree incline before stopping. This is a classic physics problem involving conservation of energy.
02

Identify Relevant Formulas

Since the disk rolls without slipping, both translational and rotational energy must be considered. The initial total mechanical energy of the disk is given by the sum of its translational kinetic energy and rotational kinetic energy: \[E_i = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\] where \(I = \frac{1}{2}mr^2\) (moment of inertia for a solid disk) and \(\omega = \frac{v}{r}\) (relation between linear and angular velocity).
03

Setup the Conservation of Energy Equation

The final mechanical energy of the disk when it stops is all potential energy. Setting initial and final energies equal under conservation, \[\frac{1}{2}mv^2 + \frac{1}{4}mv^2 = mgh\] Simplifying gives \[\frac{3}{4}mv^2 = mgh\], where \(h\) is the height the disk travels up on the ramp.
04

Solve for Height

Solve for \(h\) using: \[h = \frac{3}{4}\frac{v^2}{g}\] With \(v = 3.60 \, \text{m/s}\) and \(g = 9.81 \, \text{m/s}^2\), we find \[h = \frac{3}{4} \frac{(3.60)^2}{9.81} \approx 0.993 \, \text{m}\]
05

Calculate Distance Up the Ramp

The distance up the ramp, \(d\), is related to \(h\) by \[h = d \sin(30^{\circ})\]. Therefore, \[d = \frac{h}{\sin(30^{\circ})} = \frac{0.993}{0.5} = 1.986 \, \text{m}\].
06

Explain Independence from Mass and Radius

The result is independent of both mass \(m\) and radius \(r\) because these variables cancel out in the conservation of energy equation. Specifically, they are present in each type of energy (translational and rotational kinetic energy and potential energy) but do not appear in the final expression for \(h\) and \(d\).

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Key Concepts

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

Rotational Kinetic Energy
When a solid disk is in motion, its rotational kinetic energy is an important component to consider. This energy is associated with the disk's spin around its axis. Mathematically, it is expressed as \( KE_{rot} = \frac{1}{2} I \omega^2 \), where \( \omega \) is the angular velocity. The formula highlights how both the distribution of mass in the disk and its rotational speed contribute to the energy.
- **Rotational Kinetic Energy** is crucial in objects rolling or spinning.
- Unlike regular kinetic energy, which depends on the velocity of an object moving in a straight line, rotational kinetic energy depends on how the mass is distributed relative to the axis of rotation.
For a solid disk, the moment of inertia \( I \) is \( \frac{1}{2}mr^2 \), demonstrating how both mass \( m \) and radius \( r \) impact the rotational kinetic energy but remarkably cancel out in specific scenarios, such as when energy conservation principles are applied, permitting one to determine factors like height climbed or distance traveled without needing those specific values.
Moment of Inertia
The moment of inertia, often referred to as the rotational equivalent of mass, is crucial when analyzing rotational motion. For a solid disk, it is given by \( I = \frac{1}{2}mr^2 \).
- **Moment of Inertia** combines both mass and the radius squared, illustrating how distribution affects rotation.
- Larger radii or increased mass lead to greater resistance to changes in rotational motion.
This physical quantity doesn't just influence an object’s propensity to rotate but also impacts how it rolls without slipping. In problems like the inclined plane exercise, the moment of inertia, despite its initial critical role, elegantly cancels out along with mass in energy conservation formulas. It shows the beauty and simplicity hidden in physics equations, despite the complex sounding nature of these concepts.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, which in this problem is at \(30^{\circ}\). It allows us to explore how gravitational potential energy changes when objects move upwards along the surface.
- **Inclined planes** help in converting kinetic energy to potential energy as an object climbs up.
- Often used in physics to simplify gravitational force calculations, when an object moves up, its height h is determined by distance times the sine of the angle it makes with the horizontal plane.
In the given exercise, the inclined plane's angle and distance enable the determination of energy conversion. By knowing the angle and using trigonometric relationships, we can find how far the disk moves upwards before stopping, which illustrates how mechanical energy is conserved, knowing it originally came from the disk's speed.
Translational Kinetic Energy
Translational kinetic energy represents the energy of motion along a path or a straight line, distinct from rotational. It is calculated using \( KE_{trans} = \frac{1}{2}mv^2 \), where \( m \) is the mass, and \( v \) is the velocity.
- **Translational Kinetic Energy** applies to the center of mass moving in straight lines.
- Important in determining how objects convert velocity into distance traveled under forces such as friction or gravity, especially on inclined planes.
In the context of our problem, the disk's initial speed allowed us to compute its translational kinetic energy. Like rotational kinetic energy, it plays a vital part in energy conservation equations. Together with rotational kinetic energy, it contributes to the total mechanical energy that transforms into potential energy as the disk climbs the ramp. By conserving energy, calculations show how translational movements transition into potential gains in height and distance up the inclined plane, without the influence of mass or radius in the outcome.

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Most popular questions from this chapter

A demonstration gyroscope wheel is constructed by removing the tire from a bicycle wheel 0.650 \(\mathrm{m}\) in diameter, wrapping lead wire around the rim, and taping it in place. The shaft projects 0.200 \(\mathrm{m}\) at each side of the wheel, and a woman holds the ends of the shaft in her hands. The mass of the system is 8.00 \(\mathrm{kg}\) ; its entire mass may be assumed to be located at its rim. The shaft is horizontal, and the wheel is spinning about the shaft at 5.00 \(\mathrm{rev} / \mathrm{s}\) . Find the magnitude and direction of the force each hand exerts on the shaft (a) when the shaft is at rest; (b) when the shaft is rotating in a horizontal plane about its center at \(0.050 \mathrm{rev} / \mathrm{s} ;\) (c) when the shaft is rotating in a horizontal plane about its center at 0.300 \(\mathrm{rev} / \mathrm{s}\) . (d) At what rate must the shaft rotate in order that it may be supported at one end only?

A gyroscope is precessing about a vertical axis. Describe what happens to the precession angular speed if the following changes in the variables are made, with all other variables remaining the same: (a) the angular speed of the spinning flywheel is doubled; (b) the total weight is doubled; (c) the moment of inertia about the axis of the spinning flywheel is doubled; (d) the distance from the pivot to the center of gravity is doubled. (e) What happens if all four of the variables in parts (a) through (d) are doubled?

A block with mass \(m=5.00 \mathrm{kg}\) slides down a a surface inclined \(36.9^{\circ}\) to the horizontal (Fig. P10.70). The coefficient of kinetic friction is \(0.25 .\) A string attached to the block is wrapped around a flywheel on a fixed axis at \(O .\) The flywheel has mass 25.0 \(\mathrm{kg}\) and moment of inertia 0.500 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.200 \(\mathrm{m}\) from that axis. (a) What is the acceleration of the block down the plane? (b) What is the tension in the string?

A 42.0 -cm-diameter wheel, consisting of a rim and six spokes, is constructed from a thin, rigid plastic material having a linear mass density of 25.0 \(\mathrm{g} / \mathrm{cm} .\) This wheel is released from rest at the top of a hill 58.0 \(\mathrm{m}\) high. (a) How fast is it rolling when it reaches the bottom of the hill? (b) How would your answer change if the linear mass density and the diameter of the wheel were each doubled?

A \(50.0\)-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. P10.57). The coefficient of kinetion 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 \(\mathrm{rev} / \mathrm{min}\) in 9.00 \(\mathrm{s} ?\) (b) After the grindstone attains an angular speed of 120 \(\mathrm{rev} / \mathrm{min}\) , what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 \(\mathrm{rev} / \mathrm{min}\) ? (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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