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Chapter 11: Problem 5

From an initial height \(h\), a solid ball rolls smoothly down one side of a U-shaped ramp and then moves up the other side, which is frictionless. What maximum height does the ball reach?

Short Answer

Expert verified
The ball reaches the same height \(h' = h\) on the other side of the ramp.

Step by step solution

01

Understanding the Problem

A ball rolls down one side of a U-shaped ramp, converting its potential energy into kinetic energy. As it rolls up the other side, this kinetic energy is converted back into potential energy. Our goal is to determine the maximum height the ball reaches on the frictionless side of the ramp.
02

Analyzing Energy Conservation

Since there is no friction on the ramp, mechanical energy is conserved. Initially, the ball has potential energy and negligible kinetic energy at height \( h \): \( PE_{initial} = mgh \), where \( m \) is the mass of the ball and \( g \) is the acceleration due to gravity. At the maximum height on the other side, the ball will again have potential energy: \( PE_{final} = mg \cdot h' \), where \( h' \) is the final height.
03

Setting Up the Energy Conservation Equation

Since energy is conserved, we have \( PE_{initial} = PE_{final} \). Therefore, \( mgh = mg \cdot h' \). We can simplify this equation by canceling \( m \) and \( g \) on both sides, leading to \( h = h' \).
04

Solving for Maximum Height

From the energy conservation equation \( h = h' \), we conclude that the maximum height \( h' \) on the frictionless side is equal to the initial height \( h \). Thus, the ball reaches the same height on the other side of the ramp.

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Key Concepts

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

Potential Energy
Potential energy refers to the stored energy an object possesses due to its position or state. For example, when a ball is held at a certain height, it has potential energy due to the gravitational force acting on it. The higher the ball is, the more potential energy it has. This energy is expressed by the formula:
  • \( PE = mgh \)
where:
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \) on Earth),
  • \( h \) is the height above a reference point.
As the ball starts at height \( h \) on the ramp, this is its initial potential energy. When the ball descends, it loses potential energy but not permanently. The potential energy is transferred into kinetic energy, enabling the ball to gain speed as it rolls down.Understanding potential energy is crucial as it signifies the capability of an object to perform work due to its positioning in a gravitational field.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. When the ball rolls down the U-shaped ramp, its potential energy is converted into kinetic energy, causing it to move faster. The kinetic energy of a moving object is given by:
  • \( KE = \frac{1}{2}mv^2 \)
where:
  • \( m \) is the mass of the object,
  • \( v \) is the velocity of the object.
As the ball reaches the lowest point of the ramp, all initial potential energy is now kinetic energy, assuming there is no energy lost to friction. This maximum kinetic energy depends solely on the initial potential energy it had at the height \( h \). Then, as the ball ascends the frictionless side, kinetic energy is reconverted back to potential energy.
Frictionless Motion
Frictionless motion is an idealized concept in physics. It assumes that an object moves without any resistance from surfaces it comes into contact with. In the context of the U-shaped ramp, the absence of friction means no energy is lost between the ball and the ramp surface.The lack of friction plays a crucial role in the conservation of energy. Since the ramp is frictionless, the ball retains its mechanical energy between potential and kinetic forms throughout the motion. No work is done against friction, allowing the ball's energy to solely transition between these forms without loss.In this exercise, frictionless motion ensures that as the ball rolls up the other side of the ramp, it reaches the same height \( h \) as its initial height. This is because all energy converted to kinetic is eventually converted back to potential, illustrating the principle that mechanical energy remains conserved in an ideal system devoid of friction.

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Most popular questions from this chapter

A top spins at \(25 \mathrm{rev} / \mathrm{s}\) about an axis that makes an angle of \(30^{\circ}\) with the vertical. The mass of the top is \(0.50 \mathrm{~kg}\), its rotational inertia about its central axis is \(5.0 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2}\), and its center of mass is \(4.2 \mathrm{~cm}\) from the pivot point. If the spin is clockwise from an overhead view, what are the (a) precession rate and (b) direction of the precession as viewed from overhead?

A hollow sphere of radius \(0.15 \mathrm{~m}\), with rotational inertia \(I=0.048 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about a line through its center of mass, rolls without slipping up a surface inclined at \(30^{\circ}\) to the horizontal. At a certain initial po-sition, the sphere's total kinetic energy is \(20 \mathrm{~J}\). (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved \(1.0 \mathrm{~m}\) up the incline from its initial position, what are (c) it total kinetic energy and (d) the speed of its center of mass?

A uniform disk of mass \(10 m\) and radius \(3.0 r\) can rotate freely about its fixed center like a merry-go-round. A smaller uniform disk of mass \(m\) and radius \(r\) lies on top of the larger disk, concentric with it. Initially the two disks rotate together with an angular velocity of \(20 \mathrm{rad} / \mathrm{s}\). Then a slight disturbance causes the smaller disk to slide outward across the larger disk, until the outer edge of the smaller disk catches on the outer edge of the larger disk. Afterward, the two disks again rotate together (without further sliding). (a) What then is their angular velocity about the center of the larger disk? (b) What is the ratio \(K / K_{0}\) of the new kinetic energy of the two-disk system to the system's initial kinetic energy?

In unit vector notation, what is the torque about the origin on a particle located at coordinates \((0,-4.0 \mathrm{~m}, 3.0 \mathrm{~m})\) if that torque is due to (a) force \(\vec{F}_{1}\) with components \(F_{1 x}=2.0 \mathrm{~N}, F_{1 y}=F_{1 z}=0\), and (b) force \(\vec{F}_{2}\) with components \(F_{2 x}=0, F_{2 y}=2.0 \mathrm{~N}, F_{2 z}=4.0 \mathrm{~N} ?\)

Figure 11-45 shows an overhead view of a ring that can rotate about its center like a merry-goround. Its outer radius \(R_{2}\) is \(0.800 \mathrm{~m}\), its inner radius \(R_{1}\) is \(R_{2} / 2.00\), its mass \(M\) is \(6.00 \mathrm{~kg}\), and the mass of the crossbars at its center is negligible. It initially rotates at an angular speed of \(8.00 \mathrm{rad} / \mathrm{s}\) with a cat of mass \(m=M / 4.00\) on its outer edge, at radius \(R_{2}\). By how much does the cat increase the kinetic energy of the cat-ring system if the cat crawls to the inner edge, at radius \(R_{1}\) ?
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