/*! 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 84 A child rolls a 0.600 -kg basket... [FREE SOLUTION] | 91Ó°ÊÓ

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A child rolls a 0.600 -kg basketball up a long ramp. The basketball can be considered a thin-walled, hollow sphere. When the child releases the basketball at the bottom of the ramp, it has a speed of 8.0 \(\mathrm{m} / \mathrm{s} .\) When the ball returns to her after rolling up the ramp and then rolling back down, it has a speed of 4.0 \(\mathrm{m} / \mathrm{s}\) . Assume the work done by friction on the basketball is the same when the ball moves up or down the ramp and that the basketball rolls without slipping. Find the maximum vertical height increase of the ball as it rolls up the ramp.

Short Answer

Expert verified
The maximum height increase of the ball is approximately 3.40 meters.

Step by step solution

01

Analyze the Energy at the Bottom of the Ramp

At the bottom of the ramp, the basketball has kinetic energy due to its translational motion and rotational motion. The translational kinetic energy is given by \( KE_{trans} = \frac{1}{2} m v^2 \), where \( m = 0.600 \text{ kg} \) and \( v = 8.0 \text{ m/s} \). The rotational kinetic energy for a hollow sphere is \( KE_{rot} = \frac{1}{2} I \omega^2 \), where \( I = \frac{2}{3}mr^2 \) for a thin-walled hollow sphere and \( \omega = \frac{v}{r} \). Thus, the total energy at the bottom is \( E_{bottom} = KE_{trans} + KE_{rot} \).
02

Calculate Initial Energies

Calculate the translational kinetic energy:\[ KE_{trans} = \frac{1}{2} \times 0.600 \times 8.0^2 = 19.2 \text{ J} \]Calculate the rotational kinetic energy:\[ KE_{rot} = \frac{1}{2} \times \frac{2}{3} \times 0.600 \times 8.0^2 = 12.8 \text{ J} \]The total energy at the bottom is:\[ E_{bottom} = 19.2 + 12.8 = 32.0 \text{ J} \]
03

Analyze the Energy when the Ball Returns

When the ball returns, it has a reduced speed, which means its energy is also reduced due to work done by friction. The translational kinetic energy is \( KE'_{trans} = \frac{1}{2} m v'^2 \) with \( v' = 4.0 \text{ m/s} \). The rotational kinetic energy is \( KE'_{rot} = \frac{1}{2} I (\omega')^2 \) with \( \omega' = \frac{v'}{r} \).
04

Calculate Return Energies

Calculate the translational kinetic energy when the ball returns:\[ KE'_{trans} = \frac{1}{2} \times 0.600 \times 4.0^2 = 4.8 \text{ J} \]Calculate the rotational kinetic energy when the ball returns:\[ KE'_{rot} = \frac{1}{2} \times \frac{2}{3} \times 0.600 \times 4.0^2 = 3.2 \text{ J} \]The total energy when the ball returns is:\[ E_{return} = 4.8 + 3.2 = 8.0 \text{ J} \]
05

Calculate Work Done by Friction

The work done by friction is the difference between the total mechanical energies:\[ W_{friction} = E_{bottom} - E_{return} = 32.0 - 8.0 = 24.0 \text{ J} \]
06

Calculate Potential Energy at Maximum Height

When the ball reaches its maximum height, it temporarily stops, so all the initial energy (less the work done by friction while going up the ramp) gets converted to potential energy:\[ PE_{max} = E_{bottom} - W_{friction \text{ (up)}} = 32.0 - \frac{1}{2} \times 24.0 = 32.0 - 12.0 = 20.0 \text{ J} \]
07

Determine Maximum Height

The potential energy at maximum height is given by \( PE_{max} = mgh \), where \( h \) is the maximum height. Using \( m = 0.600 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \), solve for \( h \):\[ h = \frac{PE_{max}}{mg} = \frac{20.0}{0.600 \times 9.8} \approx 3.40 \text{ m} \]

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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 that an object possesses due to its motion. To understand kinetic energy more thoroughly, consider it as the work needed to accelerate an object of a given mass from rest to a certain velocity. There are two types of kinetic energy to consider in this scenario:

  • Translational Kinetic Energy: As the basketball rolls up and down the ramp, its motion in a straight line profile is captured by translational kinetic energy. This can be calculated with the formula: \( KE_{trans} = \frac{1}{2} m v^2 \), where \(m\) is the mass and \(v\) is the velocity.
  • Rotational Kinetic Energy: Since the basketball also spins as it rolls, it has rotational kinetic energy. This form of energy depends on both the moment of inertia and the angular velocity, calculated as: \( KE_{rot} = \frac{1}{2} I \omega^2 \), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.
By understanding these principles, students can appreciate how kinetic energy manifests in different forms during motion.
Potential energy
Potential energy is the stored energy that an object possesses due to its position or state. In this specific problem, it is the gravitational potential energy that comes into play as the basketball moves up the ramp.

As the basketball ascends the ramp, kinetic energy is converted into potential energy until the ball reaches a momentary stop at its highest point. At this point, all of the converted kinetic energy is stored as potential energy, given by the formula: \( PE = mgh \), where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(h\) is the height.

This concept aids in understanding how energy shifts between kinetic and potential forms, maintaining the total mechanical energy in an isolated system, except where non-conservative forces like friction are involved.
Rotational motion
Rotational motion refers to the spinning or rotation of an object about an axis. For a basketball, both translational and rotational motions are crucial to understanding its dynamics on the ramp.

As a dynamic behavior, rotational motion can be characterized by the moment of inertia (\( I \)) and angular velocity (\( \omega \)). The moment of inertia depends on both the mass and shape of the object, and for a thin-walled hollow sphere like the basketball, it is calculated as \( I = \frac{2}{3}mr^2 \). The angular velocity, \( \omega = \frac{v}{r} \), describes how fast the ball spins.

The combination of these principles allows students to grasp why, while climbing a ramp, a rolling object behaves differently from one that is solely sliding.
Frictional work
Frictional work is the energy lost due to the force of friction acting over a distance. In this context, the basketball rolling up and down the ramp experiences friction, which does work to dissipate some of its mechanical energy.

This work can be computed as the difference between the total energy at the beginning and end of the journey, given by the equation: \( W_{friction} = E_{bottom} - E_{return} \). As the basketball rolls up the ramp, this frictional work is a non-conservative force that reduces the mechanical energy available, effectively working against the ball’s motion.

Highlighting this concept shows how important it is to consider environmental forces when solving problems in physics, as well as their effect on the energy of systems.
Thin-walled hollow sphere
A thin-walled hollow sphere is a specific type of geometric shape significant especially when discussing objects like basketballs. Such a shape influences various physical properties including inertia.

The moment of inertia for a thin-walled hollow sphere is defined as \( I = \frac{2}{3}mr^2 \). This specific shape means the mass is evenly distributed along the surface, affecting its rotational dynamics. Understanding this characteristic is crucial when applying formulas related to rotational motion and kinetic energy.

When students recognize the importance of an object's shape, they can better predict the outcome of mechanical problems, such as those involving different kinds of rolling motions.

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Most popular questions from this chapter

A large 16.0 -kg roll of paper with radius \(R=18.0 \mathrm{cm}\) rests against the wall and is held in place by a bracket attached to a rod through the center of the roll (Fig. P10.69). The rod turns without friction in the bracket, and the moment of inertia of the paper and rod about the axis is 0.260 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) The other end of the bracket is attached by a frictionless hinge to the wall such that the bracket makes an angle of \(30.0^{\circ}\) with the wall. The weight of the bracket is negligible. The coefficient of kinetic friction between the paper and the wall is \(\mu_{\mathrm{k}}=0.25 .\) A constant vertical force \(F=60.0 \mathrm{N}\) is applied to the paper, and the paper unrolls. (a) What is the magnitude of the force that the rod exerts on the paper as it unrolls? (b) What is the magnitude of the angular acceleration of the roll?

A uniform, \(0.0300-\mathrm{kg}\) rod of length 0.400 \(\mathrm{m}\) rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.0200 \(\mathrm{kg}\) , are mounted so that they can slide along the rod. They are initially held by catches at positions 0.0500 \(\mathrm{m}\) on each side of the center of the rod, and the system is rotating at 30.0 rev/min. With no other changes in the system, the catches are released, and the rings slide outward along the rod and fly off at the ends. (a) What is the angular speed of the system at the instant when the rings reach the ends of the rod? (b) What is the angular speed of the rod after the rings leave it?

Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of a neutron star is roughly \(10^{14}\) times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was \(7.0 \times 10^{5} \mathrm{km}\) (comparable to our sun); its final radius is 16 \(\mathrm{km} .\) If the original star rotated once in 30 days, find the angular speed of the neutron star.

The solid wood door of a gymnasium is 1.00 \(\mathrm{m}\) wide and 2.00 \(\mathrm{m}\) high, has total mass \(35.0 \mathrm{kg},\) and is hinged along one side. The door is open and at rest when a stray basketball hits the center of the door head-on, applying an average force of 1500 \(\mathrm{N}\) to the door for 8.00 \(\mathrm{ms} .\) Find the angular speed of the door after the impact. [Hint: Integrating Eq. \((10.29)\) yields \(\Delta L_{z}=\int_{t_{1}}^{t_{2}}\left(\Sigma \tau_{z}\right) d t=\left(\sum \tau_{z}\right)_{\mathrm{av}} \Delta t .\) The quantity \(\int_{t_{1}}^{t_{2}}\left(\Sigma \tau_{z}\right) d t\) is called the angular impulse.]

Atwood's Machine. Figure Pl0.67 illustrates an Atwood's machine. Find the linear accelerations of blocks \(A\) and \(B,\) the angular acceleration of the wheel \(C,\) and the tension in each side of the cord if there is no slipping between the cord and the surface of the wheel. Let the masses of blocks \(A\) and \(B\) be 4.00 \(\mathrm{kg}\) and 2.00 \(\mathrm{kg}\) , respectively, the moment of inertia of the wheel about its axis be \(0.300 \mathrm{kg} \cdot \mathrm{m}^{2},\) and the radius of the wheel be 0.120 \(\mathrm{m} .\)

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