/*! 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 82 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 bas- ketball 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 ball reaches a maximum height of approximately 1.96 meters.

Step by step solution

01

Understand the Problem

We need to determine the maximum height reached by the basketball while rolling up a ramp. The basketball initially has a speed of 8.0 m/s and when returning has a speed of 4.0 m/s. We will apply principles of energy conservation and account for energy lost due to friction.
02

Apply Conservation of Energy

The initial mechanical energy (sum of kinetic and potential energies) when the child releases the ball is only kinetic: \[ K_i = \frac{1}{2}mv_i^2 + \frac{1}{2}I\omega_i^2 \] where \( v_i = 8.0 \: \mathrm{m/s} \), \( m = 0.600 \: \mathrm{kg} \), and \( I = \frac{2}{3}mr^2 \) for a thin-walled hollow sphere. The kinetic energy is both translational and rotational.
03

Initial Kinetic Energy Calculation

Calculate the translational kinetic energy: \[ \frac{1}{2}mv_i^2 = \frac{1}{2} \times 0.600 \times (8.0)^2 = 19.2 \: \mathrm{J} \]Calculate the rotational kinetic energy: Since \( I = \frac{2}{3}mr^2 \) and \( \omega_i = \frac{v_i}{r} \), \[ \frac{1}{2}I\omega_i^2 = \frac{1}{2} \left( \frac{2}{3} \times 0.600 \times r^2 \right) \left( \frac{8.0}{r} \right)^2 = 6.4 \: \mathrm{J} \]Total initial kinetic energy \( E_i = 19.2 + 6.4 = 25.6 \: \mathrm{J} \).
04

Determine Final Energy

When the ball returns, its speed is 4.0 m/s. Calculate the final mechanical energy:\[ K_f = \frac{1}{2}mv_f^2 + \frac{1}{2}I\omega_f^2 \] where \( v_f = 4.0 \: \mathrm{m/s} \), thus\[ \frac{1}{2}mv_f^2 = \frac{1}{2} \times 0.600 \times (4.0)^2 = 4.8 \: \mathrm{J} \]\[ \frac{1}{2}I\omega_f^2 = \frac{1}{2} \left( \frac{2}{3} \times 0.600 \times r^2 \right) \left( \frac{4.0}{r} \right)^2 = 1.6 \: \mathrm{J} \]Total final kinetic energy \( E_f = 4.8 + 1.6 = 6.4 \: \mathrm{J} \).
05

Analyze Energy Loss

The total mechanical energy loss from point A (start of ramp) to point B (end of ramp) is due to gravitational work and friction. The energy lost \( \Delta E = E_i - E_f = 25.6 - 6.4 = 19.2 \: \mathrm{J} \).This energy loss is equal to the potential energy at the maximum height plus work done by friction.
06

Calculate Maximum Height

The energy loss \( \Delta E \) is split between the change in gravitational potential energy and the work done against friction for both up and down journeys:\[ mgh + 2W_f = \Delta E \]Ignoring friction, solving, \( mgh = g \times 2 \times 9.8 \times 9.8 + 2W \), \[ h = \frac{19.2}{9.8}\] Thus, maximum vertical height increase \( h = \frac{19.2}{9.8} \approx 1.96 \:\mathrm{m} \).The friction is symmetric in our equation.

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Key Concepts

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

Conservation of Energy
In rolling motion, like the scenario with the basketball, conservation of energy is an essential principle. It states that the total mechanical energy of a closed system remains constant, as long as only conservative forces, such as gravity, are acting on it. In the problem, the basketball is rolling up and down the ramp, transforming between kinetic and potential energy. Because friction is considered, energy is not perfectly conserved and some of it is transformed into other forms, like heat. Understanding this principle helps us analyze how energy transitions within the system.
The core idea here is that the energy the basketball starts with is changed into different types as it moves. This energy can be summed up as:
  • Initial kinetic energy when the ball is moving fast at the start.
  • Potential energy when it reaches its highest point on the ramp.
By using the conservation of energy, we can figure out the maximum height the basketball reaches by knowing how much energy it had at the start and the end.
Kinetic Energy
Kinetic energy is the energy an object has due to its motion. In the context of the rolling basketball, kinetic energy is present in two unique forms:
First, there is the translational kinetic energy, which comes from the entire basketball moving up the ramp. This energy depends on the mass of the basketball and its velocity squared, expressed via the formula: \[ \frac{1}{2}mv^2 \]Secondly, there's rotational kinetic energy because the ball is also spinning as it rolls. This rotational aspect calls for understanding the ball's moment of inertia, which for a thin-walled hollow sphere is represented by \( I = \frac{2}{3}mr^2 \). The rotational kinetic energy can be calculated using: \[ \frac{1}{2}I\omega^2 \]
These two components together make up the initial kinetic energy, reflecting the full motion abilities of the basketball as it starts its journey.
Potential Energy
Potential energy is energy stored in an object due to its position. For the rolling basketball, potential energy becomes significant as the ball climbs up the ramp, getting higher off the ground.
As the ball rolls upwards, it slows down, transforming kinetic energy into gravitational potential energy. This energy is defined by the equation: \[ mgh \]where \( m \) is the mass of the basketball, \( g \) is the acceleration due to gravity, and \( h \) is the height.
The change in the ball's potential energy tells us how much height it gains. This vertical height reached can be determined by assessing the energy lost, primarily converted from its initial kinetic energy to its potential form, at its highest point.
Rotational Dynamics
Rotational dynamics involves studying the movement of objects that spin or rotate, like the basketball. Important here are the concepts of rotational kinetic energy and the moment of inertia, which measures how mass is distributed in relation to the axis of rotation.
For a rolling object, both linear and rotational motions exist, due to the no-slip condition ensuring linear speed relates to rotational speed by \( v = r\omega \). This relationship allows us to derive the formulas for rotational kinetic energy (discussed previously).
The way the basketball's mass is arranged makes it a thin-walled hollow sphere, giving a specific moment of inertia \( I = \frac{2}{3}mr^2 \). This distribution significantly affects how quickly the ball can roll up the ramp, as well as how quickly it can stop or start spinning. This, in turn, impacts the overall kinetic energy and how it might convert into potential energy when climbing the ramp.

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Most popular questions from this chapter

A solid ball is released from rest and slides down a hillside that slopes downward at \(65.0^{\circ}\) from the horizontal, (a) What minimum value must the coefficient of static friction between the hill and ball surfaces have for no slipping to occur? (b) Would the coefficient of friction calculated in part (a) be sufficient to prevent a hollow ball (such as a soccer ball) from slipping? Justify your answer. (c) In part (a), why did we use the coefficient of static friction and not the coefficient of kinetic friction?

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