/*! 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 69 A solid uniform spherical stone ... [FREE SOLUTION] | 91Ó°ÊÓ

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A solid uniform spherical stone starts moving from rest at the top of a hill. At the bottom of the hill the ground curves upward, launching the stone vertically a distance \(H\) below its start. How high will the stone go (a) if there is no friction on the hill and (b) if there is enough friction on the hill for the stone to roll without slipping? (c) Why do you get two different answers even though the stone starts with the same gravitational potential energy in both cases?

Short Answer

Expert verified
(a) Without friction, height is \( H \). (b) With friction, height is \( \frac{5}{7}H \). (c) Energy splits into rotation and translation in the friction case.

Step by step solution

01

Understand Potential and Kinetic Energy

Initially, the stone has gravitational potential energy at the top of the hill, which will be converted into kinetic energy as it moves down. We need to compare how this energy is transformed in both frictionless and with-friction scenarios.
02

Analyze the Frictionless Case

In the absence of friction, the stone slides down the hill. This means all gravitational potential energy gets transformed into translational kinetic energy. Use the energy conservation equation \( mgh = \frac{1}{2}mv^2 \) to solve for speed \( v \) at the bottom of the hill.
03

Solve for Maximum Height Without Friction

Since all kinetic energy at the bottom becomes potential energy when the stone reaches maximum height without friction: \( \frac{1}{2}mv^2 = mg(H - h) \). Solve for \( h \) to get: \( h = H \) because potential energy converts fully.
04

Analyze the With-Friction Case

When there is friction, the stone rolls without slipping. The initial gravitational potential energy is converted into both translational and rotational kinetic energy: \( mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \), where \( I \) is the moment of inertia (for a sphere, \( I = \frac{2}{5}mr^2 \)) and \( \omega = \frac{v}{r} \).
05

Calculate Energy Distribution in With-Friction Case

Substitute the expressions for \( I \) and \( \omega \) in the energy equation: \( mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v^2}{r^2}) \). Simplify to find that \( \frac{7}{5}mv^2 = mgh \).
06

Solve for Maximum Height With Friction

Set the kinetic energy at the bottom equal to potential energy at the highest point: \( \frac{1}{2}mv^2 = mg(H - h) \). Thus, \( mgh - \frac{2}{7}mgh = mg(H - h) \); solve for \( h \) to get \( h = \frac{5}{7}H \).
07

Compare Both Cases

In the no friction case, \( h = H \); for the rolling case, \( h = \frac{5}{7}H \). The difference is because in the rolling case, energy is shared between translational and rotational motion, reducing the height reached.

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is the energy stored due to an object's height above the ground. It is calculated using the formula:
  • \( U = mgh \)
where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height. This form of energy depends on the position and is maximal at the top of a hill.
As the stone begins its descent, this stored energy is gradually converted into other forms of energy, such as kinetic.
In cases where two scenarios need comparison, like sliding versus rolling down a hill, understanding the initial potential energy is crucial. This potential energy is entirely based on the starting height \( H \) before the stone begins its movement.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. There are two types: translational and rotational. For translational kinetic energy, the formula is:
  • \( KE_{trans} = \frac{1}{2}mv^2 \)
In the frictionless scenario, the spherical stone converts all gravitational potential energy into translational kinetic energy as it slides down the hill.
Since all energy is used in this motion, the stone eventually reaches a height \( h = H \) again after being vertically launched.
However, when friction is present and the stone rolls, part of this potential energy becomes rotational kinetic energy, too. This means not all of it is directed into upward motion.
Rotational Motion
Rotational motion occurs when the stone rolls down the hill instead of sliding. This involves another type of kinetic energy:
  • Rotational kinetic energy: \( KE_{rot} = \frac{1}{2}I\omega^2 \),
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
For a solid sphere, \( I = \frac{2}{5}mr^2 \) and \( \omega = \frac{v}{r} \).
Rolling without slipping ensures energy is split into both translation and rotation.
As a result, less energy remains for height gain, hence the stone reaches a smaller height \( h = \frac{5}{7}H \). This division explains why the total energy is dispersed differently compared to sliding.
Friction
Friction is the force that resists motion between two surfaces in contact. In the context of the rolling stone, friction provides the necessary torque for rotation without slipping.
While it might seem like an energy drain, friction fosters rolling, converting some potential energy into rotational kinetic energy.
This interaction is described by how potential energy transitions into both types of kinetic energy in the equation:
  • \( mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \)
The presence of friction ensures that energy isn't entirely funneled into translational movement. Consequently, this causes a reduced maximum height \( h = \frac{5}{7}H \) when compared to a frictionless case.
It highlights the importance of friction to initiate and sustain rolling motion, altering how high the stone travels.

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