Ch 10: Dynamics of Rotational Motion

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 10: Dynamics of Rotational Motion

Problem 28b

Chapter 10, Problem 28b

A bicycle racer is going downhill at 11.0 m/s when, to his horror, one of his 2.25-kg wheels comes off as he is 75.0 m above the foot of the hill. We can model the wheel as a thin-walled cylinder 85.0 cm in diameter and ignore the small mass of the spokes. How much total kinetic energy does the wheel have when it reaches the bottom of the hill?

Verified step by step guidance

First, identify the types of energy involved. The wheel has both translational kinetic energy due to its linear motion and rotational kinetic energy due to its spinning.

Calculate the initial potential energy of the wheel at the top of the hill using the formula:

E = m g h

, where

m

is the mass of the wheel,

g

is the acceleration due to gravity, and

h

is the height above the foot of the hill.

Determine the translational kinetic energy using the formula:

E = \frac{1}{2} m v^{2}

, where

v

is the linear velocity of the wheel.

Calculate the rotational kinetic energy using the formula:

E = \frac{1}{2} I ω^{2}

, where

I

is the moment of inertia of the wheel and

ω

is the angular velocity. For a thin-walled cylinder,

I

can be calculated as

I = m r^{2}

, where

r

is the radius of the wheel.

Sum the translational and rotational kinetic energies to find the total kinetic energy of the wheel when it reaches the bottom of the hill.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion, calculated using the formula KE = 0.5 * m * v^2, where m is mass and v is velocity. In this scenario, the wheel's kinetic energy includes both translational and rotational components as it rolls downhill.

Rotational Kinetic Energy

Rotational kinetic energy is the energy due to an object's rotation, given by KE_rot = 0.5 * I * ω^2, where I is the moment of inertia and ω is the angular velocity. For a thin-walled cylinder, I = m * r^2, where r is the radius, which is crucial for calculating the wheel's energy as it rolls.

Conservation of Energy

The conservation of energy principle states that energy cannot be created or destroyed, only transformed. As the wheel descends, its potential energy converts into kinetic energy, allowing us to calculate the total kinetic energy at the bottom by considering both translational and rotational forms.

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A 2.20-kg hoop 1.20 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.60 rad/s. Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom.

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A playground merry-go-round has radius $2.40$\text{ m}$$ and moment of inertia $2100$\text{ kg m}$^2$ about a vertical axle through its center, and it turns with negligible friction. A child applies an $18.0$\text{ N}$$ force tangentially to the edge of the merry-go-round for $15.0$\text{ s}$$ . If the merry-go-round is initially at rest, how much work did the child do on the merry-go-round?

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Textbook Question

A 2.20-kg hoop 1.20 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.60 rad/s. Find the velocity vector for each of the points in part (c), but this time as viewed by someone moving along with the same velocity as the hoop.

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A playground merry-go-round has radius $2.40 m$ and moment of inertia $2100 kg m squared$ about a vertical axle through its center, and it turns with negligible friction. A child applies an $18.0 cap N$ force tangentially to the edge of the merry-go-round for $15.0 s$ . If the merry-go-round is initially at rest, what is its angular speed after this $15.0 s$ interval?

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