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Chapter 10: Problem 39

What fraction of a solid disk's kinetic energy is rotational if it's rolling without slipping?

Short Answer

Expert verified
The fraction of a solid disk's kinetic energy that is rotational when it is rolling without slipping is \( \frac{1}{3} \).

Step by step solution

01

Defining the Components of Kinetic Energy

Kinetic energy of a moving solid disk can be divided into translational and rotational parts. The translational kinetic energy is given by \( \frac{1}{2} mv^2 \), where \( m \) is the mass of the disk and \( v \) is its velocity. The rotational kinetic energy is given by \( \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia of the solid disk and \( \omega \) is the angular velocity.
02

Applying the Condition of Rolling Without Slipping

For an object rolling without slipping, the linear velocity \( v \) and the angular velocity \( \omega \) are related: \( v = r \omega \), where \( r \) is the radius of the disk.
03

Simplifying the Rotational Kinetic Energy

Using the equation from step 2, the rotational kinetic energy becomes \( \frac{1}{2} I \omega^2 = \frac{1}{2} I (\frac{v}{r})^2 \). The moment of inertia \( I \) for a solid disk is \( \frac{1}{2} mr^2 \). Plugging this in yields a rotational kinetic energy of \( \frac{1}{4} mv^2 \).
04

Calculating the Fraction of Rotational Kinetic Energy

The fraction of the total kinetic energy that is rotational is calculated by the rotational kinetic energy (\( \frac{1}{4} mv^2 \)) divided by the total kinetic energy (\( \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \)). Thus, the fraction is \( \frac{1/4}{3/4} = \frac{1}{3} \).

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Most popular questions from this chapter

A thick ring has inner radius \(\frac{1}{2} R,\) outer radius \(R,\) and mass \(M\) Find an expression for its rotational inertia. (Hint: Consult Example \(10.7 .)\)

As an automotive engineer, you're charged with improving the fuel economy of your company's vehicles. You realize that the rotational kinetic energy of a car's wheels is a significant factor in fuel consumption, and you set out to lower it. For a typical car, the wheels' rotational energy is \(40 \%\) of their translational kinetic energy. You propose a redesigned wheel with the same radius but \(10 \%\) lower rotational inertia and \(20 \%\) less mass. What do you report for the decrease in the wheel's total kinetic energy at a given speed?

Is it possible to apply a counterclockwise torque to an object that's rotating clockwise? If so, how will the object's motion change? If not, why not?

A 25 -cm-diameter circular saw blade spins at 3500 rpm. How fast would you have to push a straight hand saw to have the teeth move through the wood at the same rate as the circular saw teeth?

In bicycling, each foot pushes on the pedal for half a rotation of the pedal shaft; that foot then rests and the other foot takes over. During each half- cycle, the torque resulting from the force of the active foot is given approximately by \(\tau=\tau_{0} \sin \omega t,\) where \(\tau_{0}\) is the maximum torque and \(\omega\) is the angular speed of the pedal shaft (in \(\mathrm{s}^{-1},\) as usual). A particular cyclist is turning the pedal shaft at \(70.0 \mathrm{rpm},\) and at the same time \(\tau_{0}\) is measured at \(38.5 \mathrm{N} \cdot \mathrm{m}\). Find (a) the energy supplied by the cyclist in one turn of the pedal shaft and (b) the cyclist's average power output.
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