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Chapter 10: Problem 34

A uniform solid cylinder of mass \(M=5.00 \mathrm{~kg}\) is rolling without slipping along a horizontal surface. The velocity of its center of mass is \(30.0 \mathrm{~m} / \mathrm{s}\). Calculate its energy.

Short Answer

Expert verified
Answer: The total energy of the uniform solid cylinder is 6,750 J.

Step by step solution

01

Write down the given values

Mass of the cylinder, \(M = 5.00 \, kg\). Velocity of the center of mass, \(v = 30.0 \, m/s\).
02

Relate linear velocity to angular velocity

Since the cylinder is rolling without slipping, its linear velocity \(v\) is related to its angular velocity \(\omega\) by: \(v = R\omega\) We will need this relation later.
03

Find the moment of inertia of the cylinder

The moment of inertia of a uniform solid cylinder rotating about its axis is given by the formula: \(I = \frac{1}{2}mR^2\) Note that we have not been given the radius of the cylinder, so we will leave the moment of inertia in terms of \(R\).
04

Calculate the rotational kinetic energy

The rotational kinetic energy is given by the formula: \(KE_{rot} = \frac{1}{2}I\omega^2\) Replacing \(I\) with the expression found in step 3, we get: \(KE_{rot} = \frac{1}{2} \times \frac{1}{2}mR^2 \times \omega^2\) Using the relation from step 2, \(\omega = \frac{v}{R}\), we can replace \(\omega\) in the expression for \(KE_{rot}\): \(KE_{rot} = \frac{1}{2} \times \frac{1}{2}mR^2 \times \left(\frac{v^2}{R^2}\right)\) Simplifying, we get: \(KE_{rot} = \frac{1}{4}m v^2\)
05

Calculate the translational kinetic energy

The translational kinetic energy is given by the formula: \(KE_{trans} = \frac{1}{2}mv^2\)
06

Find the total energy

Now we can find the total energy by adding the translational and rotational kinetic energies: \(E_{total} = KE_{trans} + KE_{rot} = \frac{1}{2}mv^2 + \frac{1}{4}m v^2\) Combining terms, we get: \(E_{total} = \frac{3}{4}mv^2\)
07

Calculate the energy using the given values

Finally, substitute the given values for mass and velocity into the expression for \(E_{total}\): \(E_{total} = \frac{3}{4}(5.00\, kg)(30.0\, m/s)^2\) After performing the calculations, we get: \(E_{total} = 6,750\, J\) The total energy of the uniform solid cylinder rolling without slipping along a horizontal surface is \(6,750\, J\).

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