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Chapter 10: Problem 35

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

Short Answer

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

Step by step solution

01

Define Translational and Rotational Kinetic Energy

The kinetic energy of a rolling solid disk is divided into translational kinetic energy and rotational kinetic energy. Translational kinetic energy, \(KE_t\), is given by \(\frac{1}{2}mv^2\), where \(m\) is the object's mass and \(v\) is its velocity. Rotational kinetic energy, \(KE_r\), is given by \(\frac{1}{2}Iw^2\), where \(I\) is the object's moment of inertia and \(w\) its angular velocity.
02

Convert Angular Velocity to Linear Velocity

The moment of inertia for a solid disk rotating about its central axis is given by \(I = 0.5mR^2\), where \(R\) is the radius of the disk. Also, when the disk is rolling without slipping, the linear velocity \(v\) and angular velocity \(w\) are related by \(v = wR\). Thus, \(w = v/R\). We substitute this into \(KE_r\), obtaining \(KE_r = 0.5 * 0.5 * m * R^2 * (v/R)^2 = 0.25mv^2\). Hence, the rotational kinetic energy is \(\frac{1}{4}\) of the translational kinetic energy.
03

Calculate the Fraction

The total kinetic energy of the disk is the sum of the translational and rotational kinetic energy, i.e., \(KE_t + KE_r\). Substituting values, we get \(0.5mv^2 + 0.25mv^2 = 0.75mv^2\). Now, the fraction of the disk's total kinetic energy that is rotational is obtained by dividing the rotational kinetic energy by the total kinetic energy, i.e., \(\frac{KE_r}{KE_t + KE_r}\). Substituting values, we get \(\frac{0.25mv^2}{0.75mv^2} = \frac{1}{3}\).

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