A bicycle of mass 25 kg has wheels of mass 5.0 kg and radii of 30 cm each (assume the wheels are uniform disks). The part of the kinetic energy that is rotational is

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Quizplus · Accepted Answer

The rotational kinetic energy is associated with the wheels, while the translational kinetic energy is associated with the entire bicycle. The moment of inertia (I) for a uniform disk is given by $I = \frac{1}{2}MR^2$, where M is the mass of the disk and R is its radius. Since there are two wheels, the total moment of inertia for both wheels is $I_{total} = 2 \times \frac{1}{2}M_{wheel}R^2 = M_{wheel}R^2$. The rotational kinetic energy (K.E._rot) is given by $\frac{1}{2}I\omega^2$, and the translational kinetic energy (K.E._trans) for the entire bicycle (including the wheels) moving at a velocity v is $\frac{1}{2}M_{total}v^2$, where $M_{total}$ is the total mass of the bicycle and the wheels. The angular velocity $\omega$ and the linear velocity $v$ are related by $\omega = \frac{v}{R}$. Therefore, the ratio of rotational to total kinetic energy is:$$\frac{K.E._rot}{K.E._total} = \frac{\frac{1}{2}M_{wheel}R^2\left(\frac{v}{R}\right)^2}{\frac{1}{2}M_{total}v^2 + \frac{1}{2}M_{wheel}R^2\left(\frac{v}{R}\right)^2} = \frac{M_{wheel}R^2v^2}{R^2(M_{total}v^2 + M_{wheel}v^2)} = \frac{M_{wheel}}{M_{total} + M_{wheel}}$$Given that $M_{wheel} = 5.0 \, \text{kg}$ (for both wheels together) and $M_{total} = 25 \, \text{kg}$, the ratio becomes:$$\frac{5}{25 + 5} = \frac{5}{30} = \frac{1}{6}$$Converting this fraction to a percentage gives approximately 16.67%, which rounds to 17%, making option B the correct answer.

A Bicycle of Mass 25 Kg Has Wheels of Mass