A bicycle is traveling north at 5.0 m/s.The mass of the wheel,2.0 kg,is uniformly distributed along the rim,which has a radius of 20 cm.What are the magnitude and direction of the angular momentum of the wheel about its axle? A) 2.0 kg∙m²/s towards the west B) 5.0 kg∙m²/s vertically upwards C) 2.0 kg∙m²/s towards the east D) 5.0 kg∙m²/s towards the east E) 5.0 kg∙m²/s towards the west

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Quizplus · Accepted Answer

The angular momentum $L$ of a rotating object is given by $L = I\omega$, where $I$ is the moment of inertia and $\omega$ is the angular velocity. For a wheel with mass $m$ uniformly distributed along its rim of radius $r$, the moment of inertia $I = mr^2$. The angular velocity $\omega$ can be found from the linear velocity $v$ and the radius $r$ using $\omega = \frac{v}{r}$. Substituting the given values, $I = 2.0 \, \text{kg} \times (0.2 \, \text{m})^2 = 0.08 \, \text{kg} \cdot \text{m}^2$ and $\omega = \frac{5.0 \, \text{m/s}}{0.2 \, \text{m}} = 25 \, \text{s}^{-1}$. Thus, $L = 0.08 \, \text{kg} \cdot \text{m}^2 \times 25 \, \text{s}^{-1} = 2.0 \, \text{kg} \cdot \text{m}^2/\text{s}$. The direction of angular momentum is perpendicular to the plane of rotation, following the right-hand rule. Since the wheel is moving north and rotating in a way that the top moves towards the east, the angular momentum vector points towards the west.

A Bicycle Is Traveling North at 5