Question

A wheel of mass 0.80kg and radius 0.30m is rolling without slipping up a plane 15 degrees to the horizontal . at some instant it has an angular speed of 12rad/s, and it comes momentarily to rest after rolling a further 2.5 revolutions up the plane. 1 Use the conservation of energy to find the moment of inertia of the wheel.

Answer 1

Answer on Question #52424, Physics, Mechanics | Kinematics | Dynamics

A wheel of mass $0.80\mathrm{kg}$ and radius $0.30\mathrm{m}$ is rolling without slipping up a plane 15 degrees to the horizontal. at some instant it has an angular speed of $12\mathrm{rad/s}$, and it comes momentarily to rest after rolling a further 2.5 revolutions up the plane. 1 Use the conservation of energy to find the moment of inertia of the wheel.

Solution:

Conservation of energy gives:

$$PE_{gravity} = KE_{translational} + KE_{rotational}$$ $$mgh = \frac{1}{2}mv^2 + \frac{1}{2}l\omega^2$$

For rolling without slipping, $\omega = v/r$.

The height is

$$h = L\sin\theta = 2\pi r N\sin\theta = 2\pi * 0.30 * 2.5 * \sin 15{}^\circ = 1.22\ \mathrm{m}$$

Thus, for the moment of inertia of the wheel

$$\frac{1}{2}l\omega^2 = m\left(gh - \frac{1}{2}v^2\right)$$ $$I = \frac{2m}{\omega^2}\left(gh - \frac{1}{2}v^2\right) = \frac{2m}{\omega^2}\left(gh - \frac{1}{2}\omega^2 r^2\right)$$ $$I = \frac{2 * 0.80}{12^2}\left(9.8 * 1.22 - \frac{1}{2} * 12^2 * 0.3^2\right) = 0.061\ \mathrm{kg \cdot m^2}$$

Answer: $0.061\ \mathrm{kg \cdot m^2}$

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