Question

A spoked wheel with a radius of \(40.0 \mathrm{cm}\) and a mass of $2.00 \mathrm{kg}$ is mounted horizontally on friction less bearings. JiaJun puts his \(0.500-\mathrm{kg}\) guinea pig on the outer edge of the wheel. The guinea pig begins to run along the edge of the wheel with a speed of $20.0 \mathrm{cm} / \mathrm{s}$ with respect to the ground. What is the angular velocity of the wheel? Assume the spokes of the wheel have negligible mass.

Short answer

Answer: The angular velocity of the wheel is approximately 0.167 s⁻¹.

Step-by-step solution

Calculate the initial angular momentum of guinea pig

First, we need to find the initial angular momentum of the guinea pig, which is given by: $$ L_{gp} = m_{gp}v_{gp}r $$ where \(L_{gp}\) is the angular momentum of the guinea pig, \(m_{gp}\) is its mass (0.5 kg), \(v_{gp}\) is its linear velocity (20 cm/s, or 0.20 m/s), and \(r\) is the radius of the wheel (40 cm, or 0.40 m). Using the given values, we have: $$ L_{gp} = (0.5\,\text{kg})(0.20\,\text{m/s})(0.40\,\text{m}) $$

Calculate the initial angular momentum of guinea pig

Let's now plug the values into the equation to find the initial angular momentum of the guinea pig. $$ L_{gp} = (0.5)(0.20)(0.40) $$ $$ L_{gp} = 0.04\, \text{kg m}^2\text{/s} $$

Apply conservation of angular momentum

Now, we will apply the conservation of angular momentum: the initial angular momentum of the guinea pig equals the final angular momentum of the system (guinea pig + wheel). Let \(\omega\) be the final angular velocity of the wheel. We have: $$ L_{gp} = I_{wheel}\omega + I_{gp}\omega $$ where \(I_{wheel}\) and \(I_{gp}\) are the moments of inertia of the wheel and guinea pig, respectively.

Calculate the moments of inertia

For a solid disk (wheel) with negligible mass in its spokes, the moment of inertia is given by: $$ I_{wheel}=\frac{1}{2}M_{wheel}r^2 $$ where \(M_{wheel}\) is the mass of the wheel (2.00 kg). For the guinea pig, we can approximate it as a point mass, and its moment of inertia is given by: $$ I_{gp}=m_{gp}r^2 $$ Now, plug in the given values to find the moments of inertia: $$ I_{wheel}=\frac{1}{2}(2.00\,\text{kg})(0.40\,\text{m})^2 $$ $$ I_{gp}=(0.5\,\text{kg})(0.40\,\text{m})^2 $$

Solve for the final angular velocity

Plug the calculated moments of inertia and the initial angular momentum of the guinea pig into the conservation of angular momentum equation: $$ 0.04\,\text{kg m}^2\text{/s} = \left(\frac{1}{2}(2.00)(0.40)^2\right)\omega + (0.5)(0.40)^2\omega $$ Solve for \(\omega\): $$ 0.04\,\text{kg m}^2\text{/s} = 0.24\,\text{kg m}^2\omega $$ $$ \omega = \frac{0.04}{0.24}\,\text{s}^{-1} $$

Find the angular velocity

Now, we can find the angular velocity of the wheel: $$ \omega = \frac{0.04}{0.24}\,\text{s}^{-1} = 0.167\,\text{s}^{-1} $$ So the angular velocity of the wheel is approximately \(0.167\,\text{s}^{-1}\).