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Chapter 8: Problem 101

A 2.0 -kg uniform flat disk is thrown into the air with a linear speed of \(10.0 \mathrm{m} / \mathrm{s} .\) As it travels, the disk spins at 3.0 rev/s. If the radius of the disk is \(10.0 \mathrm{cm},\) what is the magnitude of its angular momentum?

Short Answer

Expert verified
The magnitude of the angular momentum of the spinning disk is approximately \(2.19 \mathrm{kg\cdot m^2/s}\).

Step by step solution

01

Calculate the moment of inertia

To begin, we need to calculate the moment of inertia for the disk. The moment of inertia I of a uniform disk about an axis perpendicular to its plane and passing through its center of mass is given by the formula: $$I =\dfrac{1}{2}mR^{2}$$ Where m is the mass of the disk (2.0 kg) and R is its radius (0.1 m). Now, we can substitute the values into the formula to find the moment of inertia: $$I =\dfrac{1}{2} (2.0 \mathrm{kg}) (0.1 \mathrm{m})^{2}$$ $$I = 0.01 \mathrm{kg\cdot m^2}$$
02

Calculate angular velocity

Given that the disk spins at 3.0 rev/s, we need to convert this value to radians per second to be able to use it in our calculations. We can do that by multiplying the number of revolutions per second by \(2\pi\): $$\omega = 3.0 \frac{\text{rev}}{\text{s}} \times 2\pi \frac{\text{rad}}{\text{rev}} = 6\pi \frac{\text{rad}}{\text{s}}$$
03

Calculate rotational angular momentum

Next, we need to calculate the angular momentum caused by the rotational motion of the disk. The rotational angular momentum L_rot can be calculated using the following formula: $$L_{\text{rot}} = I \omega$$ Now, substituting the values for I and 蠅 from the previous steps, we get: $$L_{\text{rot}} = (0.01 \mathrm{kg\cdot m^2}) (6\pi \frac{\text{rad}}{\text{s}}) = 0.06\pi \mathrm{kg\cdot m^2/s}$$
04

Calculate translational angular momentum

For the translational motion, we can calculate the angular momentum L_trans as: $$L_{\text{trans}} = mvr$$ Where m is the mass of the disk (2.0 kg), v is its linear speed (10.0 m/s), and r is its radius (0.1 m). Now, we can substitute the values into the formula to find the translational angular momentum: $$L_{\text{trans}} = (2.0 \mathrm{kg}) (10.0 \mathrm{m/s}) (0.1 \mathrm{m})$$ $$L_{\text{trans}} = 2.0 \mathrm{kg\cdot m^2/s}$$
05

Calculate total angular momentum

Finally, we can find the total angular momentum of the disk by adding the rotational and translational angular momenta: $$L_{\text{total}} = L_{\text{rot}} + L_{\text{trans}}$$ $$L_{\text{total}} = 0.06\pi \mathrm{kg\cdot m^2/s} + 2.0 \mathrm{kg\cdot m^2/s}$$ $$L_{\text{total}} \approx 2.19 \mathrm{kg\cdot m^2/s}$$ Thus, the magnitude of the angular momentum of the spinning disk is approximately \(2.19 \mathrm{kg\cdot m^2/s}\).

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