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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 ω 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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Most popular questions from this chapter

A 68 -kg woman stands straight with both feet flat on the floor. Her center of gravity is a horizontal distance of \(3.0 \mathrm{cm}\) in front of a line that connects her two ankle joints. The Achilles tendon attaches the calf muscle to the foot a distance of \(4.4 \mathrm{cm}\) behind the ankle joint. If the Achilles tendon is inclined at an angle of \(81^{\circ}\) with respect to the horizontal, find the force that each calf muscle needs to exert while she is standing. [Hint: Consider the equilibrium of the part of the body above the ankle joint.]
A solid sphere is released from rest and allowed to roll down a board that has one end resting on the floor and is tilted at \(30^{\circ}\) with respect to the horizontal. If the sphere is released from a height of \(60 \mathrm{cm}\) above the floor, what is the sphere's speed when it reaches the lowest end of the board?
A bowling ball made for a child has half the radius of an adult bowling ball. They are made of the same material (and therefore have the same mass per unit volume). By what factor is the (a) mass and (b) rotational inertia of the child's ball reduced compared with the adult ball?
A merry-go-round (radius \(R\), rotational inertia \(I_{\mathrm{i}}\) ) spins with negligible friction. Its initial angular velocity is \(\omega_{i}\) A child (mass \(m\) ) on the merry-go-round moves from the center out to the rim. (a) Calculate the angular velocity after the child moves out to the rim. (b) Calculate the rotational kinetic energy and angular momentum of the system (merry-go-round + child) before and after.
A stone used to grind wheat into flour is turned through 12 revolutions by a constant force of \(20.0 \mathrm{N}\) applied to the rim of a 10.0 -cm-radius shaft connected to the wheel. How much work is done on the stone during the 12 revolutions?
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