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Chapter 8: Problem 69

A turntable of mass \(5.00 \mathrm{kg}\) has a radius of \(0.100 \mathrm{m}\) and spins with a frequency of 0.550 rev/s. What is its angular momentum? Assume the turntable is a uniform disk.

Short Answer

Expert verified
Answer: The angular momentum of the turntable is approximately 0.0864 kg·m²/s.

Step by step solution

01

Find the moment of inertia of the uniform disk

To find the moment of inertia, we will use the formula for the moment of inertia of a uniform disk, which is given by: \(I=\frac{1}{2}MR^2\), where M is the mass and R is the radius of the disk. In our case, the mass is 5.00 kg and radius is 0.100 m, so plug the values into the formula.\(I=\frac{1}{2}(5.00\,\text{kg})(0.100\,\text{m})^2\)
02

Calculate the moment of inertia

Now, we can simplify the formula to find the moment of inertia.\(I=\frac{1}{2}(5.00\,\text{kg})(0.0100\,\text{m}^2) = 0.0250\,\text{kg}\cdot\text{m}^2\)
03

Convert the frequency to the angular velocity

Since we are given the frequency in rev/s, we'll need to convert it to angular velocity in rad/s. The conversion factor is \(2\pi\) radians per revolution because there are \(2\pi\) radians in one revolution. Therefore, multiply the given frequency by \(2\pi\) to find the angular velocity. \(\omega = 0.550\,\text{rev/s}\times 2\pi\,\text{rad/rev}\)
04

Calculate the angular velocity

Now, we can simplify the formula to find the angular velocity. \(\omega = 0.550\,\text{rev/s}\times 2\pi\,\text{rad/rev} \approx 3.455\,\text{rad/s}\)
05

Find the angular momentum

Now we have everything we need to find the angular momentum, L. We'll use the formula \(L = I\omega\), where I is the moment of inertia, and \(\omega\) is the angular velocity. Plug in the values we calculated in Step 2 and Step 4: \(L = (0.0250\,\text{kg}\cdot\text{m}^2)(3.455\,\text{rad/s})\)
06

Calculate the angular momentum

Now, we'll simply multiply the moment of inertia and the angular velocity to get the final answer for the angular momentum. \(L = (0.0250\,\text{kg}\cdot\text{m}^2)(3.455\,\text{rad/s}) \approx 0.0864\,\text{kg}\cdot\text{m}^2\text{/s}\) The angular momentum of the turntable is approximately 0.0864 kg·m²/s.

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Most popular questions from this chapter

The radius of a wheel is \(0.500 \mathrm{m} .\) A rope is wound around the outer rim of the wheel. The rope is pulled with a force of magnitude $5.00 \mathrm{N},$ unwinding the rope and making the wheel spin CCW about its central axis. Ignore the mass of the rope. (a) How much rope unwinds while the wheel makes 1.00 revolution? (b) How much work is done by the rope on the wheel during this time? (c) What is the torque on the wheel due to the rope? (d) What is the angular displacement \(\Delta \theta\), in radians, of the wheel during 1.00 revolution? (e) Show that the numerical value of the work done is equal to the product \(\tau \Delta \theta\)
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