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\bullet A light rope is wrapped several times around a large wheel with a radius of 0.400 $\mathrm{m} .$The wheel rotates in frictionless bearings about a stationary horizontal axis, as shown in Figure 10.48The free end of the rope is tied to a suitcase with a mass of 15.0 kg. The suit-case is released from rest at a height of 4.00 $\mathrm{m}$ above the ground. The suitcase has a speed of 3.50 $\mathrm{m} / \mathrm{s}$ when it reaches the ground. Calculate (a) the angular velocity of the wheel when the suitcase reaches the ground and ( b) the moment of inertia of the wheel.

a) $ 8.75 \mathrm{rad} / \mathrm{s}$b) 13 $\mathrm{kg} \cdot \mathrm{m}^{2}$

Physics 101 Mechanics

Chapter 10

Dynamics of Rotational Motion

Newton's Laws of Motion

Rotation of Rigid Bodies

Equilibrium and Elasticity

University of Michigan - Ann Arbor

Hope College

University of Sheffield

McMaster University

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so the free bar, the diagram of the suitcase is very simple. We have a force going up, force tension. And then, of course, going straight down is the force of gravity mg. Now we know that the radius is going to be equal to point Ah 400 meters. We know that the mass of the suitcase is 15.0 kilograms. We know that the suitcase takes a fall of 4.0 meters and we know that the final velocity of the suitcase right before hits the ground is 3.50 meters per second. Party is very simple to find the angular velocity of the suitcase. Right before he hits the ground, it will simply be equal to the velocity divided by the radius. So 3.50 divided by 0.4. And this is equal to 8.75 radiance per second. Ah, Now, for Part B, this is a bit more complicated. We need to find the moment of inertia. So let's let's first try to find the ah, angular, the linear acceleration. So we can say that V y final squared equals V Y initial squared plus two times a doubter, Why and so a sub. Why equals V Y Final squared, divided by two times Delta. Why, given that the initial wide velocity is zero and we have that a sub, Why equals 3.5 squared, divided by two times for and this is giving us 1.5 three meters per second squared to find the angular acceleration. This will simply be equal to the linear acceleration divided by the radius. So 1.53 divided by 0.4 and this is giving us 3.8 to radiance per second squared. Let's find the tension for So if we were to apply this sum of forces in the white direction for the suitcase, the tension force is going to be equal to the mass times gravity minus the acceleration in the right direction. This will be equal to 15 9.8, minus 1.53 and this is giving us 124.5 Newton's. Now we know that for the wheel we have a tension along the side. Of course, going straight down from the center of Mass is the force of gravity and then going straight up is actually forced normal, given that this is the axis. Um so we can say that the sum of torque is going to be equal to the Forest times, the radius and this will be equal toothy force, tension, times the radius. And we know that the sum of torque also equals the moment of inertia, times the angular acceleration. So this is how we're going to find the moment of inertia. The moment of inertia will simply be equal to the tension force times the radius of the wheel divided by the angular acceleration. And this is giving us 124.5 times the radius of 0.400 meters, divided by the angular exploration that we found previously. So 3.8 to radiance per second squared. And we find that the moment of inertia is going to be equal to approximately 13.0 kilogram meters squared. So this will be our second answer. Their answer for part B. That is the end of the solution. Thank you for watching

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