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Repeat Example 10.15 in which the disk strikes and adheres to the stick 0.100 $\mathrm{m}$ from the nail.

(a) $w ^ { \prime } = 0.156$ rad/s$\begin{array} { l l } { \text { (b) } K E = 22.5 \mathrm { J } } & { \text { KE } = 0.012 \mathrm { J } } \\ { \text { (c) } p = 1.50 \mathrm { kg } + \mathrm { M } / \mathrm { s } } & { - \quad p ^ { \prime } = 0.188 \mathrm { kg } \cdot \mathrm { th } / \mathrm { s } } \end{array}$

Physics 101 Mechanics

Chapter 10

Rotational Motion and Angular Momentum

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

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here, we're gonna apply the conservation of angular momentum all equals out. Prime Al would be the angular momentum of the disk before the collision. L prime would be the angular momentum of the disc after the collision. And so we can say that here the moment. The angular momentum before the collision would be going to the moment of inertia, of the disc times, the angular velocity plus the mass of the disc times the velocity of the disc multiplied by our are not where are not is the perpendicular distance of the disk from the pivoted point just before the collision. And we can then say that the moment of inertia for the disc this has given us the mask of the the mass of the disc, multiplied by r squared the radius of the disc divided by two I know the difference. R Squared R Squared is the radius of the disc and are not as the perpendicular distance from the disc to the pivot joint just before the collision, and so we can find the moment of inertia for the disk. This would be equal to mass of a disc 50 grams so 500.50 kilograms multiplied by R squared 1.5 centimeters or 0.15 meters quantity squared and then divided by two. And we find that the moment of inertia for the disk 5.6 to 5 times 10 to the negative sixth kilograms meters squared. We then find the angular velocity. This would be 1000 revolutions per minute, multiplied by two pi radiance for every revolution multiplied by 60 seconds for every minute on this would be equaling my apologies. This would be actually one minute for every 60 seconds. My apologies. This is equaling 104.71 radiance for second. And so we're going to then find the angular momentum. The angular momentum would be equaling 5.6 to 5 times 10 to the negative sixth again kilograms meters squared, multiplied by 104.71 radiance per second. Plus the mass of the disc we know to be again 0.50 kilograms multiplied by the velocity of the disc. 30.0 meters per second, multiplied by the perpendicular distance from the disc to the pivot point right before the collision, 1.20 meters and we find that the angular velocity of the disc correct before the collision, would be equaling angular momentum of the describe it for the collision one point aids, they're out kilograms meters squared for a second. And so we can say that the system of the moment of inertia of the system right after the collision, we can say that the moment of inertia of the system would be equal to the moment of inertia of the disc pivot, plus the moment of inertia of the stick. So this would be equaling the mass of the disc times l squared, plus the mass of the disc Times are squared, divided by two plus we can say then the mass of the stick capital m times the length of the sticks squared over three and weakens. Then say that these moments of inertia for the system we can solve. This would be 0.50 kilograms multiplied by 1.20 meters quantity squared, plus the mass of the disc 0.50 kilograms times the radius of the disc we know to be point point five centimeters or 50.15 meters quantity squared. Aah! This would be divided by two plus the mass of the stick. We know to be 2.0 kilograms the length of the stick again, one point 20 meters quantity squared, divided by three. And we find that the moment of inertia for the system is going to be equaling 1.3 kilograms meters squared. This would be the moment of inertia of the system after the collision. And so we can then say that the angular momentum of the system would be equal to the moment of inertia, of the system, times, the angular velocity of the system and so the any the momentum is conserved. So this would be one 0.80 kilograms meters squared per second. This would be equal to the moment of inertia of the system 1.3 kilograms meter squared times the angular moment, angular velocity of the system. So we know that then the angular velocity of the system is 1.80 divided by 1.3 1.74 radiance per second. So the angular velocity of the system after the collision is 1.74 ratings per second. That would be our answer. Four part, eh? No, For part B, uh, we're trying to find the final kinetic energy of the system. And so we can first say that the total kinetic energy of the disk before the collision this would be like before the collision would be 1/2 times the mass of the disc, times of loss, the linear velocity of the disc squared. So the translational kinetic energy, plus the rotational kinetic energy 1/2 times the moment of inertia for the disc times the angular velocity of the disc quantity squared. Now we can actually solve. This would be equaling, we could say 1/2 times the mass of the disc 0.50 kilograms times the velocity of the disk, 30.0 meters per second quantity squared, plus 1/2 the moment of inertia for the disk we found already 5.655 point 6 to 5 times 10 to the negative six kilograms meter squared times the angular velocity of the disc 100 4.71 radiance per second Quantity squared, and we find that the total kinetic energy of the disk before the collision is equaling 22.5 jewels. Now we can then say that the kinetic energy of the system after the collision would be equal to on Lee the rotational kinetic energy 1/2 times the moment of inertia for the system times the angular velocity of the system quantity squared, and we have solved this already. So we can say that the kinetic energy of the system would be equaling to 1/2 times the angular, the moment of inertia for the system. 1.3 again kilograms meter squared, multiplied by 1.74 radiance for a second quantity squared. This is equaling 1.57 jewels. So the final wrote the final rotational kinetic energy of the system after the collision. 1.57 jewels. This would be your answer for part B and then finally four parts. See, we can say that the linear momentum before the collision we can say P initial would be equaling to the mass of the disc times the linear velocity of the disc and so this should be equaling 0.50 kilograms multiplied by 30.0 meters per second. This is equaling 1.5 kilograms meters per second and we can say that this would be the linear momentum before the collision. Now for the linear momentum of the system after the collision, we could say the moment piece of f Millennium Momentum final would be equaling the mass of the disc plus the massive the system multiplied by the linear velocity of the systems center of mass. And so here we can say that the linear velocity of the systems center of mass after the collision Ah would be the angular velocity of the system times the length of the stick over too. And so we can then say that the final linear momentum of the system would be 0.50 kilograms plus 2.0 kilograms multiplied by 1.74 radiance. Her second multiplied by 1.20 meters over too. And we find that the final and finer final linear momentum after the collision, 2.20 kilograms meters per second. This would be our second final answer for part C. That is the end of the solution. Thank you for one

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