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Twin skaters approach one another as shown in Figure 10.39 and lock hands. (a) Calculate their final angular velocity, given each had an initial speed of 2.50 $\mathrm{m} / \mathrm{s}$ relative tothe ice. Each has a mass of $70.0 \mathrm{kg},$ and each has a center of mass located 0.800 $\mathrm{m}$ from their locked hands. You may approximate their moments of inertia to be that of point masses at this radius. (b) Compare the initial kinetic energy and final kinetic energy.

(a) $\omega = 3.13 \mathrm { rnd } / \mathrm { s }$(b) They are the same. $K E = K E ^ { \prime } = 437.5 \mathrm { J }$

Physics 101 Mechanics

Chapter 10

Rotational Motion and Angular Momentum

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Rutgers, The State University of New Jersey

University of Washington

Hope College

University of Sheffield

Lectures

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inertia, Al could be calculated as the mass times, the velocity, the linear velocity, times, the radius or the moment of inertia times the angular velocity, the angular we have. We have to assume a massive two skaters. So now the moment of inertia, times the angular acceleration. This becomes two times the mass of one skater times the r. And so we can say that the moment of inertia simply equals two m r squared. And so we can then say that to m R. Squared Omega is equaling two m the r, of course, to cancels out the mass cancels out and one of the radio cancel out. And so we find that of course, the angular velocity is equaling the linear velocity divided by our This would be equaling 2.50 meters per second, divided by the radius or the distance between the two skaters 20.800 meters. And so the angular velocity of the two skaters would be 3.1 to 5 radius radiance. Rather 1st 2nd this would be our final answer for part a four part B. Ah, we know that the kinetic energy between the two skaters would be essentially, we could stay like translational. Kinetic energy would be equaling two skaters plus times 1/2 M. V squared. This is simply gonna be equal to M V squared so we can solve this would be 70 0.0 kilograms multiplied by 2.50 meters per second quantity squared and we have that the kinetic, the translational, kinetic energy equaling 438. Jules, this would be our one answer for perch B and then the final kinetic energy we can say K e final would be equaling 1/2 times the moment of inertia times omega squared. And so this would be equal in 1/2 times to m. R squared, multiplied by the, uh, angular velocity which is simply the linear velocity divided by the radius multiplied by rather race to the second power. This is equaling again M v squared, which is exactly what we have here. So the final kinetic energy or the final rotational kinetic energy is equaling again. 438 jewels, which simply means that all of the translational kinetic energy was transformed into rotational kinetic energy. After the ice skaters essentially joined hands, that is the end of the solution. Thank you for

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