Problem

Suppose a 0.250 -kg ball is thrown at 15.0 $\math…

07:15

Problem 45 Hard Difficulty

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 to
the 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.

Answer

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

Related Courses

Physics 101 Mechanics

College Physics for AP® Courses

Chapter 10

Rotational Motion and Angular Momentum

Discussion

You must be signed in to discuss.

Top Physics 101 Mechanics Educators

Christina K.

Rutgers, The State University of New Jersey

Watch More Solved Questions in Chapter 10

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Video Transcript

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

We have video lessons for 95.57% of the questions in this textbook