Follow-up problems

11-30. A potter's wheel, with rotational inertia 6.40 kg·m2, is spinning freely at 19.0 rpm. The potter drops a 2.70-kg lump of clay onto the wheel, where it sticks a distance of 46.0 cm from the rotation axis. What is the subsequent angular speed of the wheel?

Solution:

Since we know the clay sticks to the potter's wheel, we are dealing with a totally inelastic collision, in which the angular momentum is conserved, but not the angular kinetic energy. First we must find the angular momentum of the potter's wheel, given by

Li
=
Iwheel  wwheel
=
(6.40  kg ·m2) æ
ç
ç
ç
ç
ç
ç
è 		19.0  rpm ·
2 p  rad
rev
60   sec
min
ö
÷
÷
÷
÷
÷
÷
ø
Li
=
12.74  kg ·m2/s
Now we must determine the new rotational inertia of the [clay + potter's wheel] system. This is given by
Iboth
=
Iwheel + Iclay
=
Iwheel + mclayr2
=
(6.40  kg ·m2) + (2.7  kg)(0.46  m)2
Iboth
=
6.97  kg ·m2
Finally, we apply the principle of conservation of angular momentum to find the final angular speed of both the wheel and the clay.
Li
=
Lf
=
Iboth  wboth
(12.74  kg ·m2/s)
=
(6.97  kg ·m2) wboth
wboth
=
1.83  rad/s

11-36. A skater's body has rotational inertia 4.2 kg·m2 with his fists held to his chest and 5.7 kg·m2 with arms outstretched. The skater is twirling at 3.0 rev/s while holding 2.5-kg weights in each outstretched hand; the weights are 76 cm from his rotation axis. If he pulls his hands in to his chest, how fast will he be turning?

Solution:

This problem again appears to be one of conservation of angular momentum. We start by finding the total rotational inertia of the skater holding weights in his outstretched arms:

Iout
=
Iarms  out + Iweight1 + Iweight2
=
Iarms  out + mweight1r2 + mweight2r2
=
Iarms  out + 2mweightr2
=
(5.7  kg ·m2) + 2 (2.5  kg)(0.76  m)2
Iout
=
8.6  kg ·m2
Now we can determine the skater's initial angular momentum:
Li
=
Iout  wi
=
(8.6  kg ·m2) æ
ç
è 		3.0  rev/s · 2 p  rad
rev
 ö
÷
ø
Li
=
161.9  kg ·m2/s
When he brings his hands in to his chest, the distance between the axis of rotation and the weights becomes zero, so his new rotational inertia becomes:
Iin
=
Iarms  in + Iweight1 + Iweight2
=
Iarms  in + 0 + 0
Iin
=
4.2  kg ·m2
Finally, we apply the principle of conservation of angular momentum to find the final angular speed of the skater with his arms in to his chest:
Li
=
Lf
=
Iin  wf
(161.9  kg ·m2/s)
=
(4.2  kg ·m2) wboth
wf
=
38.5  rad/s

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Solutions translated from TEX by TTH, version 1.57.