A skater spins with an angular speed of 10.7 rad/s with her arms outstretched. H

Question

A skater spins with an angular speed of 10.7 rad/s with her arms outstretched. Her moment of inertia then is 2.60 kg m2. When she lowers her arms, she decreases her moment of inertia by a factor of 7.30 .

a) What will be her moment of inertia in kg m2 once she lowers her arms?

b) Ignoring friction on the skates, what is her angular momentum about her center of mass in kg m2/s then?

c) What will be her angular speed in rad/s then?

d) What is in Joules her kinetic energy about her center of mass with her arms outstretched?

e) What is in Joules her kinetic energy about her center of mass with her arms lowered?

For the assignment: Homework12-10
The question is:

A playground merry-go-round of radius 2.20 m has a moment of inertia 381 kg m2 and is rotating at 13.8 rev/min. A child with mass 42.9 kg jumps on the edge of the merry-go-round.

a) Assuming that the boy's initial speed is negligible, what is in kg m2/s the new angular momentum of the merry go round-child combination?

b) What is in kg m2 the new moment of inertia of the boy-merry go round combination?

c) What is in rev/min the new angular speed of the merry-go-round?

For the assignment: Homework12-11
The question is:

A wheel is rotating freely with an angular speed of 5200rad/s on a shaft whose moment of inertia is negligible. A second wheel, initially at rest and with 5 times the rotational inertia of the first is suddenly coupled to the same shaft.

a) What is the angular speed of the resultant combination of the shaft and two wheels? answer in units of rad/s.

b) What is the ratio of the final kinetic energy to the initial kinetic energy of the system?

Explanation / Answer

Skater spins at 10.7 rad/s
Moment of inertia= 2.60 kg/sq.m when arms outstretched

a) Moment of inertia when arms lowered = 2.60/7.30 = 0.356 kgm2

b) Her angular momentum = Iw = 10.7*2.60 = 27.82 kgm2rad/s

c) By conservation of angular monentum, final angular speed = 27.82/0.356 = 78.14 rad/s

d) Rotational KE = 0.5*I*w*w = 0.5*2.60*10.7*10.7 = 148.8 J

e) Rotational KE = 0.5*0.356*78.142 = 1086.84 J

Merry go round question

a)
New angluar momentum = old angular momentum = 381*13.8*2pi/60 = 550.31 kgm2/s

b) New moment of inertia = 381 + 42.9*2.20*2.20 = 588.636 kgm2

c) New angular speed = 550.31/588.6 = 0.93 rad/s = 0.93*60/2pi = 8.88 rev/min

question rotating wheel

Let moment of inertia of first wheel be I
second wheel = 5I
angular momentum is conserved
a) I*5200 = 6I*w (w = final ang velocity)
or w = 5200/6 = 866.6 rad/s

b) FKE/IKE = 6*866.6*866.6/(5200*5200) = 1/6
Required ration 1:6

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