A solid disk (I=1/2mr 2 ) with an initial velocity of 5m/s rolls without slippin
ID: 1553430 • Letter: A
Question
A solid disk (I=1/2mr2) with an initial velocity of 5m/s rolls without slipping up an incline of height h=.75m.
12) What is the velocity of the disk at the top of the incline?
v = 1.0 m/s
v = 2.1 m/s
v = 3.9 m/s
v = 4.3 m/s
A person is holding two 3kg weights at a distance .75m from the center of her body. She is initially rotating at .3 rad/s.
13) If she pulls the weights in so that they are only .25m from the center of her body, What is her new angular velocity (Ignore her moment of inertia)?
w = 1.0 rad/s
w = 1.4 rad/s
w = 2.7 rad/s
w = 3.2 rad/s
14) How much work did she do in pulling the weights inward?
1.2 J
2.4 J
3.6 J
4.8 J
6.0 J
15) She now lets the weights return to their initial position. What is the final angular momentum of the weights?
.51 kg-m2/s2
.68 kg-m2/s2
.82 kg-m2/s2
1.0 kg-m2/s2
1.2 kg-m2/s2
Vi F 5 m/s h 0.75 mExplanation / Answer
12.
As Per the law of conservation of energy
[Kinetic Energy+Potential Enegy]initial = [Kinetic Energy+Potential Enegy]final
= 1/2mu2 + 1/2I12 + 0 = 1/2mv2 + 1/2I22 + mgh
= 1/2mu2 + 1/2 x 1/2mr2 x (u/r)2 = 1/2mv2 + 1/2 x 1/2mr2 x (v/r)2 + mgh
= 2u2 + u2 = 2v2 + v2 + 4gh
=3u2 = 3v2 + 4gh
=3v2 = 3 x (5)2 - 4 x 9.8 x 0.75
=v = 15.2
v = 3.90 m/s
13.
Since there is no external torque applied to the system, the angular momentum is conserved, this is
L = Io wo = If wf
where Io,f and wo,f are the initial/final moments of inertia and angular velocities of the system. Solving for wf we have
wf = Io wo / If = (2X3X(0.75)2 / (2X3X(0.25)2 )X(0.3) = 2.7 rads/sec
14.
The work done by the skater is the difference between the final and initial rotational energies, this is
W = 1/2 If wf2 - 1/2 Io wo2 = 1/2X2X3X(0.25)2X(2.7)2 - 1/2X2X3X(0.75)2X(0.3)2 = 1.215 J
15.
The angular momentum is always the same during in the entire motion done by the skater, this is
L = Io wo = If wf = 2X3X(0.75)2 X (0.3) = 1.0125 kg m2/s
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