A thin, 80.0g disk with a diameter of 7.00cm rotates about an axis through its center with 0.260J of kinetic energy.What is the speed of a point on the rim?
2007-11-04 18:21:12 · 7 answers · asked by smart-crazy 4 in Science & Mathematics ➔ Physics
Use this formula:
KE = 1/2 I ω²
I is the moment of inertia. You'll have to look up the formula for this. It is different for every shape. It should be in your textbook. I don't remember it. ω is the angular velocity. The speed of a point at the edge is v = ωr.
2007-11-04 18:36:38 · answer #1 · answered by Demiurge42 7 · 1⤊ 0⤋
The kinetic energy of the disk is : K = (1/2) I w^2
I = (1/2) m r^2 = 40*10^-3 *(3.5*10^-2)^2= 4.9*10^-5 Kg m^2.
w^2 = 2K/I ----> w = 103 rad/s.
The speed requested is : v = w r = 3.6 m/s.
2007-11-04 19:47:16 · answer #2 · answered by Luigi 74 7 · 0⤊ 0⤋
The total kinetic energy is the sum of linear kinetic energy and rotational kinetic energy. so use the relation
KE = mv^2/2 + Iw^2/2 I = mr^2/2 and w = v/r so
KE = mv^2/2 + mr^2/2(v/r)^2/2 = mv2
Just be sure the moment of inertia of a disk.
2007-11-04 19:12:42 · answer #3 · answered by Beku 1 · 0⤊ 1⤋
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2016-10-15 02:17:47 · answer #4 · answered by ? 4 · 0⤊ 0⤋
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2007-11-04 18:32:40 · answer #5 · answered by Anonymous · 0⤊ 7⤋
J=1/2 m r^2
J=moment of inertia
r=radius=3.5*10^-2 m
m=mass=80*10^-3 kg
E=1/2 J w^2 --->w=Sqrt[2E/J]
w=angular velocity
w=v/r
--->v/r=Sqrt[2E/J] --->v=r*Sqrt[2E/J]=r*Sqrt[2*2E/mr^2]
v=3.61 m/s
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2007-11-04 19:25:04 · answer #6 · answered by Xenophon 3 · 0⤊ 0⤋
kinetic energy=1/2mw^2(w is angular speed)
1/2*.08*w^2=.26
so w=2.55rad/s
again linear speed=angular speed*radius.
so linear speed=2.55*.035 m/s=.0895m/s=8.95cm/s(approx)
2007-11-04 18:29:29 · answer #7 · answered by Anonymous · 0⤊ 2⤋