A Spinning Top and Rotation Dynamics

  • #1

Homework Statement



A spinning top has a moment of inertia of 0.0004 kg*m^2 and is initially at rest. It is free to rotate about a stationary axis. A string, wrapped around a peg along the axis of the top, is pulled in such a manner as to maintain a constant tension of 5.57 N in the string.

If the spring does not slip while wound around the peg, what is the angular speed of the top after 80 cm of string has been pulled off the peg?

Homework Equations



v = rω

τ = Fl = Iα

Rotation kinetic energy = 1/2Iω^2

The Attempt at a Solution



τ = Fl = Iα
(5.57 N)(0.80 m) = (0.0004)α
α = 11140 rad/s^2

I don't know where to go from here.
 

Answers and Replies

  • #2
Doc Al
Mentor
45,262
1,617
τ = Fl = Iα
(5.57 N)(0.80 m) = (0.0004)α
Careful: You are not given the radius of the peg. (The 80 cm is the length of string pulled off, not the radius of the peg.) Hint: Just call that radius "r" and continue.

Once you find α, use some kinematics to find the angular speed.
 
  • #3
I can't think of any equation that can give me angular speed from angular acceleration without time.
 
  • #4
(radius)(angular acceleration) = (radius)(angular speed)^2

^^^^Is that it?
 
  • #5
Doc Al
Mentor
45,262
1,617
(radius)(angular acceleration) = (radius)(angular speed)^2

^^^^Is that it?
No. Think of a kinematic formula relating speed, distance, and acceleration. The rotational analog of that formula is the one you want.
 
  • #6
The only equation I can think of is: w^2 = wi^2 + 2(alpha)(theta)

so:
w^2 = 2(5570 rad/s2)(.80)
w = 94.4 rad/s

BUT that is not the right answer!
 
  • #7
LowlyPion
Homework Helper
3,097
5
Is there a radius given for the peg, or are you supposed to consider that the work of pulling the string, is what develops the rotational kinetic energy?

Is what you are looking for then ...

F * d = 1/2* I * ω2
 
  • #8
YES!!!

Thank you very much. That equation has been in front of my eyes this whole time.
 
  • #9
Doc Al
Mentor
45,262
1,617
The only equation I can think of is: w^2 = wi^2 + 2(alpha)(theta)

so:
w^2 = 2(5570 rad/s2)(.80)
w = 94.4 rad/s

BUT that is not the right answer!
That equation is fine, but you used the wrong values for alpha and theta. Done right, this is equivalent to what LowlyPion suggested. (But setting work equal to KE directly is much easier! :wink:)
 

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