A Spinning Top and Rotation Dynamics

In summary, a spinning top has a moment of inertia of 0.0004 kg*m^2 and is initially at rest. 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, the top will have an angular speed of 11140 rad/s^2.
  • #1
PrideofPhilly
37
0

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.
 
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  • #2
PrideofPhilly said:
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
PrideofPhilly said:
(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
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
PrideofPhilly said:
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:)
 

1. What is a spinning top?

A spinning top is a toy that typically consists of a pointed stem with a disc or ball attached at the top. When spun, the top rotates rapidly around its axis due to its angular momentum.

2. How does a spinning top work?

A spinning top works by converting the force of gravity into rotational energy. As the top spins, the force of gravity pulls the top towards the ground, causing it to rotate around its axis.

3. What is the role of rotation dynamics in a spinning top?

Rotation dynamics is the study of how objects rotate and the factors that affect their rotation. In the case of a spinning top, rotation dynamics helps us understand the forces and torque involved in keeping the top spinning.

4. What factors affect the stability of a spinning top?

The stability of a spinning top is affected by multiple factors, such as the shape and weight distribution of the top, the surface it is spinning on, and the speed at which it is spinning. These factors influence the center of mass and the torque acting on the top, which can affect its stability.

5. Can a spinning top ever defy gravity?

No, a spinning top cannot defy gravity. While it may appear to be defying gravity by spinning on its tip, it is actually being kept stable by the force of gravity pulling it towards the ground. If the top stops spinning, it will eventually fall due to the force of gravity.

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