A Spinning Top and Rotation Dynamics

Click For Summary

Homework Help Overview

The problem involves a spinning top with a specified moment of inertia and a constant tension in a string that affects its rotation. The goal is to determine the angular speed after a certain length of string has been pulled, requiring an understanding of rotational dynamics and kinematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between torque, angular acceleration, and angular speed, with some attempting to derive angular speed from angular acceleration without time. Others suggest using kinematic equations for rotational motion.

Discussion Status

There is an ongoing exploration of the correct equations to relate the variables involved. Some participants have offered guidance on using work-energy principles, while others are questioning the assumptions about the radius of the peg and the values used in calculations.

Contextual Notes

Participants note the absence of a specified radius for the peg, which is critical for solving the problem. There is also mention of the need to consider the work done in pulling the string as it relates to the rotational kinetic energy of the top.

PrideofPhilly
Messages
37
Reaction score
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.
 
Physics news on Phys.org
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.
 
I can't think of any equation that can give me angular speed from angular acceleration without time.
 
(radius)(angular acceleration) = (radius)(angular speed)^2

^^^^Is that it?
 
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.
 
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!
 
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
 
YES!

Thank you very much. That equation has been in front of my eyes this whole time.
 
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:)
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K