Pullin a string to add energy to a wheel

In summary, to find the final rotational speed of a bicycle wheel after a string has been pulled through a distance L with a constant force F, you can use the energy method by calculating the initial rotational kinetic energy, the work done by the applied force, and adding them together to find the final kinetic energy. From there, you can solve for the final rotational speed using the equation: {KE}_{final} = 1/2 I \omega_{final}^2.
  • #1
thinktank75
19
0
Suppose that after a certain time t_L, the string has been pulled through a distance L. What is the final rotational speed w_final of the wheel?
Express your answer in terms of L, F, I, and w_0

Extra Info:
w_0 = angular velocity
k = dimensionless constant
m = mass
r = radius

Moment of Inertia = Ikmr^2

QUESTION 1
w_final = ?

QUESTION 2
What is the instantaneous power P delivered to the wheel via the force (F) at time t = 0?

P = ?



I have no clue how to approach the question, help would be much appreciated :smile:
 
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  • #2
Post the problem exactly as given.

Is the string pulled with a constant force? If so, calculate the work done by that force. (Or calculate the angular acceleration using Newton's 2nd law for rotation.)
 
  • #3
Doc Al said:
Post the problem exactly as given.

Is the string pulled with a constant force? If so, calculate the work done by that force. (Or calculate the angular acceleration using Newton's 2nd law for rotation.)

Sorry:
A bicycle wheel is mounted on a fixed, frictionless axle, with a light string wound around its rim. The wheel has moment of inertia I = k m r^2, where m is its mass, r is its radius, and k is a dimensionless constant between zero and one. The wheel is rotating counterclockwise with angular velocity w_0, when at time t=0 someone starts pulling the string with a force of magnitude F. Assume that the string does not slip on the wheel.

How exactly do I calculate the angular acceleration? If the velocity is constant, isn't accleration = 0? :blushing:
 
  • #4
thinktank75 said:
How exactly do I calculate the angular acceleration?
Use Newton's 2nd law for rotation. (Look it up! It relates torque to rotational inertia and angular acceleration.) Even easier would be to calculate the change in rotational KE of the wheel by figuring out the work done on it.

If the velocity is constant, isn't accleration = 0?
Sure, but the angular velocity is not constant!
 
  • #5
Doc Al said:
Use Newton's 2nd law for rotation. (Look it up! It relates torque to rotational inertia and angular acceleration.) Even easier would be to calculate the change in rotational KE of the wheel by figuring out the work done on it.


Sure, but the angular velocity is not constant!

Thank you soo much:smile: I'm going to try it out right now! It's making more sense now :smile:
 
  • #6
I got:

W_final: FL + (I*W^2 / 2)

P = Fr * w

but they're both wrong, I don't know what's wrong with the responses... please help...
 
  • #7
Not sure what those equations are, but try one of these:

Using energy methods: The work done by the force equals the change in rotational KE.

Using forces: The torque produces an angular acceleration.

Using Impulse: The angular impulse equals the change in angular momentum.
 
  • #8
my book says:
for the w_final:

i need to find the initial KE of rotation
Iw^2
------
2

and I need the work done by the applied force:

w = F*L
then they said to find the final kinetic energy, you add those 2 both together.

now that I have the final KE how to I find the final rotational speed?

thanks
 
  • #9
thinktank75 said:
my book says:
for the w_final:

i need to find the initial KE of rotation
Iw^2
------
2

and I need the work done by the applied force:

w = F*L



then they said to find the final kinetic energy, you add those 2 both together.
All good. This is exactly what I suggested when I recommended the "energy method".

now that I have the final KE how to I find the final rotational speed?
You know how rotational KE relates to angular speed--you've written that expression twice:
[tex]{KE}_{final} = 1/2 I \omega_{final}^2[/tex]

Just solve for [tex]\omega_{final}[/tex].
 

Related to Pullin a string to add energy to a wheel

1. How does pulling a string add energy to a wheel?

When you pull a string attached to a wheel, you are exerting a force on the wheel. This force transfers energy to the wheel, causing it to rotate and storing potential energy in the wheel's rotational motion.

2. What type of energy is added to the wheel when pulling a string?

The energy added to the wheel when pulling a string is potential energy. This is because the force applied to the wheel causes it to rotate, and the stored energy is in the form of potential energy due to its position and motion.

3. Does the length of the string affect the amount of energy added to the wheel?

Yes, the length of the string does affect the amount of energy added to the wheel. The longer the string, the more force is required to pull it and the more energy is transferred to the wheel.

4. Can pulling a string add energy to a wheel indefinitely?

No, pulling a string cannot add energy to a wheel indefinitely. Eventually, the string will reach its maximum length and the force applied will not be enough to overcome the friction and resistance of the wheel. The energy added will decrease until it reaches equilibrium.

5. How can the added energy be used from pulling a string on a wheel?

The added energy from pulling a string on a wheel can be used to power a variety of mechanisms, such as a toy car or a clock. It can also be converted into other forms of energy, such as electrical energy through a generator, for practical use.

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