Pullin a string to add energy to a wheel

[thinktank75]

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

 

[Doc Al]

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.)

 

[thinktank75]

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?

 

[Doc Al]

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!

 

[thinktank75]

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 I'm going to try it out right now! It's making more sense now

 

[thinktank75]

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...

 

[Doc Al]

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.

 

[thinktank75]

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

 

[Doc Al]

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:
$${KE}_{final} = 1/2 I \omega_{final}^2$$

Just solve for $$\omega_{final}$$.