# Relating SHM and Rotational Motion

1. Mar 27, 2015

### laurenm02

1. The problem statement, all variables and given/known data
A spring of stiffness k is attached to a wall and to the axle of a wheel of mass m, radius R, and moment of inertia I = βmR^2 about its frictionless axle. The spring is stretched a distance A and the wheel is released from rest. Assume the wheel rolls without slipping.

At some moment, the horizontal component of the spring force on the wheel is Fx, fid the magnitude and direction of:

The friction force
The acceleration of the wheel's center of mass
The angular acceleration of the wheel about its CM

2. Relevant equations
F = -kx
τ = Iα
E = (1/2)kx^2

3. The attempt at a solution
I'm really not even sure how to start this problem and how to set it up. I've always struggled with rotational motion intuitively, so I'm having trouble relating it to SHM.

2. Mar 28, 2015

### ehild

Draw a picture. What forces and torque act on the wheel?

What is the condition for rolling?

3. Mar 28, 2015

### laurenm02

For the forces, I have the force from the spring starting from the center of the wheel and extending horizontal.
I have the weight of the wheel starting from the center of the wheel and extending downward.
I have the force of friction at the point of contact between the wheel and the floor (but which direction would the force extend since the wheel changes direction with the spring's oscillations?
I have the normal force from the floor and extending upwards.

Because the force of the spring passes through the axle of the wheel, I figured that there was no external torque. Is this a correct assumption?

4. Mar 28, 2015

### ehild

The friction opposes the realtive motion of the surfaces in contact.

And what about the force of friction?

5. Mar 28, 2015

### laurenm02

Ahh, right, I forgot about the friction in torque.

So Στ = F(friction) = Iα
F(friction) = βmR^2 * α

But how do I connect that back to the Fx the question is asking for?

6. Mar 28, 2015

### Suraj M

At the point where force by spring is kx.
the energy stored in the spring is equal to the sum of rotational energy and KE of the wheel(CM)... get an equation first.

7. Mar 28, 2015

### ehild

he torque.
Iα is equal to the torque, not to the force. What is the torque of the friction?

You have to apply the condition of rolling. What is it?

8. Mar 28, 2015

### ehild

Suraj, you know, don't you, that it is wrong.

9. Mar 28, 2015

### laurenm02

Is it F(friction)*R = βmR^2 * α
So F(friction) = BmRα?

Then what?

10. Mar 28, 2015

### Suraj M

yeah, i realised when you guys were talking about friction. I didn't read that part of that question. Sorry.. the floors yours. :)

11. Mar 28, 2015

### ehild

What is the condition for rolling?

12. Mar 28, 2015

### laurenm02

Do you mean rolling without slipping? Where VCM = R*ω and aCM = R*α ?

13. Mar 28, 2015

### ehild

Yes, that is.

Write up the equation for the acceleration of the CM and that for the angular acceleration, then use the relation aCM = R*α

14. Mar 28, 2015

### laurenm02

So aCM = -kx/m, and because x = A in this problem, then my answer for the wheel's center of mass acceleration is simply aCM = -kA/m

And because aCM = R*α, then the angular acceleration of the wheel α = aCM / R, so α = (-kA)/(mR)

And substituting this relationship into the Ffr = βmRα equation, then Ffr = βm*aCM, which simplified is Ffr = -βkA?

Are these correct?

15. Mar 28, 2015

### ehild

No, you forgot the friction. And the wheel moves, so x is not A always.

16. Mar 28, 2015

### laurenm02

Where would I incorporate the friction?

17. Mar 28, 2015

### ehild

Among the forces. Newton's second law. The sum of forces = mass times acceleration.

18. Mar 28, 2015

### laurenm02

If the friction affects both the torque equation and the second law equation, how do I incorporate both into my final answer?

19. Mar 28, 2015

### ehild

Write both equations and you will see.

20. Mar 28, 2015

### laurenm02

so Στ = Ffr * R = I*α
and ΣF = Fx - Ffr = macm
solving for acm = (Fx - Ffr) / m

and plugging this into the torque equation Ffr = βmacm = β(Fx - Ffr)

so Ffr = (βFx)/(1 + β)

?