Relating SHM and Rotational Motion

[laurenm02]

Homework Statement

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

Homework Equations

F = -kx
τ = Iα
E = (1/2)kx^2

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.

 

[ehild]

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

What is the condition for rolling?

 

[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?

 

[ehild]

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?

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

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?

And what about the force of friction?

 

[laurenm02]

The friction opposes the realtive motion of the surfaces in contact.And what about the force of friction?

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?

 

[Suraj M]

The acceleration of the wheel's center of mass

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.

 

[ehild]

he torque.

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?

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?

 

[ehild]

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.

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

 

[laurenm02]

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?

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

Then what?

 

[Suraj M]

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

 

[ehild]

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

Then what?

What is the condition for rolling?

 

[laurenm02]

What is the condition for rolling?

Do you mean rolling without slipping? Where V_CM = R*ω and a_CM = R*α ?

 

[ehild]

Do you mean rolling without slipping? Where V_CM = R*ω and a_CM = R*α ?

Yes, that is.

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

 

[laurenm02]

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

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

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

Are these correct?

 

[ehild]

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

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

 

[laurenm02]

Where would I incorporate the friction?

 

[ehild]

Where would I incorporate the friction?

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

 

[laurenm02]

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

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

 

[ehild]

Write both equations and you will see.

 

[laurenm02]

Write both equations and you will see.

so Στ = F_fr * R = I*α
and ΣF = F_x - F_fr = ma_cm
solving for a_cm = (F_x - F_fr) / m

and plugging this into the torque equation F_fr = βma_cm = β(F_x - F_fr)

so F_fr = (βF_x)/(1 + β)

?

 

[ehild]

so Στ = F_fr * R = I*α
and ΣF = F_x - F_fr = ma_cm
solving for a_cm = (F_x - F_fr) / m

and plugging this into the torque equation F_fr = βma_cm = β(F_x - F_fr)

so F_fr = (βF_x)/(1 + β)

?

It looks all right now.