Relating SHM and Rotational Motion

In summary, a spring with stiffness k is attached to a wall and a wheel with mass m, radius R, and moment of inertia I = βmR^2. The spring is stretched a distance A and the wheel is released from rest, rolling without slipping. At a given moment, the horizontal component of the spring force on the wheel is Fx. The friction force, acceleration of the wheel's center of mass, and angular acceleration of the wheel about its center of mass can be found by applying Newton's second law and the condition for rolling without slipping. Taking into account the effects of friction, the equations for these quantities are: Ffr = (βFx)/(1 + β), acm = (Fx - Ffr)
  • #1
laurenm02
21
0

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.
 
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  • #2
Draw a picture. What forces and torque act on the wheel?

What is the condition for rolling?
 
  • #3
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
laurenm02 said:
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.

laurenm02 said:
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?
 
  • #5
ehild said:
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?
 
  • #6
laurenm02 said:
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.
 
  • #7
he torque.
laurenm02 said:
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?
 
  • #8
Suraj M said:
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.
 
  • #9
ehild said:
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?
 
  • #10
yeah, i realized when you guys were talking about friction. I didn't read that part of that question. Sorry.. the floors yours. :)
 
  • #11
laurenm02 said:
Is it F(friction)*R = βmR^2 * α
So F(friction) = BmRα?

Then what?
What is the condition for rolling?
 
  • #12
ehild said:
What is the condition for rolling?
Do you mean rolling without slipping? Where VCM = R*ω and aCM = R*α ?
 
  • #13
laurenm02 said:
Do you mean rolling without slipping? Where VCM = R*ω and aCM = 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 aCM = R*α
 
  • #14
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
laurenm02 said:
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

No, you forgot the friction. And the wheel moves, so x is not A always.
 
  • #16
Where would I incorporate the friction?
 
  • #17
laurenm02 said:
Where would I incorporate the friction?
Among the forces. Newton's second law. The sum of forces = mass times acceleration.
 
  • #18
ehild said:
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?
 
  • #19
Write both equations and you will see.
 
  • #20
ehild said:
Write both equations and you will see.

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)

?
 
  • #21
laurenm02 said:
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)

?
It looks all right now.
 

1. How are simple harmonic motion (SHM) and rotational motion related?

Simple harmonic motion and rotational motion are related in that both involve periodic motion around a fixed axis or point. In SHM, an object oscillates back and forth along a straight line, while in rotational motion, an object rotates around a fixed axis. Both types of motion can be described using similar equations and principles.

2. Can an object exhibit both SHM and rotational motion simultaneously?

Yes, an object can exhibit both SHM and rotational motion simultaneously. This is known as a compound pendulum, where an object is suspended from a pivot point and exhibits both oscillatory and rotational motion.

3. How can the equations for SHM be applied to rotational motion?

The equations for SHM can be applied to rotational motion by using analogies between linear and rotational quantities. For example, displacement in SHM can be related to angular displacement in rotational motion, and velocity in SHM can be related to angular velocity in rotational motion. By using these analogies, we can apply the equations for SHM to describe rotational motion.

4. What is the relationship between frequency and period in SHM and rotational motion?

In both SHM and rotational motion, frequency and period are inversely proportional. This means that as frequency increases, the period decreases, and vice versa. This relationship can be seen in the equations f=1/T and ω=2π/T, where f is frequency, T is period, and ω is angular frequency.

5. Can the concept of resonance be applied to rotational motion?

Yes, the concept of resonance can be applied to rotational motion. Just like in SHM, when a system is driven at its natural frequency, it experiences resonance. In rotational motion, this can be seen in systems like spinning tops or gyroscopes, where the applied torque causes them to rotate at their natural frequency, resulting in resonance.

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