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AP Mechanics DiffEQ

  • #1
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Homework Statement



A stick of length 2L and negligible mass has a point mass m affixed to each end. The stick is arranged so that it pivots in a horizontal plane about a frictionless vertical axis through its center. A spring of force constant k is connected to one of the masses. The system is in equilibrium when the spring and stick are perpendicular. The stick is displaced through a small angle theta.
a) Determine the restoring torque when the stick is displaced from equilibrium through the small angle theta
b) Determine the magnitude of the angular acceleration of the stick just after it has been released
c) Write the differential equation whose solution gives the behavior of the system after it has been released
d) Write the expression for the angular displacement theta of the stick as a function of time t after it has been released from rest


theta = displacement angle, phi=angle between stick and spring (complement)
Torque = f X d
f=kLsin(theta)
a) torque = [kLsintheta]*Lsin(phi) = kL2sin(theta)cos(theta)
b) alpha = torque/moment of inertia = [kL2sin(theta)cos(theta)]/[2mL2]
alpha=ksin(theta)cos(theta)/[2m] = k/[4m]*sin(2theta)

Im pretty sure about this, but now what do they mean by "behavior of the system" (displacement?, velocity?) and how do I get things in terms of t?

My attempt:
c) alpha=k/[4m]*sin(2theta) = d2theta/dt2
d2theta/sin(2theta) = k/[4m]*dt2
Am I supposed to integrate that twice? It seems way too intense for an AP to integrate ln[abs(tantheta)]dtheta. Is there a mistake anywhere?

Homework Equations


Torque = f X d
alpha = torque/moment of inertia

The Attempt at a Solution



theta = displacement angle, phi=angle between stick and spring (complement)

f=kLsin(theta)
a) torque = [kLsintheta]*Lsin(phi) = kL2sin(theta)cos(theta)
b) alpha = torque/moment of inertia = [kL2sin(theta)cos(theta)]/[2mL2]
alpha=ksin(theta)cos(theta)/[2m] = k/[4m]*sin(2theta)

Im pretty sure about this, but now what do they mean by "behavior of the system" (displacement?, velocity?) and how do I get things in terms of t?

My attempt:
c) alpha=k/[4m]*sin(2theta) = d2theta/dt2
d2theta/sin(2theta) = k/[4m]*dt2
Am I supposed to integrate that twice? It seems way too intense for an AP to integrate ln[abs(tantheta)]dtheta. Is there a mistake anywhere?
 

Answers and Replies

  • #2
169
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Just one question: spring is in the horizontal plane, or is it vertical?
 
  • #3
312
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Horizontal plane.

For anyone interested, I looked up the solutions, and they were able to simplify it greatly with the small angle approximations sin(theta)=theta and cos(theta)=1

Then the torque is -kL2theta, the acceleration is k/[2m]. Then they solved it in one magic step to get theta=theta0cos[sqrt(k/(2m))t]
 
  • #4
169
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Well, I think you have understood the Physics part, it is just the mathematical part which is troubling you. Assuming that here is a brief of the solution:
Using, [tex]\vec{t}[/tex] = [tex]\vec{R}[/tex]X[tex]\vec{F}[/tex],

[tex]\tau[/tex]restoring = l*(k*l*SinΘ)*Sin(90+Θ) = -kl²SinΘCosΘ

For small Θ, SinΘ [tex]\approx[/tex] Θ and CosΘ [tex]\approx[/tex] 1

Therefore, [tex]\tau[/tex]restoring [tex]\approx[/tex] -kl²Θ

Now, [tex]\tau[/tex] = Ia, where I is moment of inertia of the system and a is angular acceleration.
Here, I = 2mL²

Thus, a = -(k/2m)Θ

Note that a = d²Θ/dt²

If you have gone through SHM, you would be knowing that Sine and Cosine functions satisfy this differential equation.

Thus, Θ = ΘoSin(wt + Φ)

where, w² = k/2m, and Φ is phase difference and can be determined from boundary conditions. Here, at t = 0, Θ = Θo, wherefrom Φ = 90

Thus, Θ = ΘoCos(wt)

Basically, it's not magical, it's procedural. :cool:
 
Last edited:

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