What is the maximum speed of the pendulum?

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  • #1
ctwokay
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Homework Statement



A simple pendulum with mass m = 1.7 kg and length L = 2.42 m hangs from the ceiling. It is pulled back to an small angle of θ = 8.6° from the vertical and released at t = 0.

Qn: What is the maximum speed of the pendulum?

Homework Equations



θ=θmax*cos(ωt+)

The Attempt at a Solution



i derivative the equation i got θ=θmax*ω*sin(ωt)
the ω outside i know is angular velocity √(g/L) but inside ω is √(k/m)
where do i find k constant in this question?
 

Answers and Replies

  • #2
Doc Al
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i derivative the equation i got θ=θmax*ω*sin(ωt)
the ω outside i know is angular velocity √(g/L) but inside ω is √(k/m)
where do i find k constant in this question?
ω is the angular frequency. Your first expression is correct for a simple pendulum. The second is for a mass on a spring. Both are examples of SHM.

But none of that is needed to answer the question. Hint: Consider the mechanical energy of the system.
 
  • #3
ctwokay
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using PE and KE? i can't use that expression? as the prof said need to use that to find
 
  • #4
Doc Al
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using PE and KE?
Sure.
i can't use that expression? as the prof said need to use that to find
Sure you can. (I missed that you found an expression for angular speed.)
i derivative the equation i got θ=θmax*ω*sin(ωt)
That's really an expression for [itex]\dot{\theta}[/itex], not θ. Once you have the max value of [itex]\dot{\theta}[/itex], use it to find the max speed.
 
  • #5
ctwokay
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ya its θ˙ first derivative i can't find the symbol but the problem is how do i find ω inside of sin? as ω outside of sin is angular velocity which is dθ/dt
 
  • #6
Doc Al
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ya its θ˙ first derivative i can't find the symbol but the problem is how do i find ω inside of sin? as ω outside of sin is angular velocity which is dθ/dt
That's not true. Both ω's are the same: The angular frequency (not angular speed) of a pendulum. ω is given by the formula you posted for a pendulum.
 
  • #7
ctwokay
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is there a phase angle?
i put 8.6* √(9.81/2.42)*sin(√(9.81/2.42)) is that right or i have to include t also?
 
  • #8
netgypsy
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I can't quite follow what you're doing so a few pointers. You can find the period of the pendulum from the angular velocity omega. Remember the period is one full back and forth oscillation so you find the time for your equation. Be careful with your angles and angular velocity. I have no idea what is done now but we always used theta radians (not degrees - degrees times 2pi), omega radians per sec, and so on with omega being frequency (cycles per second) times 2pi and of course frequency is the inverse of the period of the pendulum. And intuitively you should know that the maximum angular velocity and linear velocity is at the bottom where the angle of displacement is zero. You can prove this of course but it's good to know it already.
 
  • #9
Doc Al
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is there a phase angle?
If you measure time from when it was let go, then you don't need a phase angle.
i put 8.6* √(9.81/2.42)*sin(√(9.81/2.42)) is that right or i have to include t also?
No, not quite right. You have sin(ω) instead of sin(ωt). You could include t, but you want the maximum speed. What's the maximum value of sinωt?

Also, be sure your angles are in radians.
 
  • #10
ctwokay
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ok i converted the θ to rad,
the equation i put 8.6*2*pi*√(9.81/2.42)*sin(√(9.81/2.42)t) so sinωt is the max which is at the period of oscillation?
i have the period but it is release at t=0 i can't use t=0 as it is a reference when it is release, so t i have to use some other equation?
 
  • #11
Doc Al
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ok i converted the θ to rad,
the equation i put 8.6*2*pi*√(9.81/2.42)*sin(√(9.81/2.42)t) so sinωt is the max which is at the period of oscillation?
i have the period but it is release at t=0 i can't use t=0 as it is a reference when it is release, so t i have to use some other equation?
You want the maximum angular speed. Since the angular speed is proportional to sinωt, the maximum speed will be when sinωt is maximum. What's the maximum value of sinωt?
 
  • #12
ctwokay
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You want the maximum angular speed. Since the angular speed is proportional to sinωt, the maximum speed will be when sinωt is maximum. What's the maximum value of sinωt?

Erm is it ω=2∏/T=2∏f?
I'm sorry i am very bad at this.
 
  • #13
Doc Al
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Erm is it ω=2∏/T=2∏f?
No, nothing to do with that.

If you plotted the sine function (sinθ), what maximum value will it have?
 
  • #14
ctwokay
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either 1 or -1?
 
  • #15
Doc Al
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either 1 or -1?
Good! +1 is the maximum value of the sine function.

So, the maximum value of sinωt will also be +1. Use that in your formula for dθ/dt to find its maximum value.

And once you have the maximum angular speed, how will you find the linear speed?
 
  • #16
netgypsy
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And you can check yourself by using the energy equivalence to calculate the same value and see if they agree.
 
  • #17
ctwokay
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linear speed is v=ωr
 
  • #18
Doc Al
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linear speed is v=ωr
Good. In this case we're already using ω to stand for angular frequency. So rewrite this one as: v = (dθ/dt)r.
 

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