Solving Pendulum Dynamics on a Rotating Platform

In summary, the tension in the cord is pulling down on the bob, while the weight of the bob is pushing up on the cord. The vertical component of the weight is constantly pulling down on the bob, while the horizontal component of the weight is constantly pushing up on the bob. The pendulum keeps swinging back and forth because the net force (the total of the tension and weight) is always pushing it in a straight line.
  • #1
pendulum! help!

Homework Statement


A pendulum 2 meters long with a mass of 1kg is mounted on a circular platform on the Earth's surface that's spinning at constant angular velocity. The pendulum is mounted on a pole that's perpendicular to the platform at a distance of 5 meters from the center of rotation. The equilibrium angle the pendulum makes with the pole is 30 degrees.

1) What's the angular velocity at equilirium?

2) If it's displaced for its equilibrium position, what will be the period of the pendulum?


Homework Equations


a=F/m
a=v2/r
F=mv2/r
http://answerboard.cramster.com/Answer-Board/Image/cramster-equation-2007422035166331114291678125006410.gif [Broken]

P.S. My physics teacher said that the diagrams below explains what's happening:

FROM A BIRD'S EYE VIEW-
http://answerboard.cramster.com/Answer-Board/Image/200743163856633112151365000000302.jpg [Broken]

FROM A NORMAL VIEW-
http://answerboard.cramster.com/Answer-Board/Image/2007431724446331121788484375005079.jpg [Broken]

The Attempt at a Solution



1) R=5+r1
R=5+1= 6

h= l(1-Cos θ)
= 2(1-Cos 30degrees)
= .2679

http://answerboard.cramster.com/Answer-Board/Image/cramster-equation-20074317292633112181428593750754.gif [Broken]
v=2.29 m/sec

ω=V/r
ω =2.29/1= 2.29 rad/sec

2) Now, my teacher said something along the lines of that I have to replace the acceleration due to gravity (g) in the period equation with an acceleration. Ergo, the equation would appear like this: http://answerboard.cramster.com/Answer-Board/Image/cramster-equation-2007431735306331121853068750002401.gif [Broken].[/URL] Now I know that it doesn't seem right, but take in mind that the circular platform is moving, not the pendulum its self.
So, I am not sure which acceleration I should use since I somehow have 3:
1)F=mv2/r a=f/m
F=1(2.29)2/6= .874 N a=.874 m/sec2
2)a1: Sin θ= opposite/hyp.
Sin 30degrees(9.8)= 4.9 m/sec2
3)a2: Cos θ=adjacent/hyp.
Cos 30degrees(9.8)= 8.49 m/sec2

Do I even use an acceleration in place of g or just use g??
 
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  • #3
ataglance05 said:
1) R=5+r1
R=5+1= 6
This I understand. But the rest I don't.

h= l(1-Cos θ)
= 2(1-Cos 30degrees)
= .2679

http://answerboard.cramster.com/Answer-Board/Image/cramster-equation-20074317292633112181428593750754.gif [Broken]
v=2.29 m/sec

ω=V/r
ω =2.29/1= 2.29 rad/sec
What does "h" represent? That equation relating speed to height has to do with the energy of a falling body--not relevant here!

Instead, analyze the forces acting on the pendulum bob. There are only two: The tension in the cord and the weight of the bob. Hint: Analyze vertical and horizontal components separately. Apply Newton's 2nd law.
 
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