# Simple pendulum part

1. Jan 26, 2004

### Rockdog

A simple pendulum of mass 15 kg with a period of 1.8 sec is displaced an angle of 10 degrees from the vertical. What would its acceleration be in the vertical (y) direction as it reachs the lowest point on its swing?
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Ok, I understand that it is a simple pendulum, but how do I figure out the acceleration?

Some help in the right direction would be helpful. I bet this is a simplistic problem, and I'm gonna hit myself in the head when I figure out what it is.

2. Jan 26, 2004

### Staff: Mentor

Hint: the pendulum is centripetally accelerated. Start by figuring out the length of the pendulum.

3. Jan 26, 2004

### Rockdog

Ok. R=2*pi sqrt(L/g)
L=.805m

I know that centripetal acceleration equals a = V^2/R

or can be rewritten as a= w^2*R

w equals 2*pi/T
w=3.49
T is the period or 1.8 seconds.

so a=3.49^2*.805 =>9.80 m/sec^2

Does that appear right? The computer says no, but all my calculations say yes.

4. Jan 26, 2004

### Staff: Mentor

Good.
Good.
No! That would only be true for circular motion. The speed (or w) changes as the mass falls. Find the speed at the bottom. (Use conservation of energy.)

5. Jan 27, 2004

### Rockdog

Ok, so use conservation of energy.
It starts off with potential energy, and then it gets converted to kinetic energy.

So it be mgh=.5mv^2
mass cancel out

so I'm left with gh=.5v^2
h=.805 m or h=.805m/cos 10

and I get v=3.972m/s

Then use a= V^2/R ?

Is this the right approach?

6. Jan 27, 2004

### Staff: Mentor

Yes, right approach. But be sure to use the correct value for h, measured from the lowest point.

7. Feb 2, 2004

### Rockdog

Ok, maybe I'm missing something obvious. When the pendulum is directly vertical, isn't h going to be same thing as the length of the string?

8. Feb 2, 2004

### Staff: Mentor

No. "h" is the vertical distance the pendulum moves from initial position to final position. Think of h as &Delta;y. When the pendulum is at its lowest point (vertical) call it y = 0. Find the intial value for y.

9. Feb 2, 2004

### HallsofIvy

Staff Emeritus
And remember that the pendulum was displaced only 10 degrees initially- not vertically above the lowest point.

10. Jul 15, 2004

### RunsWithKnives

This doesn't work.. I have a similar problem to work on.. I figured out my h by using trig, and even verified it by testing it in AutoCad... I also triple checked my units...

i put it into gh*2=v^2

and then took my v^2 and put it into a=v^2/2

and I'm still getting a wrong answer... the fundamental setup seems right but I think we're missing something big

11. Jul 15, 2004

### Staff: Mentor

So far, so good.

a = v^2/R, where R will equal the length of the pendulum.

If you still can't get it, show your work and we'll check it out.

12. Jul 19, 2004

### rayjohn01

Dare I suggest that there is NO vertical accelleration at the bottom of swing , but there is a vertical force f = m.V^2 / r ( centrifugal ) + m.g ( gravity) where v is the peripheral speed m/sec , r is radius , g is gravity m/s/s , these cause tension in the rod , but it is only the peripheral speed which keeps the pendulum swinging because it has kinetic energy at this point Ek = 1/2 . m. v^2 .
You did not say where exactly the mass was i.e. at the end of the pendulum or distributed --- r in the above means the distance to the center of mass if that's at the end of the rod then r is the pendulum length.

13. Jul 19, 2004

### Staff: Mentor

There is no problem viewing the pendulum from the usual inertial frame, in which case there is a centripetal acceleration (acting vertically) at the bottom of the swing. (Only when viewed from an accelerated frame would you introduce a "centrifugal" force.)

It is certainly true that it is the speed of the pendulum at the bottom of its swing that keeps it moving.
It's a simple pendulum. The mass is at the end of the rod.

14. Jul 19, 2004

### RunsWithKnives

I was getting the right answer all along.. I actually did the problem right myself before I started searching for help... I didn't notice that it was asking for the answer in cm.. sorry guys.. thanks for the help...