Homework Help: Conical Pendulum Problem

1. Nov 21, 2007

tachu101

1. The problem statement, all variables and given/known data

Conical Pendulum Question

Mass (m) is attached to the ceiling by a String of Length (s)
The string makes an angle of ($$\theta$$) with the vertical

Compute the Period of Revolution and the Tension in the string?

Mass of Bob (m) -- 4.35kg
Length of String (s) -- 5.50 meters
$$\theta$$ -- 64 degrees

2. Relevant equations

1/2(mv^2)$$/$$radius -- Centripetal Force

W= mg -- weight

Not sure what else I need

3. The attempt at a solution

I think that the tension in the string would be

Tcos($$\theta$$)=mg ---- T=(mg)$$/$$cos($$\theta$$)

So

(4.35)(9.81)$$/$$cos(64)= ---- 97.35 N

I have no idea how to get the period though, but I think it has to do with

Tsin$$\theta$$= 1/2(mv^2)$$/$$radius

2. Nov 21, 2007

Staff: Mentor

So far, so good.

Once you find the speed you can use that to find the period. What's the radius of the bob's path?

3. Nov 21, 2007

tachu101

I think that the radius would be

string length --- (s) sin (Theta) --- so 5.5sin(64)= 4.94m

I am not sure the equation to find the period though?

I have radius, string length, tension in string, and mass of the Bob.

4. Nov 21, 2007

Staff: Mentor

Good.
The period is just the time the bob takes to make one complete revolution. Find the speed (and figure out the distance).

5. Nov 21, 2007

tachu101

Would I find the speed by using Tsin(theta)= (1/2)(mv^2)/radius ----

97.35 sin (64) = (1/2)(4.35)(v^2)/(4.94) ----- velocity = 14.097 m/sec ?

so the distance would be (2)(radius)(pi)= (2)(4.94)(pi)= 31.03m

I am stuck now on how to find the period. Is there a certain equation that is used to find the period, we just started this topic?

6. Nov 21, 2007

Staff: Mentor

How about distance = speed x time?

7. Nov 21, 2007

tachu101

So (2)(pi)(r)=(velocity)(time)

thus

31.03m=(14.1m/sec)(t) ---- so ---- period= 2.2 sec?

8. Nov 21, 2007

Staff: Mentor

Sure. It's just the time required for one revolution. You have the distance and the speed--that's all there is to it.

9. Nov 21, 2007

tachu101

thank you so much for the help, I am going to go back and check all of my work.