This, is just a concept question:(adsbygoogle = window.adsbygoogle || []).push({});

My teacher gave the Answer has something like this: A mass is moving in a horizontal circle at the end of a conical pendulum, the string making an angle with the vertical of the cone of 25 degrees. Conditions are changed so that the angle the string makes with the vertical is doubled to 50 degrees.

By what factor does the tension increase?

By what factor does the frequency increase?

I tried to find the tension increase by finding the radius' of the formed circles by setting the length of the string equal to 1, and then dividing the radius of the bigger circle by the radius of the smaller one and got 1.8, but feel like that isn't right. I have no idea how to find the frequency change factor.

However I understand most of it, except the part where it says: The general solution for this problem is as follows:

From a free body diagram about the mass we get weight m*g acing down& T tension actiung up at angle theta

Now summing forces in the vertical we get T*cos(theta) - m*g = 0 or T = m*g/cos(theta)

In the horizontal we get T*sin(theta) . This is the centripetal force whcih equals m*a = m*v^2/r

Now r = L*sin(theta) where L is the length of the string.

So we have T*sin(theta) = m*v^2/r = m*v^2/L*sin(theta)

Now sub for T ..... m*g/cos(theta)*sin(theta) = m*v^2/(L*sin(theta))

Now v = 2*pi*L*sin(theta)/t circumference divided by period

Note m drops out leaving g*tan(theta) = 4*pi^2*L*sin(theta)/t^2

So solving for T we get ... t = sqrt(4*pi^2*L*cos(theta)/g)) If L = 1 then t = 1.92s..So f = 1/t

= 0.523Hz

When theta = 50 t = sqrt(4*pi^2*cos(50)/g) = 1.61s so f = 1/t = 0.621Hz

So the frequency increases by .621/.523 = 1.19

Now the Tension = m*g/cos(theta)

T50/T25 = cos(25)/cos(50) = 1.41

What is happening in this? Where does the 4*pi^2*L*sin(theta) come from, and how is v found? why is f = 1/t? Now v = 2*pi*L*sin(theta)/t circumference divided by period

Note m drops out leaving g*tan(theta) = 4*pi^2*L*sin(theta)/t^2

So solving for T we get ... t = sqrt(4*pi^2*L*cos(theta)/g)) If L = 1 then t = 1.92s..So f = 1/t

= 0.523Hz

**Physics Forums - The Fusion of Science and Community**

# The factor of tension and frequency increase in a conical pendulum?

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: The factor of tension and frequency increase in a conical pendulum?

Loading...

**Physics Forums - The Fusion of Science and Community**