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

In summary, the tension increases by a factor of 1.41 when the angle of the string is doubled from 25 degrees to 50 degrees in a conical pendulum. The frequency also increases by a factor of 1.19 due to the change in angle. The equations used to find the velocity, period, and frequency are derived from the fact that the circumference of the circle is equal to the total distance traveled by the mass in 1 period, and the fact that the centripetal force acting on the mass is equal to m*a, or m*v^2/r.
  • #1
PhoniexGuy
26
0
This, is just a concept question:

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.

My teacher gave the Answer has something like this:

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

However I understand most of it, except the part where it says:

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

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?
 
Physics news on Phys.org
  • #2
The equation v = 2*pi*L*sin(theta)/t is derived from the fact that the circumference of the circle is equal to the total distance traveled by the mass in 1 period. Since the length of the string is 1, then the circumference of the circle will also be 1 and thus, the total distance traveled by the mass in 1 period will be 1. The equation for finding the velocity can then be derived from the equation for circumference, i.e. 2*pi*r where r is the radius of the circle. Since the radius of the circle is equal to the length of the string multiplied by the sine of the angle the string makes with the vertical of the cone, we can substitute this into the equation to get 2*pi*L*sin(theta). We then divide this by t, the period, to get the velocity. The equation g*tan(theta) = 4*pi^2*L*sin(theta)/t^2 is derived from the fact that the centripetal force acting on the mass is equal to m*a, or m*v^2/r. This can be rearranged to obtain the equation T*sin(theta) = m*v^2/r. By substituting the equation for velocity obtained earlier, we get T*sin(theta) = m*(2*pi*L*sin(theta)/t)^2/r. We can simplify this equation to get g*tan(theta) = 4*pi^2*L*sin(theta)/t^2. Finally, the equation t = sqrt(4*pi^2*L*cos(theta)/g)) is derived by substituting the equation for centripetal force, i.e. T = m*g/cos(theta), into the equation obtained earlier, i.e. g*tan(theta) = 4*pi^2*L*sin(theta)/t^2. By rearranging the equation, we get t = sqrt(4*pi^2*L*cos(theta)/g). From this equation, we can find the period (t) and hence, the frequency (f=1/t).
 

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

1. What is a conical pendulum?

A conical pendulum is a type of pendulum where the bob (weight) is suspended by a string or rod from a pivot point and moves in a circular motion instead of a back and forth motion.

2. How does the factor of tension affect the frequency of a conical pendulum?

The tension in the string or rod of a conical pendulum is directly proportional to the frequency of the pendulum. This means that as the tension increases, the frequency of the pendulum also increases.

3. What is the relationship between the angle of the conical pendulum and its frequency?

The angle of the conical pendulum does not affect its frequency. As long as the tension and length of the string or rod remain constant, the frequency will remain the same regardless of the angle of the pendulum.

4. How does increasing the frequency affect the period of a conical pendulum?

As the frequency of a conical pendulum increases, the period (time for one full revolution) decreases. This is because the period is inversely proportional to the frequency.

5. Can the factor of tension and frequency increase be used to determine the length of the string or rod in a conical pendulum?

Yes, the factor of tension and frequency increase can be used to determine the length of the string or rod in a conical pendulum. By measuring the tension and frequency, the length can be calculated using the formula: L = T/(4π²f²).

Similar threads

  • Mechanics
Replies
17
Views
425
Replies
76
Views
4K
Replies
7
Views
1K
Replies
1
Views
116
Replies
1
Views
731
  • Introductory Physics Homework Help
Replies
20
Views
1K
Replies
9
Views
1K
Replies
2
Views
1K
  • Mechanics
Replies
2
Views
992
Replies
3
Views
2K
Back
Top