1. May 3, 2010

### harkyb

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

This setup of a conical pendulum is actually at the end of crane which is rotating, however I do not think that this should effect the situation:

I need to calculate the radius so that I can then add this to the radius of the crane which will give my actual required answer.

A mass m= 250 kg is suspended from a string of length L = 2.00m. It revolves in a horizontal circle due to the rotation of the crane. Angular velocity of crane and assume this will be same for the conical pendulum situation w= 1.0 rad/s. Angle between string at vertical is not known.

2. Relevant equations

v=rw
centripetal acceleration = Ac = w^2r
centripetal force = m(Ac) = m*w^2*r

3. The attempt at a solution

Tried few different methods and haven't seemed to come to a solution. Not sure if any method is even correct.

Attempt 1: Went further to calculate period of revolution is 2pi which gave h to be 9.81 which I am sure is incorrect.

Attempt 2: Not sure if this was correct and did not know how to go on.

Appreciate all the help in advanced!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

File size:
208.4 KB
Views:
73
File size:
202.6 KB
Views:
62
2. May 3, 2010

3. May 3, 2010

### harkyb

Yes, I have followed those steps and they are similar to my previous attempt. Using the final formula on the Wiki page;

for my conical pendulum;

t = (v x 60 x 2pi) / 60
t = 1 x 2pi
t = 2 pi

Therefore using the above formula:

2pi = 2pi sqrt(2cos(theta)/9.81)
9.81 = 2cos(theta)
4.905 = cos(theta)
theta = 130.224 ???

4. May 3, 2010

### ideasrule

I don't know how you got 130, but taking the inverse cosine of 4.905 should give you "ERROR!" on the calculator because the maximum value of cos is 1.

It makes a huge difference that this pendulum is on a rotating crane. This causes an extra acceleration towards the center of the crane's axis of revolution that you need to consider. Put any way, in the reference frame of the pendulum's pivot point, there's a centrifugal force pushing outwards.

5. May 3, 2010

### harkyb

I do get an Error when taking the inverse cosine of 4.905, I used another method and some how worked out 130 which is obviously wrong as you stated.

Taking into account that this pendulum is in a rotating crane with the values;

m1 = 250 kg
m2 = 200 kg
h = 2.0 m

I have enclosed the originial figure of the crane..

#### Attached Files:

• ###### Crane.pdf
File size:
146.2 KB
Views:
64
Last edited: May 3, 2010
6. May 3, 2010

### harkyb

Can no one help?

7. May 4, 2010

### Cleonis

You have not given enough information to address the problem. The following information is necessary: the length of the boom of the crane.

In order to proceed I will assume that the boom of the crane has a length of 2 meters. I will also assume the following: that the 250kg mass is purely co-rotating with the crane. That is, there is no swing with respect to the crane, the mass just moves in a circle around the crane pivot, along with the crane.

Let's do some explorations of extreme cases, visualizing the physics taking place.
- Imagine the crane not rotating. Then the string hangs straight down, and the tension in the string will be 250 kg.
- Imagine the crane rotating extremely fast, so fast that the string is close to horizontal.
Then the tenstion in the string will be extremely high, many times the gravitational force.
- Imagine the angular velocity of the crane is cranked up to the point where the string is at a 45 degrees angle with the vertical.
You can decompose the force that the string exerts upon the mass in two component: a vertical component and a horizontal component.
The vertical component of the string tension is providing the force that is required to lift the mass off the ground. The horizontal component of the string tension is providing the required centripetal force. You can calculate how fast the crane must rotate to raise the string to a 45 degree angle.

More generally: at any angle of the string the vertical component of string tension lifts the mass of the ground, and the horizontal component of string tension provides the required centripetal force.

The total radial distance is the length of the crane boom plus the amount that the mass swings out.

8. May 4, 2010