# Engineering Dynamics problem

1. Aug 27, 2013

### sharkNC

1. The problem statement, all variables and given/known data
A ball of 10kg mass is suspended by a cable that passes through a hole in a floor. The ball rotates in a circular orbit, A, that is in a plane that is 3m below the floor. If the orbit has a radius of 1m, calculate;

a) Tension in the cable
b) speed va of the ball in orbit A
c) angular momentum about the vertical axis

The motor then draws in the cable until the ball rotates in orbit B. If the radius of orbit B is 0.5m, calculate;

d)The new speed vb of the ball
e) The angle between the cable and vertical axis
f) The height, h, gained by the ball

2. Relevant equations
θa - tan-1(1/3) = 18.43

3. The attempt at a solution
I am able to a-c by working out the tension as T = 10x9.81/cos18.43 = 103.4 N
the velocity √1x103.4xsin18.43/10 = 1.808 m/s
the angular momentum as L = r x mv = 1 x 10 x 1.808 = 18.08

When I attempted to do the rest of the question I got the new speed to be half of the original speed and angle B to be smaller. However in the question the image for angle B looks to be greater than angle A so I am not sure whether its just the question or if I am doing something wrong to calculate the angle. (I find the sum of the forces in the x and y direction and make them equal to each other and then find angle B). For part f I am using the conservation of energy however when I make h the subject I find that it comes out as 3 which doesn't make sense.

Can someone please help me. I have been stuck on this question for a long time and cannot seem to find where I have gone wrong.

2. Aug 27, 2013

### voko

How did you compute the new speed of the ball?

3. Aug 27, 2013

### sharkNC

I assumed that the angular momentum was constant so 18.08 = 0.5 x 10v and v as 0.904

4. Aug 27, 2013

### voko

If the new radius is half the original radius and angular momentum is conserved, how can the new speed be half the original speed? Note that if you substitute v = 0.904 into 18.08 = 0.5 x 10v, you get 18.08 = 4.52, which can't be right.

5. Aug 27, 2013

### sharkNC

Yes you're right sorry. So assuming angular momentum is conserved then from that equation v = 3.616. so for part e to get the angle I used mv^2/rsintheta = mg/costheta to get tantheta = v^2/rg and theta as 69.43°. Does that sound correct?

6. Aug 27, 2013

### voko

Yes, that looks good.