Ball rotates around a shaft by two cords

1. Apr 8, 2014

mackeis

1. The problem statement, all variables and given/known data
A 1.0kg ball is attached to a vertical rod by two strings. Each string is 1.0m long, and they are attached to the rod a distance of 1.0m apart. The rod-ball-strings setup is rotating about the axis of the rod. As it rotates the strings are taut and form an equilateral triangle. The tension in the upper string is 25N.
a. Describe a free-body diagram showing all the forces on the ball.
b. What is the tension in the lower string?
C WHat is the speed of the ball in rev/min?

I calculated tension in the lower string to be 8.2N, 36.9 degrees down and to the left. (I used arctan .3/.4 to caluclate the angle about the radius and used trig to solve tension) But should I have it as -8.2N as it is downward direction? Can tension be negative based on direction?

Part C:
I was using ∑F=ma-v^2/r and the sum of the forces on the X axis. Question again relates to direction - Should I have a negative value for the forceX of the second cord or are they added since they are both pulling in towards the shaft?

Thanks!

2. Apr 8, 2014

mackeis

Just realized i put the wrong numbers in. THe ball is actuall attached where the radius is .4m and the string above and below are each attached .3m from the centre of the rod. Tension in the upper string is 18N. The questions are the same. Sorry about that.

3. Apr 8, 2014

jbunniii

Where did 36.9 degrees come from? If the triangle is equilateral, what does that tell you about the angle at each vertex? From that, you can determine the angles of the tension vectors.

Please carefully specify what coordinates you are using, including the directions (i.e. which way do positive $x$ and positive $y$ point?) Then specify all of the forces acting on the ball, and which directions they are pointing.

4. Apr 8, 2014

BiGyElLoWhAt

how are you getting .3 and .4? the way I'm looking at this, the angles the strings make with the rod are the same. so solve for the angle theta = arccos(1/2) =60degrees, and radius = sin(60degrees)=(sqrt3)/2m which is about .85 m

the strings are attached 1m apart, and if the lower string has any tension, then the ball is smack dab in the middle (y axis)

I'm gonna let you draw the freebody diagram, but remember gravity only affects the tension in the upper string, so the lower tension is due to the angular velocity, which is what, rw^2 ? or something? i'm not sure, it's been a while. but try doing A, then C, then B.

the N2L eq should give you the tension for the bottom. I have to go for now, I'll check back later.

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