Ball rotates around a shaft by two cords

[mackeis]

1. Homework Statement
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!

 

[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.

 

[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.

 

[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 going to 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.

 

[HalfLostAndSearching]

I would like to provide a response to this content by first acknowledging the detailed and thorough description of the setup. This level of detail is important in understanding the forces at play in this scenario.

In response to part a, a free-body diagram is a useful tool to visualize the forces acting on an object. In this case, the forces acting on the ball would include its weight (mg), the tension in the upper string (25N), and the tension in the lower string (8.2N). The weight would act downwards, while the tension in the strings would act upwards and towards the center of rotation.

In part b, the calculated tension in the lower string is correct. As for the direction, it is important to note that tension is a vector quantity, meaning it has both magnitude and direction. In this case, the tension in the lower string is acting downwards and to the left, as you have correctly calculated.

For part c, the speed of the ball can be calculated using the formula v = ωr, where ω is the angular velocity and r is the radius of rotation. In this case, the angular velocity can be calculated by dividing the speed in rev/min by 60 seconds, and the radius is 1.0m. It is important to note that the direction of the forces does not affect the calculation of speed, as long as the magnitude of the forces is correct.

In summary, it is important to pay attention to the direction of forces when using vector quantities like tension, but it does not affect the calculations for speed. Keep in mind that tension can be negative based on direction, but it does not change the magnitude of the force. Keep up the good work with your calculations and continue to pay attention to details like direction in your analysis.