Ratio of tensions

[pivoxa15]

Homework Statement

Two spheres A and B of equal mass are attached to each other by a light rod of length 2.4m as shown. The rod has negligible mass. The spheres and rod are set in motion in a circular motion about the point C. Calculate the ratio:

Tension in the rod between A and B
--------------------------------------------
Tension in the rod between A and C

Note: Although this problem mentions spheres, I think it is only 2D.

Homework Equations

Tension = m*v^2/r

V = angular velocity * r

Angular velocity = a wich is constant.

The Attempt at a Solution

Tension between A and B is worked out from working out the tension between C and B which is m(a*2.4)^2/2.4 then multiply this by (0.8/2.4) because we only want the segment between A and B.

Tension between C and A is m(a*1.6)^2/1.6

The ratio is worked out after much cancellation 0.8/1.6=1/2

However the answers suggested 3/5

Have I worked out the tension between A and B correctly?

 

[andrevdh]

The sphere at A will also experience an outwards tension, $$T_{AB}$$, which need to be "overcome" by $$T_{AC}$$ in order to give it the necessary centripetal acceleration.

 

[pivoxa15]

I don't understand you. Could you reexplain with a bit more elaboration and possibly several shorter sentences?

 

[Doc Al]

Tension between A and B is worked out from working out the tension between C and B which is m(a*2.4)^2/2.4 then multiply this by (0.8/2.4) because we only want the segment between A and B.

Not quite. Hint: Find the net force on spheres A and B, using what you know about centripetal force. Hint2: The net force on the outer sphere (B) is the tension between A and B. However two forces are exerted on inner sphere (A).

 

[andrevdh]

Tension in a string (rod) is transmitted from the one side to the other. If you pull on a rope (tension at B) it pulls on the object on the other side of the rope (A). The tension at B will therefore be transmitted to A and pull it outwards with the same tension that is experienced at B.

Also when several objects are pulling on a rope (tug of war) the tension can be different in the segments between the points where (the men are pulling) objects are connected to it. Since both A and B are pulling on the rod the tension on the inner portion of the rod (segment CA) needs to be bigger, since it has to pull both A and B around , while the outer portion, AB, needs to pull just B around.

In short - A pulls on B, B pulls back on A with the same force. C pulls on A. A therefore experiences one in- and one outwards pull. These two pulls combined provide the centripetal force on A.

 

[pivoxa15]

So for Tension between C and A, there is a ma^2*1.6 force (where a is angular velocity) to the left and ma^2*0.8 force to the right. Tension between A and B is a ma^2*0.8 force to the left. But the ratio turns out to be 1 which is incorrect. I don't think I have understood.

 

[Doc Al]

There are two tensions involved: T(ca) between C and A, and T(ab) between A and B.

Consider the net force on B. That's T(ab), since that's the only force acting on B. Since B is in uniform circular motion, apply Newton's 2nd law for centripetal acceleration.

Consider the net force on A. The net force is T(ca)-T(ab), since both tensions act on A. Since A is in uniform circular motion, apply Newton's 2nd law for centripetal acceleration.

Those two force equations should allow you to calculate the ratio of the tensions.

 

[andrevdh]

So for Tension between C and A, there is a ma^2*1.6 force (where a is angular velocity) to the left and ma^2*0.8 force to the right. ....

The right one will be with a radius of 2.4, but the left one cannot be calculate that easy.

You need to add the two tensions up in order to get the resultant force acting on the sphere at A.

This resultant force can then be calculated with ma^2r, which then enables you to calculate the left tension on A.

 

[pivoxa15]

I see. I had a hazy physical view of the situation. The rod attaches two spheres A and B, which rotate in a circle. The tension from A to B is 2.4N because if ony B was attached, that would be the tension force in the rod. However, A is also attached If there was no B, there would be a tension force of 1.6N but B exists which adds 2.4N as shown previously, tension adds vectorially. Therefore

tension in rod between A and B
------------------------------
tension in rod between A and C

= 2.4N/(1.6N+2.4N)
= 3/5

I have a general question, using Newton's third law, there must be a reaction from the tension. Or is it the case that the tension is the reaction.
i.e Ball exerts a force on rod radially outwards. Rod exerts a force on ball radially inwards which is the tension. However, we only consider the net force on the ball because it's the ball that is moving in a circle hence has centripetal acceleration. The one external force on the ball being tension in this case.

 

[andrevdh]

I pressume you are referring to T_AC. In this case both balls pulls outwards on the inner rod due to their inertia (of the balls). The rod reacts by pulling back on A. You can see that from the calculation too: 2.4 + 1.6. These tensions are the forces acting on the balls. According to N3 we know that the balls are pulling just as hard back on the rods. Therefore we say that the rods are under (or is it in?) tension. If you take T_AC again we get a similar force acting in the opposite direction at the central connecting point. The rod therefore experiences two similar forces at its ends acting in opposite directions - the rod is in tension between these two forces acting at its ends.