# Motion in a circle

#### psingh

Two wires with 2.0kg sphere is tied to a pole with 1m in between the wires. each wire makes an outward angle with the pole of 60 degress and an inward angle with the pole of 30 degrees.

for what speed is the tension the same in both wires and what is that tension.

well i figured you needed to find the tension in the vertical direction and then subsitute that into the tension in the horizontal and find the radius. but im not sure exactly how to go about doing this. so any help would be nice =] thanks

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#### KingOfTwilight

I'm not sure whether I understood your problem but I think you need these:

$$T_{vertical} = G$$ and the horizontal component is causing the rotational movement, so $$T_{horizontal} = m\frac{v^2}{R}$$

#### abercrombiems02

I would need to see a picture or have a better physical description as to whats going on here, but for the general idea of the problem. You are correct, in the fact that the sum of forces in the radial (horizontal) direction must equal the centripetal force. This is 1 equation. You are also correct that the sum of forces in the vertical direction must equal zero. Between these two equations you should have 3 unknowns, both the tensions and the velocity of the object. Your third equation is simple, its one of the constraints in the problem... t1 = t2. So you have three equations and three unknowns, the problem is solvable. I actually would find it slightly more general to write the magnitude of centripetal force as m*r*omega^2 as opposed to mv^2/r because that velocity only corresponds to a single radius. Granted we assume the radius doesnt change due to any of the deformable characteristics of the wire, I still think its more general to determine the angular velocity which is omega, having units of rad/sec.

#### psingh

oh no i forgot to attach the inital picture.. ok basically its two strings attached to a vertical pole with 1m of space between them, it looks like a triangle attached to a pole with the interior angles 30,30,120 the 30 degree angles are with the pole. so if u break that up its two triangles (30,60,90).

this is what i got so far since the interor angles 30 and 30 are the same both sides should have the same length, so i used the law of sines and found the length of the rope and since they both have the same angle they have the same length. after finding this i was able to find the radius = 0.289 by trig.

so my equations i set up were..

T(1)cos60+ T(2)cos60=mv^2/r (im using v because the answer needs to be in m/s)

T(1)sin60-T(2)sin60-mg=ma=0 T(1)sin60-T(2)sin(60)=19.62

and then they said that T(1) = T(2) and that zeros out the equation so it doesnt work?? i dunno

#### marlon

psingh said:
T(1)cos60+ T(2)cos60=mv^2/r (im using v because the answer needs to be in m/s)
One can directly conclude that this is incorrect since the centripetal motion is two dimensional, ie the r-vector has both horizontal and vertical components. Since you are writing down two equations of motion in vertical and horizontal direction, you must be consequent. The left hand side of this equation contains components of forces in ONE single direction, while the right hand side is two dimensional. You see my point ?

regards
marlon

Besides, could you add a figure ?

Last edited:

#### psingh

the picture of the problem

i finally figured out how to put the picture on

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#### psingh

o yeaa its not drawn to scalee

#### Doc Al

Mentor
The only radial forces on the sphere are the horizontal components of the tension in the wires. So those forces create the centripetal acceleration. (Use some geometry to find the radius of the circular motion.)

#### psingh

for my radius i got 0.289m

i understand that the x components are the radial components and u set that equal to centrip. accel, but what i dont get is how to find the tension

#### Doc Al

Mentor
You find the tension by writing equations for both horizontal and vertical forces on the sphere. (The vertical forces are in equilibrium.) With those two equations, you'll have all the info you need to find the speed.

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