Calculating Tension in a Revolving Sphere: Solving for T1 and T2

[klm]

Two wires are tied to the 400 g sphere shown in figure. The sphere revolves in a horizontal circle at a constant speed of 8.30 m/s

1) What is the tension in the upper wire?
2) What is the tension in the lower wire?

okay i know that m=.4 and v=8.3
i don't know r, which i thought that maybe i could find by doing the square root (1^2-.5^2) =.866
and then Force= m V^2 / R sooo.. (.4)(8.3^2)/(.866) = 31.819 . but i don't know where to go from there?

 

[klm]

okay i also think..
Fnetx= -T1 cos 26 + -T2 cos 26 = mv^2/ r
Fnety= Tsin 26 + -T2 sin 26 -Fg= mv^2/ r

 

[klm]

so my free body diagram looks like 1 wire pulling in the 2nd quad, one in the 3 quad, and Fg going straight down on the y axis

 

[Sourabh N]

yep. No motion in y-dir., so forces sum up to zero. In X- dir. they will contribute to centripetal force.

 

[klm]

but i don't understand how the tension in the 2 wires are going to be different

 

[klm]

Fnety= Tsin 26 + -T2 sin 26 -Fg= 0
T1 sin 26 - T2 sin 26 = Fg
T1 sin 26 - T2 sin 26 = 3.92

Fnetx= -T1 cos 26 + -T2 cos 26 = mv^2/ r
-T1 cos 26 + -T2 cos 26 = 31.82

 

[klm]

^ i am not sure how to solve for T1 or T2 . can you please help me with that

 

[klm]

please can someone help me!

 

[klm]

i figured out the T2, but now i can't figure out T1. shouldn't it just be plug T2 into one of the equations and it should give you T1??

 

[Doc Al]

Fnety= Tsin 26 + -T2 sin 26 -Fg= 0
T1 sin 26 - T2 sin 26 = Fg
T1 sin 26 - T2 sin 26 = 3.92

Where does the 26 degrees come from?

Fnetx= -T1 cos 26 + -T2 cos 26 = mv^2/ r
-T1 cos 26 + -T2 cos 26 = 31.82

Which way does the centripetal acceleration act?

Once you correct your equations, isolate the T1 term in each, like this:

$$T_1 \sin\theta = mg + T_2 \sin\theta$$

 

[klm]

26 came from inverse tan (.5/1)

 

[klm]

does centripetal acc act on the Horizontal force?

 

[Doc Al]

26 came from inverse tan (.5/1)

But the 1 m side is the hypotenuse. That should be inverse sin (.5/1).

 

[Doc Al]

does centripetal acc act on the Horizontal force?

My question is: What's the direction of the centripetal acceleration?

 

[klm]

umm this may be wrong, but i thought centripetal acc just means that the direction is toward the center

 

[klm]

ohh wait i know what i did. it is just the angle that is messing me up! thank you for pointing that out. i don't know why, but the angle i used still let me get the tension of the first string??!

 

[Doc Al]

umm this may be wrong, but i thought centripetal acc just means that the direction is toward the center

That's exactly right. So, what direction is that? Should it be positive or negative?

 

[klm]

negative

 

[klm]

would that be right?

 

[Doc Al]

negative

Good. So fix your equation for horizontal forces.

 

[klm]

Fnetx= -T1 cos 26 + -T2 cos 26 = - mv^2/ r

 

[Doc Al]

Fnetx= -T1 cos 26 + -T2 cos 26 = - mv^2/ r

Good. Now combine with the other equation and solve.