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Circular motion of a tether ball

  • Thread starter in10sivkid
  • Start date
36
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a .225 kg ball tethered to a tall pole on a 1.37m rope is thrown so that it travels in a horizontal circle with the rope making an angle of 40 with the pole
a) what is the speed of the ball
b) what is the tension in the rope

i am stuck on this problem because i do not have a unit time to be able to use any of the equations dealing with circular motion

i started out with knowing that the radius will be the cos40*1.37m = 1.05 m

but after that i'm always left with two unknowns for any equation. any hints on what i could be missing.....thanks!
 

Andrew Mason

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in10sivkid said:
a .225 kg ball tethered to a tall pole on a 1.37m rope is thrown so that it travels in a horizontal circle with the rope making an angle of 40 with the pole
a) what is the speed of the ball
b) what is the tension in the rope

i am stuck on this problem because i do not have a unit time to be able to use any of the equations dealing with circular motion

i started out with knowing that the radius will be the cos40*1.37m = 1.05 m

but after that i'm always left with two unknowns for any equation. any hints on what i could be missing.....thanks!
You have to use the angle to determine the relative magnitudes of the centripetal (horizontal) and gravitational (downward) forces. That will give you the centripetal force from which you can easily determine the speed.

AM
 
36
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once again could you elaborate a little bit more on that...

how can I find the centripetal force if all I know is the radius, and mass

i don't have a velocity, or time, or angular velocity....what part am I failing to see here?
 
111
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Centripetal force=mv^2/r
but you need speed. My advice to you is that you draw a representative picture of the problem. Then show the forces acting on the particle. gravitational force is vertical to the horizontol. And the centripetal force is parallel to the horizontal. Try to use tan40... You already know the radius. v will be easy to find out then...
 
36
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hmmmm...ok 2 questions i'm not following

why would i use tan40...would i multiply that by .225 kg or the length of the rope...almost there i think
 
36
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i have this equation and it involves banking angles thos

tan(angle) =v^2/gr

would that be what you are referring to???
 
111
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exactly, you can find v now...
 
36
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last question dealing with this they then ask for the tension would that simply be
Mass * acceleration
mass * v^2/r??

if so...i'm getting a wrong answer or my book is wrong

so i got V = 2.94 m/s

from tan40 = v^2/(9.8 m/s)(1.05 m)

Ac = 8.23 m/s^2 = (2.94 m/s^2)/(1.05 m)
then T = (.225 kg)(8.23 m/s^2) = 1.85 N
 
111
0
I didn't calculate v but I suppose you found it right
Anyway T is equal to the combination of the gravitational and centripetal forces... Try to use the pyhtagoras theorem... You have a right triangle.
 
36
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sooooo would that make tension the following?

T^2 = (mv^2/r)^2 + (mg)^2
 
111
0
yup! congrats man :)
 
36
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ahhhh I guess there are just typos in the book then :)
 
111
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Well I'm quite sure I gave the correct solution
 
36
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yah it seems correct
 

OlderDan

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in10sivkid said:
a .225 kg ball tethered to a tall pole on a 1.37m rope is thrown so that it travels in a horizontal circle with the rope making an angle of 40 with the pole
a) what is the speed of the ball
b) what is the tension in the rope

i am stuck on this problem because i do not have a unit time to be able to use any of the equations dealing with circular motion

i started out with knowing that the radius will be the cos40*1.37m = 1.05 m

but after that i'm always left with two unknowns for any equation. any hints on what i could be missing.....thanks!
You need to give this another try. The angle of 40 degrees is the angle the rope makes with the pole. The radius of motion is not cos40*1.37m. Furthermore, the tension cannot possibly be less than the weight of the ball, which is what you found.
 
36
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gah...you're right..I didn't even realize my calculated T was less than mass

i've been pondering over this...and i'm stumped..i can't figure out how i would calculate the radius...unless it really was 1.37m (actual length of the rope), but somehow i think thats not right.

was my line of thinking right in all the other aspects of the calculation once the radius was found?
 

OlderDan

Science Advisor
Homework Helper
3,021
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in10sivkid said:
gah...you're right..I didn't even realize my calculated T was less than mass

i've been pondering over this...and i'm stumped..i can't figure out how i would calculate the radius...unless it really was 1.37m (actual length of the rope), but somehow i think thats not right.

was my line of thinking right in all the other aspects of the calculation once the radius was found?
The 40 degree angle is between the rope and the pole. The radius of the circular path is the side opposite this angle, not adjacent to it. The vertical component of the tension has to equal the weight of the ball. With that and the correct angle you can find the tension, and then find its horizontal component. It looks like you were calculating just the horizontal component before, but you did not identify it as such. The horizontal component of the tension will be less than the weight of the ball, so that's OK, but the total tension has to be greater.
 

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