Posted by **Adam** on Wednesday, May 23, 2012 at 11:45am.

A 0.700-kg ball is on the end of a rope that is 2.20 m in length. The ball and rope are attached to a pole and the entire apparatus, including the pole, rotates about the pole's symmetry axis. The rope makes an angle of 70.0° with respect to the vertical as shown. What is the tangential speed of the ball?

I got this:counterbalanced by the horizontal component of rope tension:

m*g*sin(70)

r=2.2*sin(70)

So here's the result

using g=9.81

v=sin(70)*sqrt(2.2*9.81)

v=4.36 m/s but its wrong could someone please help me out. Thank you and sorry for reposting.

- physics -
**bobpursley**, Wednesday, May 23, 2012 at 12:02pm
If I visulize the situation,

mg*sintheta=mv^2/r.

Now if the pole movement is small, then r= length*sinTheta

mg(sinTheta)=mv^2/length*sinTheta

v= sinTheta*sqrt (g*lengthrope)

that is what you have in theequation, but not what you calculated. Recalculate

- physics -
**Adam**, Wednesday, May 23, 2012 at 1:04pm
Sorry I keep recalculating, but i keep getting 4.36 m/s and that's wrong I don't know what I am doing wrong.

- physics -
**bobpursley**, Wednesday, May 23, 2012 at 2:35pm
sqrt(g*lengthrope)=4.64

sin70deg=hmmmm.

Ok, without a picture, I am not certain of where the angle is.

If the angle is measured from the downward vertical to the slanted rope, then sin70 is correct.

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