A 0.700 kg ball is on the end of a rope that is 2.00 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 have a picture of this situation if you need it and I can email it to you if you let me know.
Thank you so much for any help. This is the only problem left that I just can't figure out.
2007-10-03 16:04:37 · 2 answers · asked by Physics 1 in Science & Mathematics ➔ Physics
Yes, the picture would be helpful.
I reconstructed it here:
http://i20.tinypic.com/9la9fr.jpg
T cosφ = mg
T = mg/cosφ
Fc = T sinφ = mg tanφ
Fc = ma = mv²/r = mv²/(Lsinφ)
Answer:
v = sinφ √(Lg/cosφ) = 7.11 m/s
2007-10-04 04:42:47 · answer #1 · answered by Alexander 6 · 0⤊ 0⤋
We need no pictures.
Fc=m ac
so ac=Fc/m
ac=V^2/R
R=Lsin(A)
Fc= mg sin(A)
so V=sqrt (ac R)
V=sqrt(R Fc/m)
V= sqrt(R mg sin(A) / m) finally since R= Lsin(A)
V= sqrt(L sin^2(A) mg / m)
V=sin(a) sqrt(L g)
V=sin(70) sqrt( 2.00 x 9.81)=
V=4.16m/s
2007-10-04 09:55:07 · answer #2 · answered by Edward 7 · 0⤊ 0⤋