8.60.
Model:
Visualize:
Solve:
Assume the particle model for a sphere in circular motion at constant speed.
(a) Newton’s second law along the r and z axes is:
mvt2
∑ Fz = T1 cos30° + T 2 cos60° − FG = 0 N
r
Since we want T1 = T2 = T , these two equations become
∑ Fr = T1 sin 30° + T2 sin 60° =
T ( sin 30° + sin 60° ) =
mvt2
T ( cos30° + cos60° ) = mg
r
Since sin 30° + sin 60° = cos30° + cos 60°,
mvt2
⇒ vt = rg
r
The triangle with sides L1 , L2 , and 1.0 m is isosceles, so L2 = 1.0 m and r = L2 cos30°. Thus
mg =
L2 cos 30° g =
(b) The tension is
T=
(1.0 m ) cos 30° g = ( 0.866 m ) ( 9.8 m/s2 ) = 2.9 m/s
( 2.0 kg ) ( 9.8 m/s
mg
=
cos30° + cos60°
0.866 + 0.5
2
) = 14.3 N