What tension must a 40.4 cm length of string support in order to whirl an attached 1,000.0 g stone in a circular path at 2.53 m/s?
- Anonymous6 months ago
It depends if the circle is in a horizontal or a vertical plane, which you haven't stated.
- billrussell42Lv 76 months ago
here is a very similar Q/A, just change a few numbers and recalculate
v = 2.59 m/s
r = 41.8 cm
there are two forces acting on the stone, both of which cause tension.
assuming this circular path is horizontal.
1. Centripetal force f = mV²/r = mrω²
ω is angular velocity in radians/sec
1 radian/sec = 9.55 rev/min
m is mass in kg
r is radius of circle in meters
V is the tangental velocity in m/s = ωr
f is in Newtons
F = (1 kg)(2.59 m/s)² / 0.418 m = 16.05 N
2. force of gravity F = mg = 1•9.8 = 9.8 N
Those two are orthogonal, therefore the sum of the two is
F = √(16.05² + 9.8²) = 18.80 N
- oubaasLv 76 months ago
T = m*V^2/r = 1.00*2.53^2/0.404 = 15.8 N