What tension must a 47.9 cm length of string support in order to whirl an attached 1,000.0 g stone in a circular path at 3.37 m/s?
- oubaasLv 711 months ago
T = m*g+m*V^2/r = 1*(9.806+3.37^2/0.479) = 33.5 N
T = m*V^2/r = 3.37^2/0.479) = 23.7 N
- electron1Lv 711 months ago
To support the weight of the stone, the string must be at angle below horizontal. Let θ be the angle. The vertical component is equal to the weight of the stone. The horizontal component is equal to the centripetal force. The length of the string is 0.479 meter.
Weight = 1 * 9.8 = 9.8 N
Fc = 1 * 3.37^2 ÷ 0.479 = 11.3569 ÷ 0.479
This is approximately 23.7 N. Since these two forces are perpendicular to each other, we can use the following equation to determine the magnitude of the tension.
T = √(Vertcal^2 + Horixontal^2)
T = √[(9.8^2) + (11.3569 ÷ 0.479)^2]
This is approximately 25.7 N. To determine the angle, use the following equation.
Tan θ = Weight ÷ Centripetal force
Tan θ = 9.8 ÷ (11.3569 ÷ 0.479) = 4.6942 ÷ 11.3569
The angle is approximately 22.5˚ below horizontal.
- billrussell42Lv 711 months ago
assuming a horizontal circle, so gravity is not relevant...
f = mV²/r = 1•3.37²/0.479 = 23.7 N
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
edit: above is wrong, gravity does play a part even in a horizontal circle