A light inextensible string AB is fixed at A and carries a particle of mass 50g at B. The particle moves in a horizontal circle of radius (square root 3)m with centre O vertically below A so that the string describes the curved surface of a cone.
If the angle between the string and the vertical is 30 degrees, find the tension in the string and the angular speed of the particle.
2007-06-12 09:39:04 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Nothing to do with friction. It is a centrifugal (centripetal) force problem.
The centrifugal force F pushes the string outward against the tension of the string, and is equal to the horizontal component of the tension.
F = MR(omega)^2
M is the mass of the object, R is the radius of the orbit and omega is angular velocity in radians/sec.
But we also know that the weight of the object is the vertical component of the tension. Therefore,
F/Mg = tan 30
F = Mg (tan 30) = MR(omega)^2
Omega = sqrt (g(tan 30)/ R)
2007-06-12 11:04:53 · answer #1 · answered by mr.perfesser 5 · 0⤊ 0⤋
this is summation of forces on a 30-60-90 right triangle. there is a vertical force due to gravity and a radial force that is based on the acceleration making the sphere rotate. the sum of forces is along the string.
so in the 30-60-90 triangle the resultant force is the hypotenuse and the force due to gravity is adjacent to the 30 deg angle.
cos 30 = force of gravity / resultant in string
string force = (.050kg x 9.8m/s^2)/ .866 = 0.566N
the acceleration can be figured from the horizontal force.
sin 30 = horizontal force / .566N
h force = sin30 x .566 = .283
and the acceleration is related through f=ma
a = h force/.050kg = 5.66 m/s^2
given the radius (sqrt 3) and the accel (5.66)
a=(ang vel )^2 x radius
(ang vel)^2 = 1.732 / 5.66 = 0.30
ang velocity = 0.55 rad/sec
2007-06-12 17:28:39 · answer #2 · answered by Piglet O 6 · 0⤊ 0⤋