Whirling Motion?

Question

A 2.3 kg stone is whirled at 55 rpm in a circle at the end of a wire of length 1.2 meters. If the plane of the circle is vertical, calculate

(1) the tension in the wire at the top of the cycle
(2) at the bottom of the cycle

Answer 1

the tension in the wire at the top of the cycle will be equal to the centripetal force produced by the rotation minus the weight force of the stone

let centripetal force be Fc
Fc = mv^2/r
we need v first
v is related to w (angular velocity by
v = rw
first we need w in radians
w = 55 x (2pi/60) = 5.76 rad (approx)
now
Fc = 2.3x(5.76^2) / 1.2
= 63.6 N
weight force = 2.3 x 9.81 = 22.56 N

Tension in wire at top of cycle = 63.6 - 22.56 = 41 N approx (2 sig. figs)
acting directly upwards towards the stone

Tension in wire at bottom of cycle = Fc + mg
because both forces are acting in same direction now
= 63.6 + 22.56 = 86.2N = 86 N approx (2 sig. figs)

Answer 2

The acceleration of the stone is:

(1): a = ω²r

where:
ω = angular speed in radians per second (convert from rpm to radians per second by multiplying rpm by 2π/60)
r = length of wire.

By Newton's 2nd Law:

(2): Fnet = ma

At the top of the circle, the 2 forces acting on the stone are:
(a) its weight (mg), pulling down;
(b) the tension (T_top), also pulling down.
Since they both act in the same direction, add them together to get the net force:

(3): Fnet = mg + T_top

Combine equations (1), (2) and (3), then solve for T_top.

At the bottom of the circle, the two forces acting on the stone are:
(a) its weight (mg), pulling down;
(b) the tension (T_bottom), pulling UP.
Since they act in _opposite_ directions, _subtract_ them to get the net force:

(4): Fnet = T_bottom − mg

Combine equations (1), (2) and (4), then solve for T_bottom.