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September 30, 2014

September 30, 2014

Posted by **Steven** on Wednesday, December 8, 2010 at 1:37pm.

- Physics -
**MathMate**, Wednesday, December 8, 2010 at 1:58pmThe force on the string is composed of two components, the weight of the ball, fy=mg, and the centripetal force due to rotation, fx=mrω².

m=0.2 kg

r=0.5m

ω=2π/2s=π rad./s

Resultant force

=√(fx²+fy²)

- Physics -
**drwls**, Wednesday, December 8, 2010 at 2:37pmI have a different interpretation of this problem. r is the radius of the circular path of what is in this case a conical pendulum, not the length of the string.

Let the string tension be T. Let the string angle from vertical be A. Let the period be P = 2s

T cos A = M g

T sin A = M V^2/R

= M*(2 pi)^2 R/P^2

T^2 = M^2 [g^2 + (2 pi)^4*(R^2/P^4)]

You can also solve for the angle A,

tan A= (2*pi)^2*R/(P^2/g)

which leads to

P = 2 pi sqrt[R/(g tanA)]

= 2 pi sqrt[R cosA/sinA]

= 2 pi sqrt(L cosA/g)

For an L = 0.5 m length string, with a period of P = 2s

L cosA/g = 1/pi^2 s^2

This leads to an impossible value for cos A, greater than 1

The problem is overdetermined and the input data are inconsistent. The period of a mass on a string 0.5 m long in a conical orbit cannot be as long as 2 seconds. The same applies to a plane pendulum of that length.

- Physics -
**MathMate**, Wednesday, December 8, 2010 at 3:20pmI agree that it is impossible.

My previous solution omitted the reduction of radius as the angular speed is decreased.

If L=length of the string,

and θ=angle of string with vertical,

then

tan(θ)=fx/fy

=m(l*sin(θ)ω²/mg

On simplicication,

cos(θ)=g/(Lω²)

=9.81/(0.5*π²)

=1.98 which is impossible

For the system to work, either the length has to be doubled, or the period reduced to 1.4 sec. or less

In short, the data from the question should be confirmed or revised.

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