a semicircular wire has length L and mass M. a particle of mass m is placed at the crntre of the circle. find the gravitational force of attractionon the particle.
2007-04-05 20:30:34 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
k = mass per unit length = M / L = constant
L = pi R so radius R = L/pi
consider an element of length dL between an angle p and p+dp having mass k*dL = k* Rdp
G force between this element mass and m will be
dF = G*m [k* Rdp] / R^2 = G*m k*dp] / R
total force will be within limit p = 0 to p= pi
F = ∫ [ 0 to pi] [G*m k*dp] / R
F = (G*m k / R) ∫ [ 0 to pi] [dp]
F = (G*m k / R) ∫ [ pi]
F = G*m (M / L) (pi / L) [ pi]
F = G M m [ pi]^2 /L^2
2007-04-05 21:55:34 · answer #1 · answered by anil bakshi 7 · 1⤊ 0⤋
The radius of this semicircular wire would be L/Ï, so if all of the mass M was at some point L/Ï from mass m, then the force would be:
F = (GMm)/(L/Ï)²
However, being that mass M is distributed in a semicircle, we gotta do the following definite integral:
F = â« ((GM(m/L) Sin((x/L)Ï))/(L/Ï)² dx from 0 to L, which gets us
F = (2/Ï) ((GMm)/(L/Ï)²)
In other words, approximately 64% of the force in the first case.
2007-04-05 21:35:33 · answer #2 · answered by Scythian1950 7 · 1⤊ 1⤋
F = G * M1 * M2/ D^2 IS THE FORMULA TO CALCULATE THE FORCE OF GRAVITATION BETWEEN 2 OBJECTS.
IN THIS CASE THE DISTANCE WOULD BE THE RADIUS OF THE SEMICIRCLE SO THE FORCE WOULD BE
G*M*m/ r^2
2007-04-05 20:56:36 · answer #3 · answered by MAGNETO 1 · 0⤊ 1⤋
Fg = (GMm)/(R^2)
Use the length to find the radius of the circle (C = 2piR)
2007-04-05 20:36:37 · answer #4 · answered by cmw88 1 · 0⤊ 0⤋