Physics Gravitational Force

Question

a. Consider a wire with mass M bent into the shape of one-quarter of a circle with radius R. What force will the wire exert on a point-mass (m) placed at its center of curvature?

b. What force would the rod exert on the point mass (M), if the point-mass were located on the y-axis at the point (0, 2L)? The rod remains on the x-axis between (0,0) and (L,0).

Hint: This will require trigonometric substitution. Make sure you can correctly set up any definite integrals that needs to be evaluated. Start with a picture and find the magnitude of the force a small mass “dm” exerts on M. Draw in the components of this force. Determine what trigonometric functions are needed to find each component

Answer 1

part aQuarter circle of radius RRR, total mass MMM

The wire is uniform, so:

λ=Marc length\lambda = \frac{M}{\text{arc length}}λ=arc lengthM​Arc length of quarter circle:

πR2\frac{\pi R}{2}2πR​So linear mass density:

λ=2MπR\lambda = \frac{2M}{\pi R}λ=πR2M​ Step 1: Small mass element

Take small element at angle θ\thetaθ

Arc element:

ds=Rdθds = R d\thetads=Rdθ dm=λds=λRdθdm = \lambda ds = \lambda R d\thetadm=λds=λRdθDistance from center = RRR

Step 2: Small force magnitude

dF=Gm dmR2dF = \frac{G m \, dm}{R^2}dF=R2Gmdm​ dF=GmλRdθR2dF = \frac{G m \lambda R d\theta}{R^2}dF=R2GmλRdθ​ dF=GmλRdθdF = \frac{G m \lambda}{R} d\thetadF=RGmλ​dθ Step 3: Components

Only the x and y components matter.

By symmetry:

Fx=FyF_x = F_yFx​=Fy​Component:

dFx=dFcos⁡θdF_x = dF \cos\thetadFx​=dFcosθ Fx=GmλR∫0π/2cos⁡θ dθF_x = \frac{G m \lambda}{R} \int_0^{\pi/2} \cos\theta \, d\thetaFx​=RGmλ​∫0π/2​cosθdθ Fx=GmλR[sin⁡θ]0π/2F_x = \frac{G m \lambda}{R} [\sin\theta]_0^{\pi/2}Fx​=RGmλ​[sinθ]0π/2​ Fx=GmλRF_x = \frac{G m \lambda}{R}Fx​=RGmλ​Substitute λ=2MπR\lambda = \frac{2M}{\pi R}λ=πR2M​

Fx=2GmMπR2F_x = \frac{2 G m M}{\pi R^2}Fx​=πR22GmM​Since Fy=FxF_y = F_xFy​=Fx​,

Step 4: Total Force Magnitude

F=Fx2+Fy2F = \sqrt{F_x^2 + F_y^2}F=Fx2​+Fy2​​ F=2FxF = \sqrt{2} F_xF=2​Fx​ F=22GmMπR2\boxed{ F = \frac{2\sqrt{2} G m M}{\pi R^2} }F=πR222​GmM​​Direction: along the line at 45° toward the a part b

Rod lies on x-axis from 000 to LLL

Point mass at (0,2L)(0, 2L)(0,2L)

Step 1: Linear density

λ=ML\lambda = \frac{M}{L}λ=LM​Small element at position xxx:

dm=λdxdm = \lambda dxdm=λdxDistance from point mass:

r=x2+(2L)2r = \sqrt{x^2 + (2L)^2}r=x2+(2L)2​ Step 2: Small force

dF=Gmdmr2dF = \frac{G m dm}{r^2}dF=r2Gmdm​ dF=Gmλdxx2+4L2dF = \frac{G m \lambda dx}{x^2 + 4L^2}dF=x2+4L2Gmλdx​ Step 3: Components

Only y-components survive after integration.

cos⁡θ=2Lr\cos\theta = \frac{2L}{r}cosθ=r2L​So:

dFy=dF2LrdF_y = dF \frac{2L}{r}dFy​=dFr2L​ dFy=Gmλ(2L)dx(x2+4L2)3/2dF_y = \frac{G m \lambda (2L) dx} {(x^2 + 4L^2)^{3/2}}dFy​=(x2+4L2)3/2Gmλ(2L)dx​ Step 4: Integral

Fy=Gmλ(2L)∫0Ldx(x2+4L2)3/2F_y = G m \lambda (2L) \int_0^L \frac{dx}{(x^2 + 4L^2)^{3/2}}Fy​=Gmλ(2L)∫0L​(x2+4L2)3/2dx​Use trig substitution:

x=2Ltan⁡θx = 2L \tan\thetax=2LtanθAfter evaluation:

Fy=GmM4L25F_y = \frac{G m M} {4L^2 \sqrt{5}}Fy​=4L25​GmM​Direction: downward toward the rod answer

(a)

F=22GmMπR2\boxed{ F = \frac{2\sqrt{2} G m M}{\pi R^2} }F=πR222​GmM​​(b)

F=GmM4L25\boxed{ F = \frac{G m M}{4L^2 \sqrt{5}} }F=4L25​GmM​​ and sovle this problem using newton's of gravitation formula

dF=Gmdm​ by r*r