b. Find the equation for the distance r from m1 to a differential mass using the

Question

b. Find the equation for the distance r from m1 to a differential mass using the x axis given.

c. Compute the magnitude of the net gravitational force F on the particle from this non-uniform rod.

e Use/understand Integral Calculus in Physics. Dynamics: Problem 13.16, page 380 In the right figure, a particle of mass mi =0.67 kg is at a length L = 3.0 m and mass M = 5.0 kg. distance d= 23 cm from one end of a uniform rod with T1 (1) What is the linear density of the rod, ? Since the rod is uniform, A1

Explanation / Answer

part a:

as x=0 at the middle of the rod, the rod stretches for -L/2 ro L/2 i.e. -1.5 m to x=1.5 m

as total mass is still 5 kg,

then integration of lambda_0*(9-4*x^2)*dx=5

with x varying from -1.5 to 1.5

==>lambda_0*(9*x-(4/3)*x^3)=5 with x varying from -1.5 to 1.5

using the limits of x, we get

lambda_0*(9*(1.5-(-1.5))-(4/3)*(1.5^3-(-1.5)^3))=5

==>lambda_0*18=5

==>lambda_0=5/18

part b:

location of m1=-1.5-d=-1.5-0.23=-1.73 m

hence if dm is located at x, then r=x-(-1.73)=x+1.73

where x can vary from -1.5 m to 1.5 m

part c:

force on m1 due to dm at location x:

dF=G*m1*dm/r^2

=G*m1*(5/18)*(9-4*x^2)*dx/(x+1.73)^2

==>dF=6.674*10^(-11)*0.67*(5/18)*(9-4*x^2)*dx/(x+1.73)^2

==>dF=k*(9-4*x^2)*dx/(x+1.73)^2

where k=6.674*10^(-11)*0.67*(5/18)=1.2421*10^(-11)

now x+1.73=r

as x varies from -1.5 to 1.5 ==> r varies from 0.23 to 3.23

==>dx=dr

and x=r-1.73

hence dF=k*(9-4*(r-1.73)^2)*dr/r^2

==>dF=k*(9-4*(r^2+2.9929-3.46*r))*dr/r^2

==>dF=k*(9-4*r^2-11.9716+13.84*r)*dr/r^2

==>dF=k*(-2.9716-4*r^2+13.84*r)*dr/r^2

integrating both sides:

F=k*((2.9716/r)-4*r+13.84*ln(r))

where ln=natural logarithm

using limits of r from 0.23 m to 3.23 m

we get F=k*36.5674 N

using k=1.2421*10^(-11)

we get F=4.542*10^(-10) N

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