Question

General Physics Integration Examples to Solve F 1. Find the total force on an acquarium window. Given that force is related to pressune area thusly: F-p. And that the pressure varies with depth below the surface according top-ped where ρ is the water density, g is acceleration due to gravity, and d is depth below the surface. (Problem: The force is not the same over the window. Hint: divide the window into pieces so that within each piece the pressure is constant. Then integrate the pieces) 2. Find the moment of inertia of a disk of mass M and radius R. moment of inertia for an object that has i point masses m, some distance ri from an axis is given by Eimr?. (Problem: The radius is not the same for all of the parts of the disk. Hint: Divide the disk into pieces so that within each piece, the radius is constant. Then integrate the pieces) Use the basic fact that 3. The energy to stretch a spring of constant k through a total distance s. You may use the fact that the force needed to stretch a spring of constant k by a distance x is given by F-kx. Also a force F when acting on an object that moves through a distances does an amount of work given by W- Fs. (Probem: The force is gradually increases as the spring is stretched. Hint: divide the motion into increments so that within each increment, the force does not change. Then integrate the increments) F-ks cs

Answer 1

1.The total force on the aquarium window is given by:

$F=\int PdA=\int_{0}^{d}(p_{0}+\rho gh)wdh= \int_{0}^{d}p_{0}wdh+\int_{0}^{d}\rho gwhdh=p_{0}wd+\rho gwd^{2}/2.$

Here p₀ is the pressure at the top of the aquarium and w is the width of the aquarium.

2.The moment of inertia of the circular disk is given by:

$I=\int r^{2}dm=\int r^{2}\rho dA= \int r^{2}\rho 2\pi rdr=\rho *2\pi \int r^{3}dr=M/\pi R^{2}*2\pi R^{4}/4=MR^{2}/2.$

Here $\rho$ is the density of the disk and dm is the mass of a differential element of the disk of area dA which is at a distance r from the center of the disk.

3.The potential energy to stretch the spring by a distance s is given by:

$U=-W=-\int dW=-\int_{0}^{s}-Fdx=\int_{0}^{s}kxdx=1/2ks^{2}.$ where dW is the work done in stretching the spring by a distance dx.The total work done is the integral or summation of the differential elements of work over the entire distance.