Find the total energy of a sphere with radius $$R$$ and a uniform charge density $$\rho$$.

The electrical potential energy of a test charge $$dq$$ in the electric field of a charge $$Q$$ is given by $$dU = \frac{1}{4 \pi \epsilon_{o}}\frac{Qdq}{r}$$ Now, to find the total energy of the sphere, we break it down into shells of radius $$r$$ containing a charge $$4 \pi r^{2} \rho dr$$ and sum up the energy contributions from each shell $$U = \int_{0}^{R} 4 \pi r^{2} \rho dr \frac{1}{4 \pi \epsilon_{o}}\frac{\frac{4}{3} \pi r^{3} \rho}{r}$$ $$ = \frac{4 \pi \rho^{2}}{3 \epsilon_{o}} \int_{0}^{R}r^{4} dr$$ $$= \frac{4 \pi \rho^{2} R^{5}}{15 \epsilon_{o}} $$

A heavy chain of mass $$m$$ and length $$L$$ is attached to a block of mass M, which lies on a frictionless table. A small piece of the chain $$l$$ dangles through a hole in the table. Find the length of chain $$x$$ underneath the table as a function of time, assuming the chain starts at rest.

To solve this problem, we consider the forces acting on the piece of chain. Because the length of chain does not change, its acceleration (and the acceleration of the block) will equal $$\ddot x$$. We have two forces to consider: gravity and tension Gravity only provides a net force of the length of chain underneath the table (the force of gravity on the remaining length of chain is balanced by the normal force of the table). The net force of gravity is given by $$Fg = \text{mass below table} \times g = \frac{m}{L} \times x \times g$$ Therefore, our equation of motion is: $$ m \ddot x = \frac{m}{L}gx - T$$ Where T is the tension in the chain. Because tension provides the acceleration of the block (which is $$M\ddot x$$), by Newton's third law we have: $$m \ddot x = \frac{m}{L}gx - M\ddot x $$ Re-arrange to get: $$\ddot x - \frac{m}{M+m} \frac{g}{L} x = 0 $$ If we take $$\delta = \sqrt{\frac{m}{M+m} \frac{g}{L}}$$, we will have a solution of the form $$x(t) = Ae^{\delta t} + Be^{-\delta t}$$. Now we apply the initial conditions $$x(0) = l$$ and $$\dot x(0) = 0$$: $$ A + B = l$$ $$A - B = 0$$ Therefore, the length of chain under the table is given by $$x(t) = \frac{l}{2}e^{\delta t} + \frac{l}{2}e^{-\delta t}$$, where $$\delta = \sqrt{\frac{m}{M+m} \frac{g}{L}}$$

A skydiver experiences forces due to gravity and turbulent drag as she is falling. We can model her velocity with the following equation: $$ m\dot v = mg - bv^2$$ a) Find an expression for the terminal velocity b) Find the skydiver's velocity at time $$T$$ assuming she starts at rest

a) Terminal velocity is the velocity where $$\dot v$$ is zero. We can then solve for $$v$$: $$v = \sqrt{mg/b}$$ b) To solve the differential equation we employ separation of variables. Start by solving for $$\frac{dv}{dt}$$: $$\frac{dv}{dt} = g - \frac{b}{m}v^2$$ Move all the dependent variable terms over to the left hand side and all the independent variable terms over to the right hand side, then integrate $$ \int_{0}^{v(T)}\frac{dv}{g - \frac{b}{m} v^2} = \int_{0}^{T}dt$$ The right hand side is a trivial integral. The left-hand side can be simplified by some manipulation and the substitution $$v' = \sqrt{\frac{b}{mg}}v $$ $$ \int_{0}^{v(T)}\frac{dv}{g - \frac{b}{m} v^2} = \frac{1}{g}\sqrt{\frac{mg}{b}} \int_{0}^{v(T)\sqrt{\frac{b}{mg}}}\frac{dv'}{1 - v'^2} = \sqrt{\frac{m}{bg}}ArcTanh(v(T)\sqrt{\frac{b}{mg}}) = T $$ Therefore, we have $$v(T) = \sqrt{\frac{mg}{b}}tanh(\sqrt{\frac{bg}{m}}T) $$