HW 2 Solution Key - Drexel University

1 Classical Mechanics I HW # 2 Solution KeyHW 2 Solution Key1. (5 points) shell traveling with velocity v explodes into three pieces of equal masses. One piece hasvelocity~v1=~v0and the other two have velocities that are equal in magnitude but mutuallyperpendicular. Find the two velocities and sketch the three 3m=M, andvas speed of the two unknown the outgoing shells, we get:~v2=vcos 2 i+vsin 2 j~v3=vcos 3 i+vsin 3 jThe only way for momentum in the y-direction to be conserved is if 2= 3, so sin 3= sin 2andcos 3= cos 2. Further, for them to be perpendicular:~v2 ~v3=v2(cos2 2 sin2 2) = 0which is satisfied for = /4 = 45 Thus:v2=v0 2 i+v0 2 j2. (5 points)A uniform thin sheet of metal is cut in shape of a semicircle of radius R and lies in the xyplane with its center at the origin and diameter lying along the x axis. Find the position ofthe demands that the CM must lie along the y-axis.

2 But where?First, note that the total mass of a semicircle is:M= A=piR2 2where is the density per unit :yCM= ydAM= (rsin )r dr d M= R3 /3 0sin d M= R3 /3 cos | 0M=2 R3/3M=4R3 3. (5 points)A uniform spherical asteroid of radiusR0is spinning with angular velocityomega0. It picksup more matter until its radius is R. The density remains the same and the additional matterwas originally at rest Find the final angular velocity if the radius a couple of things. First, angular momentum is conserved, so: = 0I0 ISecond, the moment of inertia of a sphere is:I=25MR2If the radius doubles,M= 8M0, andR2= 4R20, soI= 32I0. Thus: = 0324. (15 points) A planet, of mass,mis orbiting a star of mass,M. At some instant in its orbit,t0, it is adistance,r0from the star, has a radial velocity,vr,0, a tangential velocity,vt, :The algebra in the latter half of the problem is a bit hairy. You ll be fine, though.

3 (a) In terms of the numbers above, compute the mechanical energy,T+U, of the planet point of this problem is really to recognize that a few fixed numbers define the properties ofan orbit entirely. In this case:T(t0) =12m(v2r,0+v2t,0)andU(t0) = GMmr0 These combine to produce a total (conserved) energy:E=12m(v2r,0+v2t,0) GMmr0,a quantity we ll use in later parts of the problem.(b) In terms of the numbers above, compute the angular momentum,Lof the planet angular momentum, only the tangential velocity matters. Thus:L=mvt,0r0which is also means, for what it s worth that atanypoint in the orbit:vt=Lmr(c) There are two points in the orbit known asperiastronandapastronthat represent the nearestand closest approach to the star. At those times, the radial component of the velocity is conservation of mechanical energy and angular momentum, compute the distance from thestar at apastron and periastron in terms ofM,r0,vr,0, andvt, can, of course, leave all 5 variables in the expression if you like, but as we ve already computedthe conserved quantities,E, andL, I m going to write out my Solution in terms of periastron and apastron:E=12mv2t GMmrwith no radial term (and whereEis the conserved energy computed in part a).

4 Substituting fromthe angular momentum calculation ofvtgives:E=L22mr2 GMmrThis is nice! There is only one unknown,r. It is even simpler if I multiply byr2and recast theequation as:|E|r2 GMmr+L22m= 0where I cleaned up a little by noting thatE <0. At any rate, this is simply a quadratic equation,which has two solutions forr(the smaller, given by the minus sign, is perihelion and the plus signgives apastron.). So:ra,p=GMm G2M2m2 2|E|L2/m2|E|(d) What is the kinetic energy of the star at periastron and apastron? You may express your answerin terms each case:va,p=Lmra,psoT=12mv2=L22mr2a,p5. (15 points)A particle of mass m is moving on a frictionless horizontal table and is attached to a masslessstring, whose other end passes through a hole in the table, where I am holding it. Initially theparticle is moving in a circle of radiusr0with angular velocity , but I now pull the stringdown through the hole until a lengthrremains between the hole and the (a)What is the particles angular velocity now?

5 From conservation of angular momentum we have:m 0r20=m r2so = 0r20r2(b)Assuming that I pull the string so slowly that we can approximate the particles path to be circleof slowly shrinking radius, calculate the work I did pulling the much force are we applying? Well, the point of this question is that the force is the centripetalforce, which is:F=mv2r=mr 2=mr40 20r3where I ve ignored the sign (the force is inward). The work done is thus:W= Fdr= r0rmr40 20r3dr=12mr20 20(r20r2 1)where I cleaned up a bit in the last line.(c)Compare your answer to part (b) with the particles gain in kinetic kinetic energy at any given time is:T=12mv2=12mr2 2=12m 20r40r2 Thus: T=12mr20 20(r20r2 1)6. (10 points)A small frictionless puck perched at the top of a fixed sphere of radiusRis given a tiny nudgeand begins to slide down. Through what vertical height will it descend before it leaves thesurface of the sphere?

6 , note that if the top of the sphere if = 0, the puck will fall a vertical height,hsuch that:h=R(1 cos )as we did for the pendulum. But further, that descent gives us an increase in kinetic energy:12mv2=mgR(1 cos )orv2= 2gR(1 cos )To keep the sphere, the gravity must be sufficient to satisfy the centripetal force:mgcos =mv2 Rcos = 2(1 cos )cos =23soh=R(1 cos ) =R37. (5 points)Calculate the gradient fof the following functions,f(x,y,x):Sol.(a)f= ln(r) The problem was phrased rather awkwardly, but since we know how to take gradientsin spherical coordinates, these are trivial: f= f r r=1r r(b)f=rn f=nrn 1 r(c)f=g(r) g= g r r8. (20 points) Let s get a little practice taking divergence, gradient, and curl, since we re going to needto do these things later. Let s define a couple of scalar functions (one in Cartesian coordinates, andone in spherical) and a couple of vectors:f(x,y,z) =x2+y2g(r, , ) =rcos ~u=y i x j~v=y i+x jPlease calculate:(a)~u ~vSol.

7 ~u ~v= i j ky x0y x0 =2xy k(b) ~uI believe we did this one and the next one in class. It is:Sol. ~u= i j kddxddyddzy x0 = 2 k(c) ~vSol. ~u= i j kddxddyddzy x0 =0(d) ~uSol. ~u=d(y)dx+d( x)dy= 0(e) ~vPerhaps a bit ~u=d(y)dx+d(x)dy= 0(f) fSol. f= f x i+ f y i+ f z k=2x i+ 2y j(g) can actually do this in 2 ways. The simplest is to note thatg=x, and thus: g=dgdx i= iPicking the best coordinate system can be very you can also use polar coordinates: g= g r r+1r g (Note that in 2-d, we typically used rather than in the class, but it ll work fine either way).So: g= cos r sin I ll leave it to you to show that this is the same as i.(h) Could either of~uor~vdescribe a conservative force (generated by U)? ~vcould, since it is curl free. Incidentally, the potential might take the form:Uv= xy+Cbut you aren t required to compute (10 points) child toy has a shape of a cylinder mounted on top of a hemisphere.

8 The radius of thehemisphere is R and the CM of the whole toy is at a height h above the (a) Write down the gravitational potential energy when the toy is tipped to an angle from center of mass,h and the angle, can be computed as:h = (h R) cos +R=hcos +R(1 cos )Thus:U=mgh =mg(hcos +R(1 cos ))relative to the floor. If we wanted to compute relative to vertical, we d subtractmgh.(b) Find the values of R and h for which the equilibrium at is , note:dUd =mg((R h) sin )which, naturally, is zero at = 0 (the definition of an equilibrium). For it to be stable:d2Ud 2>0sod2Ud 2=mg((R h) cos )which is stable forR > (10 points)Sol.(a) Consider a electron in a circular orbit of radius r around a fixed proton. The centripetal force isgiven by the Coulomb force. Prove thatT= 1/2U,E=T+U= 1 of these statements are, of course, equivalent. We actually did this in class. The force is:|~F|=ke2r2=mv2rfor a particle in a central potential and a circular orbit.

9 The potential energy is:U= ke2r(negative because it s attractive.)From the force equation, we get:mv2= 2T=ke2r= Uwhich proves the result.(b) Consider the inelastic collision of an electron with a hydrogen atom. Write down the total energyof the is actually just a one-liner. Since the total energy of the bound atom isU/2, then ifT1isthe energy of the electron prior to collision:E=U2+T1(c) (c)Find out the final kinetic energy of electron 1,T 1. Energy is conserved, so:T 1=T1+12U 12U The electron can gain or lose energy by transferring it to the atom.