Solutions Assignment 2

Transcription

1 Solutions Assignment.9 (a) The EOM for a particle in a vacuum are = and y = g: Integrating these equations for a particle starting from the origin with initial velocities v o and v yo yields The trajectory, y (), is then (t) = v o t and y (t) = gt + v yo t: y (t = =v o ) = v yo v o (b) Equation.37 in the tet is y = v yo + v ter + v ter ln v o g v o : : v o In the limit of small resistance both and v ter = g become very large. This enables us to epand the log function as ln = v o v o + vo vo : Substituting this epansion into the equation for the trajectory yields y ' v yo + v ter v ter v o v o + vo vo 3 3 y ' v yo g + 3 g 3 v o 3 vo 3 = y vac 3 vo 3 : v o The leading order term agrees with the vacuum epression, and the rst order correction shows that the height is reduced for even small drag e ects..39 (a) The EOM and its integral for a cyclist are Z v m = cv f f + cv = Z v c f=c + v = m Z t = t m : As a change in variables let v= p f=c tan! f=c + v = (f=c) = cos and = p f=c= cos : The integral of the EOM then becomes t m = r c tan p c=fv tan p c=f c f m t = p tan p c=f tan p c=fv : fc

2 (b) For c = :N= (m= sec) m = 8kg and f = 3N the times to slow are given by the epression t = 3:3 tan 5:64 tan :58v : For an initial speed = m=s the time required to slow to 5m=s, m=s, 5m=s, and m=s are v (m=s) 5 5 t (s) 6:3 8:4 48:3 4:5 :.4 For a baseball thrown vertically upward with a velocity of the EOM is m = mg cv = mg + v =vter = g + v =v ter where v ter = mg=c: Since = = v (=dy) we can write Z v v + v =vter = g Z y dy For the maimum height v = : This leads to or v ter Z + v =vter d v =vter = y ma = v ter g ln + =vter : vter ln + =vter = gyma In a vacuum v ter! : Epanding the natural log we nd y ma = v ter g ln + =vter v = ter g y ma = v + g v ter v =v ter v 4 =v 4 ter + As epected the maimum height is reduced in the presence of drag. For a baseball thrown upward with a velocity of = m= sec and a terminal velocity of v ter = mg=c = p :5 9:8= (:5 49) = 35m=sthe maimum height is y ma = 7:m as compared to v=g = :4m in a vacuum.

3 .4 For a baseball dropped from the elevation y ma of problem.4 (y is now positive going down) the EOM and its integral are Z V m = mg cv = mg v =v ter = v dy = g v =vter v v =vter = v Z V Z ter (v =vter) d v =vter yma = gdy v ter gy ma = ln V =vter where V is the velocity when the ball returns to the ground. Using the epression for y ma obtained in problem.4 yields v ter ln V =vter v ter = ln + =vter V =vter = = + v=v ter V =vter = = + v=v ter = v =vter= + v=v ter q q V = = + v =v ter = v ter = vter + v It is of interest to eamine this result for small drag, =v ter << and large drag, v ter = << : These results are =v ter < < : V = = q + v =vter ' vter q =v ter > > : V = v ter = + vter=v ' v ter : In the case of small drag the return velocity is slightly less than the initial velocity while for large drag the return velocity is slightly less than the terminal velocity. For the case of the baseball in problem.4 V = 35= p + 35 = 7:4m=s as compared to a return velocity of m=s in a vacuum..49 (a) From Euler s theorem Thus we can write z = e i = cos + i sin : e i = cos + i sin = e i = (cos + i sin ) e i = cos sin + i sin cos : Equating real and imaginary parts we nd cos = cos sin and sin = sin cos : v ter v 3

4 .55 (a) Now consider crossed E and B elds,! E = Eby and! B = Bbz: The EOM is m d! v = q (Eby +! v Bbz) : Separating this vector equation into its components yields =!v y y = qe m!v and z = where! = qb=m: The initial conditions are v = with all other initial component velocities vanishing. This allows us to integrate the equation for the z component and nd v z =! z = z : De ning z = the particle remains in the plane de ned by z = : (b) Note from the equation for y = that if = qe=m! = E=B our remaining EOM at t = reduce to y = =!v y: After a time t we see these equations imply v y = v y + y t = v y: From the initial condition v y = the epression for v after a time t becomes v = + t = +!v y t = : Continuing this integration process yields v y = and v = for all time. (c) To solve these equations for a general initial it is convenient to de ne v = u + qe=m! so that u is the velocity di erence between v and the drift velocity = qe=m! = E=B: The EOM of interest then reduce to du =!v y y =!u : We have already solved these EOM. For this initial condition these solutions are u = u cos!t! v = E=B + ( E=B) cos!t v y = u sin!t! v y = ( E=B) sin!t: 3.4 (a) Assuming that the velocity of the hobos relative to the atcar is u then from the conservation of momentum after both hobos jump we nd m fc v f m h (u v f ) =! v f = m h m fc + m h u: 4

5 (b) If the hobos jump one after the other (with velocity relative to the atcar remains u for both hobos) then after the rst hobo jumps the velocity of the at car vf is found from (m fc + m h ) f m h u f =! v f = m h m fc + m h u: After the second hobo jumps the conservation of momentum yields Since we have m fc v f m h (u v f ) = (m fc + m h ) vf m h (m fc + m h ) v f = m h u + (m fc + m h ) u m fc + m h m h m h v f = + u m fc + m h m fc + m h m h m fc + m h + m h m h > m fc + m h m fc + m h m h m fc + m h > m h m fc + m h and v f for process (b) is greater than v f for part (a). Algebraically you could also take the ratio of the nal velocity for process (b), v b to that for (a), v a : This results in m h m h mfc + m h v b =v a = + m fc + m h m fc + m h m h mfc + m h v b =v a = + m fc + m h = m fc + 3m h > : m fc + m h 3. (a) In a time the change in momentum from Newton s nd law is F et = dp = (m + dm) (v + ) mv + dm (u v) = m + udm where F is any eternal force. Hence m = F et (b) In a gravitational eld F et = u dm : mg and the EOM takes the form m = mg udm : Assuming that the rocket ejects mass at a constant rate, m = m kt we nd (m kt) = (m kt) g + ku: 5

6 Separating and then integrating to solve for v = g + ku kt v = u ln m m m kt gt = u ln m m (t) gt This is eactly what one would epect given the solution for the velocity of a rocket in the absence of a gravitational eld. (c) For the data m = 6 kg m ( min) = 6 kg and u = 3m= sec the approimate velocity of the shuttle after min is v = 9:8 + 3 ln = 93m=s In free space (g = ) the velocity would be v = 3 ln = 8m=s: (d) The rocket would remain on the launch pad reducing its mass until udm= mg: 3.3 From problem 3. the rocket s height is given by y (t) = Z t u ln m m kt gt = Z t y (t) = u (m =k t) ln (m kt) m gt + ut y (t) = u k m ln m m gt + ut: u ln m =k t m =k gt From the data in problem 3.7, m = 6 kg m (t = min) = 6 kg and v e = 3m= sec : after min y (t = min) = 3 6 = 6 ln 9:8 () + 3 () y (t = min) ' :5 5 m :7 5 m + 3:6 5 m = 4 m 3. The CM is found from the epression Z! R =!r dm: M For an object of uniform density, it is convenient to epress the mass element as dm = dv where is the mass per unit volume. For this eample = 3M=R 3 : The volume element in spherical coordinates is dv = drr sin drd = r dr sin dd: For a hemisphere the coordinates r range from!! = and! R respectively. Due to symmetry the and y 6

7 coordinates for the CM are both zero. The integral for the z coordinate of the CM of a hemisphere is z CM = 3 R 3 Z R Z = Z zr dr sin dd = In spherical coordinates z = r cos hence 3 R 3 Z R Z = zr dr sin d z CM = 3 Z R R 3 r 3 dr z CM = 3 R 3 R 4 4 Z = = 3 8 R: cos sin d = 3 R 3 Z R Z = r 3 dr d sin Uniform Rod After the impulse the linear momentum of the rod is F t = mv cm! v cm = F t=m: The angular momentum about its center of mass is F tl= = I! = ml!: Eliminating F t in terms of v cm we nd mv cm l= = ml!=!! = 6v cm =l: Immediately after the impulse the velocity of the end of the rod struck by the impulsive force is v = v cm +!l= = 4v cm : The velocity of the opposite end of the rod at this point is v = v cm!l= = v cm : 7