PHYS 34 HOMEWORK #3 Due : 8 Feb. 07. A unifrm chain f mass M, lenth L and density λ (measured in k/m) hans s that its bttm link is just tuchin a scale. The chain is drpped frm rest nt the scale. What des the scale measure at the mment the last link hits the scale? Slutin : This is a variable mass prblem, in that we are cmputin the frce exerted by the scale n the chain as the chain falls n the scale. As we discussed in class, part f the frce supprts the weiht f the chain n the scale, and part f the frce stps the mmentum f the fallin chain. We use the eneral frm f Newtn' s secnd law : F dp d (mv) m dv + v dm Since the chain is fallin under ravity with n air frictin (and n internal frictin between the links f the chain), we knw that dv/. Nw, we emply the chain rule and btain: v dm v dm dy dm/dy is just the mass density f the chain, λ, and dy/ is the speed f the fallin chain. Cmbinin all these, we et : dy F m + v λ where v is the speed f the last link as it hits the scale. We knw frm elementary physics that an bject fallin thruh a heiht L acquires a speed iven by: s we have: v L F m + L λ But L λ is just the mass f the entire chain, s we have finally:. Text, prblem, p. 9 F m + m 3 m. Slutin : We have a particle fallin in a ravitatinal field with quadratic air resistance. If we chse dwn as the neative directin (air frictin acts up) and ur frce equatin becmes :
phys34-07hw3s.nb F m dv - m + k m v (we' ll use the bk' s ntatin since we want t et their answer). Nw, the questin asks us t find the distance the bject falls in acceleratin frm initial t final velcity, this suests we shuld cnvert dv/ t dv/dx via : m dv m dv dx dx m dx We can separate variables (and divide ut a cmmn factr f m) t btain the differential equatin : Interatin bth sides between limits: v0 v k v - x (Since we chse up t be psitive, x > x. k v - dx x dx x - x -(x - x ) - v k ln k v - v 0 Evaluatin between limits and usin the prperties f ls ives the requested answer: s distance traveled k ln k - k v - Yu will btain the answer in the frm iven in the bk if yu set dwn as the psitive directin (and s wuld be psitive). 3. Text, prblem p. 9 Slutin : We break the prblem int tw parts, the upward prtin and the dwnward prtin. Fr the upward le, we write : F m dv -m - k m v Fllwin the pattern f the prblem abve, we cnvert dv/ t /dy and et: Interatin yields: + k v Apply the initial cnditin that v when y 0: - dy k ln + k v - y + C k ln + k C
phys34-07hw3s.nb 3 and cmbine results t et: y k ln + k v + k v We can use this equatin t find an expressin fr the maximum heiht f the prjectile. Since we knw the prjectile has zer speed at its hihest pint, we can write: y max k ln + kv Nw we cnsider the dwnward prtin. It is easiest t let dwn be the psitive directin, s that ur secnd law becmes: r This interates t: m dy m - k m v - k v dy - k ln - k v y + C Since we are chsin dwn as ur psitive directin, the trip starts at y 0 when v 0, s that we have: which yields : - k ln[] C y k ln - k v Nw, the hihest pint is cmmn t bth the upward and dwnward prtins. This means that their expressin fr the hihest pint must be the same, in ther wrds : r Slvin fr v : y max k ln + kv k ln - k v + k v - k v k k + k
4 phys34-07hw3s.nb We are asked t express this in terms f the terminal velcity. We can find the terminal velcity frm the statement f the secnd law fr the dwnward le. Terminal velcity ccurs when the velcity is cnstant, r when: dv 0 - k v 0 v t k Substitutin this int ur velcity expressin yields the desired result: v T v v T + v 4. Text, prblem 36, p. 95. Slutin : We start by writin the equatins f mtin : x (t) cs θ t y (t) h + sin θ t - t T find the rane, we slve fr the time f fliht by settin y0, and then use that time in the x(t) equatin t find rane. y (t) 0 t - sin θ t - h 0 We use the quadratic equatin t find: t sin θ + sin θ + h sin θ + sin θ + h Make sure yu can explain why we chse the plus branch f the slutin. If we substitute this fr time in the x(t) equatin, we et the rane: rane cs θ sin θ + sin θ + h Nte that if we set h 0, we recver the well knwn rane equatin fr a prjectile n level rund with n air frictin. Havin taken several semesters f calculus, yu knw that when yu are asked t maximize, yu set the first derivative f the apprpriate variable t zer. In this case, that means takin the first derivative f rane with respect t θ, which ives: d (rane) dθ
phys34-07hw3s.nb 5 -sin θ sin θ + sin θ + h + cs θ cs θ + sin θ + h sin θ cs θ 0 But this equatin is s nn-alebraic that is it nt pssible (r nt easily pssible) t slve fr the value f θ that maximizes rane fr a certain set f parameters. This is an example f a prblem that yu have t slve either by numerical r raphical means. Based n recent class wrk, yu miht think t expand the radical, but fr many values f and h, h v is nt smaller than, s that series expansin desn t help. Let s see hw raphical methds help. We already knw that if h 0 the anle that maximizes rane is 45. Let s cnsider: In[7]: Clear[rane, v0,, h, θ] 9.8;v030;h00; ranev0 Cs[θ] Sin[θ]+ Sin[θ]^ + h v0 ; Plt[rane,{θ,0,π/4}] 65 60 55 Out[0] 50 45 40 0. 0.4 0.6 0.8 Remember that the hrizntal axis is in radians, s we can see that the maximum rane fr these parameters (launch speed 30 m/s, heiht 00m) is abut 9. We can be mre exact by usin the FindMaximum cmmand: In[]: FindMaximum[rane, θ] Out[] {63.7, {θ 0.55}} And we btain that the maximum rane is 63 meters and the launch anle is 0.5 radians (9.3 ). Or we culd try t differentiate the rane expressin, set it equal t zer, and slve fr θ by usin
6 phys34-07hw3s.nb FindRt: In[3]: FindRt[D[rane, θ], {θ, π / 4}] Out[3] {θ 0.55} Technly is yur friend. 5. Suppse the mn is stpped in its rbit at a distance r frm the Earth and beins t fall inward. Determine the time it will take fr the mn t crash int the Earth. Dependin n hw yu apprach this prblem, yu miht encunter an interal where the substitutin r r sin θ is useful. Lk up the values f the apprpriate astrnmical parameters (mass f Earth, averae distance f mn frm Earth, etc.) and calculate the time it will take fr the mn t hit the Earth. Slutin : We bein, as all thins d, with Newtn' a secnd law : F m dv - G m M r where r is the instantaneus distance f the mn frm the Earth. We can et an expressin fr v in terms f r by settin dv/ /dr and: separate variables, interate, and: we knw that v 0 when r r, s and : dr v 0 G M r - G M r G M r + C + C C -G M r v G M r - r T find time f infall, we nte that v dr/, s: dr v dr G M r - r Nw we make use f ur handy substitutin: and r r sin θ dr r sin θ cs θ dθ
phys34-07hw3s.nb 7 r - r r sin θ - r - sin θ r sin θ cs θ r sin θ substitutin this int ur interal: r sin θ cs θ dθ G M cs θ r sin θ r : r3/ t sin θ dθ G M we have left ut ur limits f interatin. In r space ur limits are r and 0; these translate int θ0 and θ π/, s ur cmplete interal is: G M r 3 0π/ sin θ dθ π 4 G M r3 Substitutin values (all in MKS (SI) units): G 6.67 0 - ; M 6 0 4 k; r 3.84 0 8 m we find that the time t infall is 4.8 0 5 s r 4.84 days.