EN40: Dynamics and Vibrations. Midterm Examination Tuesday March

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EN4: Dynaics and Vibations Midte Exaination Tuesday Mach 8 16 School of Engineeing Bown Univesity NME: Geneal Instuctions No collaboation of any kind is peitted on this exaination.

1 EN4: Dynaics and Vibations Midte Exaination Tuesday Mach 8 16 School of Engineeing Bown Univesity NME: Geneal Instuctions No collaboation of any kind is peitted on this exaination. You ay bing double sided pages of efeence notes. No othe ateial ay be consulted Wite all you solutions in the space povided. No sheets should be added to the exa. Make diagas and sketches as clea as possible, and show all you deivations clealy. Incoplete solutions will eceive only patial cedit, even if the answe is coect. If you find you ae unable to coplete pat of a question, poceed to the next pat. Please initial the stateent below to show that you have ead it `By affixing y nae to this pape, I affi that I have executed the exaination in accodance with the cadeic Hono Code of Bown Univesity. PESE WRITE YOUR NME BOVE SO! 1 (19 points). (11 points) 3. (1 points) TOT (38 points)

2 1. The figue shows an inetial cawle that will spontaneously tanslate to the ight ove a vibating suface. ssue that The device and suface ae both at est at tie t=. Fo the tie inteval <t<t (whee T is a constant) the suface has velocity + Vi < t < T / v = Vi T / < t < T V is sufficiently lage that slip occus at thoughout the inteval <t<t. The contact at has a fiction coefficient µ. Cente of ass The wheel at B is assless and olls feely (so the contact at B is subjected only to a noal foce). The cente of ass of the device is idway between and B and a height h above the suface. h Base otion j B i 1. Daw a fee body diaga showing the foces acting on the cawle duing the peiod < t < T / on the figue povided below. Note that duing this phase of the otion the suface oves to the ight, and is oving faste than the cawle. j i h B µn N N B. Wite down Newton s law and the equation of otational otion fo < t < T / [3 POINTS] N i+ ( N + N g) j= ai B ( N N h N ) k = B [3 POINTS] 3. Hence, show that the acceleation of the cawle fo < t < T / is given by µ g a = i µ ( h/ ) N i+ ( N + N g) j= ai B N N h N = B N + N g N + N = g B B ( h/ ) ( ) ( ) N N h N = N 1 h/ N = B B dd : N h / = g N = a a = g [3 POINTS]

3 4. Repeat steps 1-3 fo the peiod T /< t < T (note that duing this phase of the otion the suface oves to the left) to show that duing this tie inteval µ g a = i + µ ( h/ ) New FBD j i h B µn N N B New EOM N i+ ( N + N g) j= ai B N + N h N = B New solution N + N g N + N = g B B ( + h/ ) ( ) ( ) N + N h N = N 1 + h/ N = B B dd : N + h / = g N = a a = g [4 POINTS] 5. Use the esults of (3) and (4) to find a foula fo the distance that the cawle oves duing the tie peiod <t<t. ssue that the cawle is at est at tie t=. The acceleation is constant duing both phases of otion so we can use the staight-line otion foulas: µ gt 1 1 µ gt < t < T /: v = at = x = at = µ ( h/ ) µ ( h/ ) µ gt / 1 µ gt /4 t t = T / v = x = µ ( h/ ) µ ( h/ ) 3

4 µ gt / µ g( t T /) T/ < t< T: v= v + at ( t) = µ ( h/ ) + µ ( h/ ) 1 1 µ gt /4 µ gt / 1 µ g( t T /) x= x + v( t t) + a( t t) = + ( t T /) + µ ( h/ ) µ ( h/ ) + µ ( h/ ) 1 µ gt /4 µ gt /4 1 µ gt /4 t t = T x = + + µ ( h/ ) µ ( h/ ) + µ ( h/ ) µ gt 3 1 µ gt (4+ 4 µ h / ) µ gt (1 + µ h / ) = = = 8 µ ( h/ ) + µ ( h/ ) 8(4 ( µ h/ ) ) (4 ( µ h/ ) ) [4 POINTS] 6. Find a foula fo the iniu value of V fo slip to occu at the contact point between the cawle and the suface in the tie inteval <t<t. The suface ust be oving faste than the cawle thoughout the tie inteval. This equies µ gt / V > µ ( h/ ) [ POINTS] 4

5 . The figue shows a chaged paticle with ass in a static Kingdon tap that uses an electic field to confine the paticle within a cylinde. The electic field subjects the paticle to a adial foce F F= e whee F is a constant (which depends on the electic field in the cylinde). t tie t= the paticle is located at position θ =, = R and has velocity vecto v= Ve θ j e θ ion Wie Cylinde θ i e.1 Wite down the foula fo the acceleation of the paticle in pola coodinates, in tes of tie deivatives of, θ. (Do not assue cicula otion). This is just the pola coodinate acceleation foula d d θ d θ d d θ a= e + e θ dt dt dt dt dt [ POINTS]. Hence, wite down Newton s law F=a fo the paticle, using pola coodinates. F d dθ d θ d dθ e e e dt dt dt dt dt = + + θ [1 POINT].3 Wite down the angula oentu vecto of the paticle at tie t=, in tes of V and R Fo the angula oentu foula h= v= Re Voeθ = RV k [1 POINT].4 Use angula oentu, o othewise, show that θ and ae elated by dθ VR = dt This is a cental foce poble so angula oentu is conseved. t a late tie d dθ dθ and v= e e + eθ = k dt dt dt d d VR ngula oentu consevation gives RV θ θ = = dt dt d dθ v= e + e θ, dt dt 5

6 You can also get the sae esult fo the e θ coponent of F=a (pat.1) which shows d θ 1 d d θ d d θ + = = dt dt dt dt dt This can be integated to get dθ = const dt nd we can find the constant is RV fo the initial conditions. [ POINTS].5 Use the answes to. and.4 to show that the coodinate satisfies the diffeential equation Pat. shows that ( VR) F + = 3 d dt d dθ F = dt dt dθ VR Substituting = fo pat.4 gives dt This gives the equied answe. VR F + = d dt [3 POINTS].6 Reaange the equation of otion fo in pat 3.3 into a fo that MTB can solve. We need to tansfo the second ode equation into two fist ode equations. et v d ( VR) dt v = F 3 d v =, then dt [ POINTS] 6

7 3. The figue shows a device that is intended to detect an ipulse. If the casing is subjected to an ipulse that exceeds a citical agnitude, the ass will flip fo its initial position to the ight of the pivot to a new stable position to the left of the pivot at. The goal of this poble is to calculate the citical value of ipulse fo which this will occu. Ipulse I / x(t) Α k, Casing Mass M 3.1 t tie t= the syste is at est and the sping is un-stetched. The casing is then subjected to a hoizontal ipulse I. Wite down a foula fo the speed of the casing just afte the ipulse. Note that the sping exets no foce on eithe the casing o the ass duing the ipulse. Just afte ipulse y(t) / V x(t) Α k, M MV = I V = I / M [1 POINT] y(t) V 1 3. Find expessions fo the total linea oentu and total kinetic enegy of the syste just afte the ipulse, in tes of I and the ass M of the casing. 1 1 T = MV = I M p= MV i= Ii Sping at shotest length / y(t) x(t) Α k, M [ POINTS] 3.3 Conside the syste at the instant when the sping eaches its shotest length (assue x> ). Using enegy and/o oentu consevation show that at this instant x= I km ( M ) + 4 (You can find the sping length using Pythagoas theoe) t this instant both the casing and the poof ass ove with the sae (unknown) speed v 1. Moentu consevation equies that ( + M) v1 = I Enegy consevation equies that 1 I 1 ( M v ) 1 { } 1 k x /4 M = Hee the fist te on the ight is the KE, the second is the PE of the sping. We used x + /4instead of x /4 positive) Eliinate v 1 : + fo convenience (it akes the quantity inside the { } 7

8 { } / I = k x + M M + x= I km ( M ) + 4 [5 POINTS] 3.4 Hence, find a foula fo the citical value of I that will flip the ass past x=. M Set x= in the peceding poble and solve fo I: I = km + 1 [ POINTS] 8

EN40: Dynamics and Vibrations. Midterm Examination Thursday March

EN40: Dynamics and Vibrations. Midterm Examination Thursday March EN40: Dynamics and Vibations Midtem Examination Thusday Mach 9 2017 School of Engineeing Bown Univesity NAME: Geneal Instuctions No collaboation of any kind is pemitted on this examination. You may bing