50cuTt 0 Ns vl,,oPhysics l2lB-RosenbergAutumn 2013

Prinaed Name

Exam 3November 22, 2013

Sest Numberl.ttl lirsl

I certit thrt the work I shrll lub|r|it k my owtr crcation, ||ot copied frod sry source,.odthet I shall rbid by rhe exrmirrlion procedures oollind below.

Sigtrrture Saudena ID Number

_l rould litc to pi.t up m! .rrm in cl.$ or pi.k il !p 1.t.. f.or Su'n Mill.r! oflic. (R. C116)_ Pl.rr. hold my. m i. Su3.n itlill.t! d.rl for pri'rl. pi.k{p

READ THIS ENTIRf, PAGE NOW, BEFORE THE HALF-HOUR BELL.Do !9!!opeD the exem before the hrlf-hour bll,

You will have 50 mirutes after lhc bell to complete the cramin{tion.No writing rfter the final bell.

N CELL PHONES, TEXT MSG, etc, ALLOWED AT ANY TIME

Before the eram begins:. Write )our name, student ID number and exam version on your bubble sheet. and

fill in the corresponding 'bubbles" using dark pencil marks.

DuriDg tbe erad:. lmportcnt first step: Print your name and student ID at the top of each page.. lf you are confused about a question, raise your hand and ask for an explanation-o If you cannot do one pan ofa problem, move on to the next part.. This is a closed book examination. You may not bring any information or an

equation sheel to this exam.. You may use a calculator, but you miy not use text storage capabilities, graphics

capabilities, intemet connections, phones, nor any programmable devicc.. You may use scratch paper. Do not tum in scratch paper. it will not be graded. You may not communicate with any person.

For multiplo.choice problems (those on rdhiie paper):. Fill in bubble sheets carefully and derkly. Make no stray marks. Erase carefulll.. Also circle youi choices directly on the exam paper for later reference.

For hand gnded problems (those on colored p.per):. If )ou need more space fian is available to answer any part ofa problem, use the

brck side ofthe seme pagc to complete your answer. Clearly indicate to thegrader that you used the back side. Do 491us scratch paper.

. Sho\,\'your work in ercugh detail so that the grader can follow your reasoning andyour method ofsolution. Circle your answers, and sta(e units ifappropriate. Fornumerical ansxers significant figures should match the number of significantligures in the numerical values given in the problem (usually 2 or 3).

Autumtr Qusrrcr 2013l2lB (homrs) Eran 3 Equltion8

F-nn^6 dt

i -nf; g-9.8N/kg_fr_frv-- a--dt dt- | t - _ - \rcv - il\4+ m2rz+...+ mNrN )F^.Tfr"^E - Kr+ Kr + Kr+...+rr (ar)+v (ar)+v (ru)+ -.

. . t -p2r-Ktr -:nnt' -:-i K- -:Ia'22m'2dK-F.dP-tF.fl-l4:l*vvl

' ' ldt l la l=du,r, - --;hdnBrior

Y,(zl-nvv (r) - h%lzi & - 8.99 x loeJ. tr/ct; lrl - r.662 t ro-'t"

v(r)--cn\bt

n@-cffrrk)--r"t-'.)

!.vl)':k,lr-r,fu'Y - u,Y,+u "Y"+u,Y, -

w cosonag 0 xl - UY sn?0,1-*(u,rt,-u"vt)+i(u,y,-u,v,)+i(u,v,-u,v,)

G - 6.67 xlllt ! .m*igz

L * r x I' - mr-u (C[Cular mOIlOn)-dL-=

dt- tdot^ va-l_lai a4_

ldt l rloop

= 16 (azimuthally symmeaic)

$.d

lMr'

ln.ra- Mcqrnadarad*

lMr*

r=\- ,2

VR

U-^ - a Nk^T: t - - 1.38xl0-"J/K

dIJM-McdTi s- ln t@l

dU *^ - m L; hrcnt h6t io v+o.iz or F.hRockets with no externalforces. -d, -= dMM_=y.-

dt-qr-w

dt dt

llia{h.tl('lft!

,,.Dumlonn crrculat monon: ,t trotrurf form circular motion:

- dv^ v2^dt .R

v ( t l=q t+v^

4.'"u- =P4. "l-Fd,"E=;CpAv'

x(t) - Ls,t2 +vo,t + xo

Name Studetrt lD Score

L CoID o|r t[rDtablc. A small coio is placed on a llat, borizontal tumtable. The lumtablcmSkes 3 clockwise revolutiotrs i, r secolds.

l. (4 pls) Which answer is closest to the spccd ofthe coin as it rides a distancc 5-0 cm ftom thelErter4t{he tumtable?( A.0 r s.)>-rnd/rC. 20 m,/s.D, 30 m,/s.E. 40 m/s.

2. (4 pts) Which alswer bst describes the direction ofthe acceleration ofthe coin?A-+rQially outwads from the centcr.

Q$dially inwards towards the center.C. One componenl is radially inwards lowards thc center, atrothe! component is in the directionoffte coil's velocity.D. Orc comporcnt is radially outwards, another component is opposite the direclion of thccoin's velocity.E, In tie direction ofthe coin's !elocity.

3. (4 pts) Which aNwer is closest to the magtritude ofthe acceleration ofthe coitr?A. 0 m,is'.

-!,-t!si.

Nrme Studtrt lD Scofe

II. Loop-ihc-loop.A small block ofmass In slides along the frictionless"loop-thcloop ' track shown.

5. (4 pts) If the block starts from rest at P (at height 5R), which answer best describes thcdirection ofthe total force acting on it at Q.A. Upwads.B. Towards the right.C. Towards the left.rr Ope componenl is upwards. another coroponeot is lo lhe right.

1 E. OnDcomponenl is do\ Twards, another componeot is to thc lefr.

6. (4 pts) If thc block srans fiom rest at P (at height 5R), which answer is closest to themagdtude of the total force acting on it at Q.A. ON8.

-gC. Ji msn. J36 me(1}s''t7. (4 pts) which answer is closcst to the height abovc the ground where thc block should berclcascd so that the forcc thc block exerts on tbc track at the top of the loop equals its weiSht?A.ORB. IRc. JtRD. JJTIR

Physics 12I, Autum l l Eram I, page 2

Stud.rt ID Scorc

tII. Work-Errcrgy. An escalator joins otr6 floor with another floor 5 m above. The escalatoris l0 o long and moves along irs length I rnls.

t. (4 pts) Which answer is closest to the power the motor must deliver ifthe escalator calries100 persons/minute, with e5ch person's average mass of 80 kg?A. 1000 Watts.

,!=l@!atts.( C. 6000 qatts.-T--Tr-ffiw"tt..E. 25,000 Watts.

9. (4 pts) A 80 kg person walks up the escaLtot in 8 seconds. Which answer is closest to howmuch work tlle moio. docs on lhat prsotr?A. 50 Joules-8. l0O Joules.C. 250 Joules.

I l-!,Y\t'"aeeE\ t&ffilf;D. 2500 Watts.E. 5000 watts.

@::fi,10. (4 pts) If this same tnan tums arouDd itr the middle ofthc escalator aad rralks downwards soas to stay at lhe samc height in space, rvhich amwer is closst to how much power thc motorwould deliver to him as he's walling?

Pb)sics l2l, Aulumn I l Eram 3, pagF 3

Nrme Student ID Scorclast frsl

IV. Strtlct. As shown in lhe figure. a thin masslcss horizoDtal bar of length I ispinnedtoavertical wall at A and suppofied at B by a thin wire that makes an aoglc e with the horizontal. Ablock ofweight P can be movcd anywhere along lhe bai and at some panicular time it's adistance x from the wall-

11. (4 pts) Suppose the weighl is positioned at the wall (x at 0). which answer is closest to the,.Jtslsiqin lhe wirc?/ 4.0l \Y\-r.-s6Nc. 120 ND. 600 NE. 900 N

12. (4 pts) Suppose the veight is still posilioned at the wall (x at 0). Mich answcr is closest tothe magnitudc of tbe horizontal and vcrtical components of the forcc exerted on the bar by thepitr at A?A. Both horizonlal and vcrtical components arc zcro.

-!,Jhe{orizontal component is P. tbe venical componcot is zero.

( C. Tbp4orizontal component is zero. the venical componcnl is P.'D..-. Boh horizonul and venical comooncnts arc P.

E. The horizontal component is % P, the vcnical comporcnt is % P.

13. (4 pts) Suppose fte weight is movcd complelely to the other crd of the rod (position B).which aDswer besl describcs the magnitude ofthc venical component of$c force exerted on thebar by the pin at A?

/A-&tlel lo l}|c Pg {mass times gravirarioDal acceleration).( B- 7.erd\tsA66ut half way between zero aod Pg.D. Just slightly less than Pg.E. Just slightly morc than zcro.

Physics l2l , Aurumn I3 Exam.]. pagc 4

Nrm Studetrt lD Score

V. lD rnd 2D notlotr. As shown in the figure, block C is placed atop block A. Block A hasmass 8 kg. Block B, the hanging weight, has mass 4 kg. Suppose the coefTicient ofboth kinetic

la. (4 pts) Which answcr is closesl to thc minimum mass ofblock C rcquircd to keep the blocksfrom sliding?A. I kg.B. 4 kg.

E,22kg.

15. (4 pts) Block C is suddenly removed. Which aDswer is closest to the magdtude oftheacceleration of block A?

_44,@41$#uD.8 m,/s'].E. l0 m/s?.

Physics l2l, Autumn ll Exan 3, pagc 5

121H Exam 3 SolutionsVI. Coupled Motion. As shown in the figure, two blocks m1 (1.65 kg) and m2 (3.30 kg) are attached

to a rigid massless rod. The rod is parallel to the surface of the inclined plane. The coecient of kineticfriction between block m1 and the plane is 0.226, the coecient of kinetic friction between block m2 and theplance is 0.113, and the inclination angle is 30.

1. Is the force on the rod a compressive or a tension force?

If the blocks were not connected by the rod, m2 would accelerate faster than m1 (because it has a lowercoecient of friction), and so the blocks would become farther and farther apart. The rod is thereforeholding them together (as opposed to keeping them apart), and so the force is a tension force.

2. Find the magnitude of compressive (or tension) force in the rod.

Taking down the slope to be the positive x direction, and away from the slope to be the positive ydirection, we can note that ay = 0 for both blocks, and so FN = mg cos for each block (the values willbe numerically dierent because the masses are dierent). Since the blocks are connected by a rigid rod,and can only move in one direction (along the slope), m1 and m2 must have the same acceleration ax,and thus the same ratio of force to mass. Letting T be the unknown tension,

a =Fnet,1m1

=Fnet,2m2

m2Fnet,1 = m1Fnet,2m2(

m1g sin + T k,1FN,1) = m1(m2g sin T k,2FN,2)(m1 +m2)T = m2k,1FN,1 m1k,2FN,2

= m1m2g cos (k,1 k,2)T =

m1m2m1 +m2

g cos (k,1 k,2) = 1.06 N

1

3. Find the magnitude of the common acceleration of the masses.

We can plug in the above value of T into the net force on either block. Picking block 1 gives

a =Fnet,1m1

=1m1

m1g sin +

m1m2m1 +m2

g cos (k,1 k,2)m1g cos k,1

= g sin m1k,1 +m2k,2m1 +m2

g cos = 3.62 m/s2

4. Describe what, if anything, would change in your answers above if the two masses were interchanged.

If the positions of m1 and m2 were swapped, then (without the rod) the top mass would acceleratefaster, and so the force from the rod would now need to be a compression force. Everything else wouldstay the same.

2

Name ______________________________________ Student ID _______________ Score________ last first

Physics 121B, Autumn 2013 Exam 3

IV. [20 points total] A bead of mass m is constrained to move on a frictionless vertical circular loop of radius R as shown. In the following experiments, the bead is at point A at initial time t0. A. [7 pts] In experiment 1, the bead is released from rest at point A.

Is the magnitude of the centripetal acceleration of the bead at point A greater than, less than or equal to that at point B? Explain. The centripetal acceleration for a point mass on a circular trajectory is given by v2/R. By conservation of energy, the bead will have a maximum velocity at point B, and thus the magnitude of centripetal acceleration is greatest at point B.

B. [7 pts] In experiment 2, the bead has an initial downward velocity of magnitude v0 at point A. Is the magnitude of the tangential acceleration of the bead at point A greater than, less than or equal to that at point B? Explain. At point A, the force of gravity on the bead points tangent to the circular loop, while at point B the force of gravity points perpendicular to the loop. Since the centripetal force on the bead at points A and B is perpendicular to the loop, and these are the only other forces, the tangential acceleration of the bead is greatest at point A.

C. [6 pts] In experiment 3, the bead has an initial upward velocity of magnitude v0 at point A. It is

observed that the bead reaches point C. Is the net force on the bead at point B in experiment 2 greater than, less than or equal to the net force on the bead at point C in experiment 3? Explain. (Hint: Use energy conservation.) As the hint suggests, it is easiest to start with conservation of energy. This will give us the velocities at point B and point C, which we can then use to compute the centripetal acceleration v2/R. Let the gravitational potential energy U be 0 at point B, and downward-pointing vectors be negative. Then Exp. 2: m v02 + m g R = m v2 v2= v02+2gR Fnet, 2= m (v02/R+2g g) j = m (v02/R+g) j Exp. 3: m v02 + m g R = m v2 +2 m g R v2= v02 2gR Fnet, 3= m ( v02/R + 2g g) j =

= m ( v02/R + g) j Hence in magnitudes: Fnet, 2 = m (v02/R+g) > Fnet, 3 = m (v02/R g)

Or equivalently: Fnet, 2 Fnet, 3 = 2mg

A

B

C

R