Energy 1 Solutions

8
-qN%,1 t J,r^' I tt,ls T = I,-, = trne.* E{"| ile ZnnL" e)E lf 1TE B-tn>rrvt 9F nE l)ff Awf FFs fD oU*csr..^€ * 2) rN€ro'fiA, 4ct F -g^ c-1- F- rtc|' f, go! 6"'"' F,-vw@, l) c*a^rr,.'1 aFrgpa* ou rrE- t T^ rr5 : (g")" uir r) a z) e. 3\g fa. l$a,=F - *j 1? + tu|^ -P" I =7, rJ. = *L? + rtr Ln T f -ro l-l . t LQrr M4= t fsar -T [9a" tPt -f a"'{ r ttl q r 2k 254= T.t- : 0,4r5*' e 'T- .q -9- :. T (.r)" Feorgsror'r F z Mq+ryt^ F = fvrl + rrlrf !t'f 'ffi

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Transcript of Energy 1 Solutions

Page 1: Energy 1 Solutions

-qN%,1

t J,r^'

Itt,ls

T = I,-,= trne.*

E{"|ile ZnnL"

e)E

lf 1TE B-tn>rrvt 9F nE

l)ff Awf FFs fD oU*csr..^€

* 2) rN€ro'f iA,4 c t

F

-g^ c-1- F-rtc|' f,

go !

6"'"'

F,-v w@,l) c*a^rr,.'1

aFrgpa* ou

rrE- t T^rr5

: (g")" uir

r) a

z) e.

3\g

fa. l$a,=F - * j

1 ?+ tu |^

-P"

I =7, rJ.= *L? + rtr Ln

Tf

-rol-l .

tLQrr

M4= t fsar -T

[9a" tP t - f

a" ' {r

t t lq

r

2k 254= T.t-

: 0,4 r5*' e 'T-

.q -9- :. T(.r)"

Feorgsror'r

F z Mq+ryt^

F = fvrl + rrlrf !t'f

'ffi

Page 2: Energy 1 Solutions

I

Mech .2 .

A solid disk of unknown mass and known radius R is used as a pulley in a lab experiment, as shown above. A

small block of mass m is attached to a string, the other end of which is attached to the pulley and wrapped

around U several tunes. The block of mass m is released from rest and takes a time t to fall the distance D to the

floor,(a) Calculate the l inear acceleration a of the fall ing block in terms of the given quantit ies.

Ax=v.f +!at"

D - !, .g.

(b) The time t is measured for various he and the da recorded in the following table.D and the data are recorded ln

D(m) t (s) z

0.5 0.68 ,41,l .04

V)L\ .40

1,02

1 . 5 l . l 9

az r . 38

i. What quantities should be graphed in order to best determine the acceleration of the block? Explaur

l ,ourreasonine D p= ,4 D l+o-- Z ' ' ' -? , . t f loatsurP 19

ii. On the grid below, plot the quantities determined in (b)i., label the axes, and draw the best-fit lin. toLtJ#

the data.

2

.31.5

I

o.(

rb

Page 3: Energy 1 Solutions

iii. Use your graph to calculate the magnitude of the acceleration.

tt.t6,l.2r 9-ope =+-

Cp1rL16) 4 -- -L

G1trtA z 7pr-Tr

"fa*nn 9-T.

=I

(c) Calculate the rotational inenia of the pulley in terms of m, R, a, and fundamental constants.

I* 'Tr ]Aa= M1 'T

(d) The value of acceleration found in (b)iii, along with numerical values for the given quantities and youranswer to (c), can be used to determine the rotational inefiia of the pulley. The pulley is removed fromSupportandi tsrotat ional iner t ia is foundtob.g '9@.! !e 'Giveoneexplanat ionfor th isdiscrepancy.

Fa*^* Be htcnonr (h,..,ot No*to !ra-s.A

^b P-as*ur flter 19 !.*{ nlrru Acua,u f\

i A ' l ^

'tD5 ratiu.^ Aaotll" o P*,*rc1 4creuD

ffic-t\tff T]tlrJsagp

lee

t7r - l&€c.pr = ,

T = t_449 r.^rt+ T

Page 4: Energy 1 Solutions

Figure I

Fiqure II I

1980M2, Ab locko fmassms l i desa tve loc i r yvoac rossaho r i zon ta l f f i c t i on lesssu r face towarda la rge

curved movable ramp of mass 3m as shown in Figure I . The ramp, init ially at rest, also can move without

friction and has a smooth circular frictionless face up which the block can easily slide. When the block

slides up the ramp, it momentarily reaches a maximum height as shown in Figure II and then slides back

down the frictionless face to the horizontal surface as shown in Figure III.

a. Find the velocity v; of the moving ramp at the instantthe blockleachesjts maximum height.:t^oUf^rt- Wftt 5,,er€ Ua-octrY :-t lNctlsos (-cit-LtstoN-

M$o + g* (o )= 4 r r (

b. To what maximum height h does the center of mass of the blocklse above its original height?

- larruelc du€ga1 Losr BEaouet f 16' r\AL

AY; - At^LV=

I t- ?l.trf

i^v- r

t^u.t3 r.V.

3 ) €6

Y+-Ei I

{ ca') v,' - ?

t4"(+r-+

c. Determine the final speed vi of the ramp and the final speed v' of the block after the block returns

to the level surface, State whether the block is moving to the right or to the left.

- t\hN No kE LosS -f C'rutst olrr ' 3 e?9n&^ l l

l l , 9'c$ al

=:.v,Iaj' Y t5 ,z,r{, Y"

E o-^^ l-t- J '/ D -r

+'-"q ! l lo

l ta

'uf; =,h

- --3n'/,t b / . -

t -

l - l / -ttJ/r-LI'l!1J

Page 5: Energy 1 Solutions

Lp

F r o u r c 1 F r g u r e I i

l98lM2. A swing seat of mass M is connected to a fixed point P by a massless cord of length L. A child

also of mass M sits on the seat and begins to swing with zero velocity at a position at which the cord makes

a 60o angle with the vertical is shown in Figure L The swing continues down untilthe cord is exactlyveftical at which time the child jumps off in a horizontal direction. The swing continues in the same

direction unti l i ts cord makes a 45o angle with the vertical as shown in Figure II: at that point it begins toswing in the reverse direction. With what velociry relative to the ground did the child leave the swing?

(cos 45o = s i n 45o : J i t 2 , s i n 30 ' : cos 60o : l / 2 , cos 30 '= s i n 60 " = J i D )

A'J|@- e{4

t t z '

.lSi 'f(adu'

q( L- | cor 6o) = {u '

i f t- .tt-) = lvrt-

/r' --'L"

' ,JI = v,-

O{AA:1 3 - '-{

" = tr3L

.;4-i:)L^n,1,,

)v,"Vs z;LLT)

z-G2 -

M osEY ltru; r t T Ur*

n-_.\,)z-{a )z*64) = M(i,'

U

ZJSt- rlsL(z=G) = v

+ut /

cL(z-v:)Z6L-

Page 6: Energy 1 Solutions

t = tlo0 Nlm 4kg 2kg

MLFigurc I

l 993Ml . Amass lesssp r i ngw i th fo rcecons tan tk :400ne \ l t onspe rme te r i s fas teneda t i t s l e f t end toaverlical wall, as shown in Figure l. Init ially, block C (mass m. : 4.0 kilograms) and block D (mass mp :

2.0 kilograms) rest on a horiiontal surface with block C in contact with the spring (but not compressing it)

and with block D in contact with block C. Block C is then moved to the left, compressing the spring a

distance of 0.50 meter, and held in place while block D remains at rest as shown in Figure 1 I . (Use g : 10

m/s2.)a. Determine the elastic energy stored in the compressed spring.

\ l 4rx" t -

t \ / \ l

= { (q@ 'J/r'I '9 nI

\,{, = 5o:

Block C is then released and accelerates to the right, toward block D. The surface is rough and the

coefficient of friction between each block and the surface is p : 0.4. The two blocks coll ide

instantaneouslg@"ther, and move to the right. Remember that the spring is not aftached to block C.

Determine each of the following.

^ l-o'5 mi

U = O OFigurc Il

b. The speed v, of block C just before it coll ides with block D

e t = Ez

t'K .,Just usrurs rr vurrru

5D = LCul vL + I

fDJ= K+ We

{- 4o(.qx.e5os" I r,nva

5Ds.l(q\r1*&w**

Thc

4.a speed v, blocks C and D iuslafter they coll ide

f l, (q)* D(z): Gv

* "u'=

6o('e) J

| ( .Ye'D'= 2 '{ J

d. The horizontal distance the blocks move before coming to rest

Lnv" = $Jt

iz^ =J

Page 7: Energy 1 Solutions

f= I t , ,s .* I^o4.D" 'e ' r r= [.errBe ***Pj z^Gd'

T- lo-v3.p i=fv l2t1999M3 As shown above, a uniform disk is mounted to an axle and is free to rotlte without friction, Athin uniform rod is rigidly attached to the disk so that it wil l rotate with the disk. A block is aftached to the

end of the rod. Properties of the disk, rod, and block are as follows.Disk: mass : 3m, radius : R, moment of inertia about center Io = I .5mR2Rod: mass : m, length : 2R, moment of inertia about one end Ip : 4/3(mR')Block: mass : 2m

The system is held in equil ibrium with the rod at an angle 0q to the vertical, as shown above, by ahorizontal string of negligible mass with one end aftached to the disk and the other to a wall. Express your

answers to the following in terms of m, R,0e, and g.a. Determine the tension in the string.

2 t ="J''"j" "i-l;{

zrg (Lc)i" 0 ̂ r"iP gr ru 0

f? = 4n32sr^/ 9'+ t52sr"' 9

f= Si^9S\^,The string is now cut, disk-rod-block system is ftee to rotate.b. Determine the following for the instant immediately after the string is cut.

i. The magnitude of the angular acceleration ofJhe disk 6 - e

/+= f " r : ? Zn7(zrz5 srog + i l tg v ' t r tv

of the rod

6= t d

C\: Zf( ft#)i i . The maenitude of the l inear aciE

Page 8: Energy 1 Solutions

13' \*

E=V

As the disk rotates, the rod passes the horizontal position shown above.

b. Determine the linear speed of the mass at the end of the rod for the instant the rod is in the

horizontal position.

€tlt't ? irnSPsruO

€u3B'''9 = t

L W Sf Pt'n' I

-g gj'n I =t3 r-

2llur

t 9

= {rtr) w'

thJ

l= l- L^J