PHYS%2013.%%General%Physics%1.%!

Exam%3%study%guide%and%useful%information%!

Instructor:!Dr.!Marc!S.!Seigar!Phone:!(218)!726<6704!

Email:!msseigar@d.umn.edu!Office:!MWAH!374!

!The!material!you!will!need!to!know!for!Exam!3!is!covered!in!Chapters!8,!9,!and!10!of!Young!&!Freedman.!!This!document!lists!some!important!information!you!will!need!to!know!to!do!well!in!the!exam.!!There!is!an!information!sheet!given!to!you!as!part!of!the!exam.!!!This!is!what!you!will!need!to!know:!!You!should!understand!the!definition!of!displacement,!velocity,!and!acceleration,!and!the!derivatives!that!relate!them.!!You!should!also!understand!the!definition!of!average!velocity!(or!speed)!and!average!acceleration.!!You!should!be!able!to!determine!velocity!from!a!graph!of!x!vs!t.!!You!should!be!able!to!determine!acceleration!from!a!graph!of!x!vs!t,!and!from!a!graph!of!v!vs!t.!!You!should!know!the!kinetic!equations:!!! vf = vi + at !

! x f = xi + vit +12at2 !

! vf2 = vi

2 + 2a Δx( ) !

! x f = xi +12vf + vi( ) t !

!Newton’s!laws!of!motion.!!Friction!!Masses!on!frictionless!slopes!!Masses!on!slopes!with!friction!!

Definition!of!kinetic!energy! k = 12mv2 !

Definition!of!potential!energy!U =mgy !!Definition!of!work!W = Fd !(work!=!force!x!distance!moved)!or!in!integral!form!!!

W =F•dr∫ !

!The!work!energy!theorem!(i.e.,!work!done!is!change!in!kinetic!energy!of!work!done!is!change!in!potential!energy).!!For!conservative!forces,!total!energy!(kinetic!energy!+!potential!energy)!is!conserved.!!Friction!is!a!non<conservative!force.!!Gravity!is!a!conservative!force.!!Definition!of!momentum,!

p =mv .!!Conservation!of!(linear)!momentum.!!Elastic!and!(perfectly)!inelastic!collisions.!!!Momentum!is!always!conserved.!!Energy!is!conserved!for!elastic!collisions.!!Impulse,! I = Δp = FΔt ,!and!its!uses.!!Rotational!kinematics,!angular!velocity,!and!angular!acceleration!definitions.!!The!kinetic!equations!for!rotational!kinematics:!!! ω f =ωi +αt !

! θ f =θi +ωit +12αt2 !

! ω f2 =ωi

2 + 2α Δθ( ) !

! θ f =θi +12ω f +ωi( ) t !

!Definition!of!torque,!τ = rF sinθ ,!or!more!generally,!

τ =r ×F .!!Torque!has!the!same!

units!as!work.!!A!body!in!equilibrium!has!Στ = 0 !and!ΣF = 0 !(static!equilibrium).!!If!not!in!equilibrium!then,!Στ = Iα !(recognize!similarity!to!ΣF =ma ).!!Work!and!energy!in!a!rotational!system.!!

Definition!of!angular!momentum,!L = r × p = rpsinφ = rmvsinφ .!

!

And!τ =dLdt!or!Στ =

dLtotdt

!!(which!is!analogous!to!ΣF = dptotdt

).!

!For!a!rotating!body,! L = Iω !(analogous!to! p =mv ).!!Conservation!of!angular!momentum.!!Energy!of!rotating!systems.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Example!questions!!1. A 1220 kg car traveling initially with a speed of 25 m/s in an

easterly direction crashes into the back of a 8300 kg truck moving in the same direction at 20 m/s. The velocity of the car right after the collision is 18 m/s to the east.

(a) What is the velocity of the truck right after the collision? (Give your answer to five significant figures.) (b) What is the change in mechanical energy of the car–truck system in the collision? (c) Account for this change in mechanical energy.

2. A 10.1-g bullet is fired into a stationary block of wood having mass m = 4.93 kg. The bullet imbeds into the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.607 m/s. What was the original speed of the bullet? (Express your answer with four significant figures.)

3. A car of mass m moving at a speed v1 collides and couples with the back of a truck of mass 4m moving initially in the same direction as the car at a lower speed v2. (Use any variable or symbol stated above as necessary.)

(a) What is the speed vf of the two vehicles immediately after the collision? (b) What is the change in kinetic energy of the car–truck system in the collision? !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

4. A 90.2-kg fullback running east with a speed of 5.10 m/s is tackled by a 95.8-kg opponent running north with a speed of 2.90 m/s.

(a) Explain why the successful tackle constitutes a perfectly inelastic collision. (b) Calculate the velocity of the players immediately after the tackle. (c) Determine the mechanical energy that disappears as a result of the collision. (d) Account for the missing energy. !!!!!!!!!!!!!!!!!!!!!!!!!!

5. A racing car travels on a circular track of radius 290 m. Suppose the car moves with a constant linear speed of 52.0 m/s. (a) Find its angular speed. (b) Find the magnitude and direction of its acceleration.

6. A discus thrower accelerates a discus from rest to a speed of 25.8 m/s by whirling it through 1.29 rev. Assume the discus moves on the arc of a circle 1.01 m in radius.

(a) Calculate the final angular speed of the discus. (b) Determine the magnitude of the angular acceleration of the discus, assuming it to be constant. (c) Calculate the time interval required for the discus to

accelerate from rest to 25.8 m/s.

7. Find the net torque on the wheel in the figure below about the axle through O, taking a = 7.00 cm and b = 21.0 cm. (Assume that the positive direction is counterclockwise.)

!

8. Calculate the net torque (magnitude and direction) on the beam in the figure below about the following axes.

(a) an axis through O, perpendicular to the page (b) an axis through C, perpendicular to the page !

9. A cylinder with moment of inertia I1 rotates about a vertical, frictionless axle with angular velocity ωi. A second cylinder; this one having a moment of inertia of I2 and initially not rotating, drops onto the first cylinder. Because of friction between the surfaces, the two eventually reach the same angular speed ωf.

(a) Calculate ωf. (Use any variable or symbol stated above as necessary.) (b) Show that the kinetic energy of the system decreases in this

interaction by calculating the ratio of the final to initial rotational energy. Express your answer in terms of ωi. (Use any variable or symbol stated above as necessary.)

10. A playground merry-go-round of radius R = 1.80 m has a moment of inertia I = 255 kg · m2 and is rotating at 12.0 rev/min about a frictionless vertical axle. Facing the axle, a 26.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round?

Solutions 1. (a) conservation of momentum, mCvCi +mTvTi =mCvCf +mTvTf

i.e.,!30500+166000 = 21960+8300vTf !Solve!for! vTf :!

! ! vTf =1745408300

= 21.029 !m/s!

!

! (b)!Change!in!KE,!Δk = 12mCvCi

2 +12mTvTi

2"

#$

%

&'−

12mCvCf

2 +12mTvTf

2"

#$

%

&' !

! ! ! ! Δk = 381250+1660000−197640−1835208 = 8402 !J!!! (c)!The!energy!has!been!transformed!into!internal!energy!and!sound.!!!2. Conservation!of!momentum,!mbulletvbullet = (mblock +mbullet )vf !

i.e.,!0.0101vbullet = (4.94+ 0.0101)0.607 !

Therefore,! vbullet =2.99860.0101

= 296.9 !m/s!

!!3. (a)!From!conservation!of!momentum,!mv1 + 4mv2 = 5mvf !

so,! vf =v1 + 4v25

!

!

(b)!Δk = 0.5mv12 + 2mv2

2 − 2.5mvf2 = 0.5mv1

2 + 2mv22 − 2.5m v1 + 4v2

5#

$%

&

'(2

!

! Δk = 25m v1

2 + v22 − 2v1v2( ) !

!!!!!!!!!!!

4. (a)!!The!opponent!grabs!the!fullback!and!does!not!let!go,!so!the!two!players!move!together!at!the!end!of!their!interaction.!!Thus!the!collision!is!perfectly!inelastic.!

!(b)!!Conservation!of!momentum:!! !x<direction,!90.2(5.1) =186vfx ,!and!so! vfx = 2.47 !m/s!! !y<direction,!95.8(2.9) =186vfy ,!and!so! vfy =1.49 !m/s!

! ! and!therefore,! v = 2.472 +1.492 = 2.9 !m/s!!(c)!Δk = 0.5(90.2)5.12 + 0.5(95.8)2.92 − 0.5(186)2.92 = 794 !J!!(d)!The!kinetic!energy!is!lost!to!internal!energy!

!!

5.! (a)!ω =vr=52290

= 0.18 !rad/s!

!

! (b)! ac =v2

r=522

290= 9.32 !m/s2!!toward!the!center!of!the!circle!

!!

6.! (a)!ω =vr=25.81.01

= 25.5 !rad/s!

!

! (b)! ac =v2

r=25.82

1.01= 659 !m/s!

!! (c)! vf = vi + at !! ! 25.8 = 659t !

! ! t = 25.8659

= 0.04 !s!

!!7.! Στ =12(0.07)−10(0.21)− 9(0.21) = −3.15 !N!m!!!8.! (a)!Στ = 25(2)sin60−10(4)sin20 = 43.3−13.7 = 29.6 !N!m!clockwise!! (b)!Στ = 30(2)sin 45−10(2)sin20 = 42.4− 6.8 = 35.6 !N!m!counterclockwise!!!!!

9.! (a)!Angular!momentum!is!conserved:!! I1ωi = (I1 + I2 )ω f !

! ! ! Therefore,!ω f =I1ωi

I1 + I2!

!

! (b)!Initial!KE,! ki =12I1ωi

2 !

!

! ! Final!KE,! k f =12(I1 + I2 )ω f

2 =12I1 + I2( ) I12ω1

2

I1 + I2( )2=

I12ωi

2

2 I1 + I2( )=

I1kiI1 + I2( )

< ki !

!!10.!! Conservation!of!angular!momentum:!

! ! ! Iωi = I +mr2( )ω f ! ! and!ω =12 rev/min = 12(2π )60

=1.256 rad/s !

! ! ! 255(1.256) = (255+ 26(1.82 ))ω f !

!

! ! Solve!for!ω f =320.28339.24

= 0.94 rad/s = 0.94(60)2π

= 9 rev/min !

!

Information Page

g = standard gravity at earth’s surface = 9.80 m/s2; 1 atmosphere (atm) = 1.013 x 105 Pa G = gravitational constant = 6.67 x 10-11 m3/(kg s2) Density of water = 1 x 103 kg/m 1 day = 8.640 x 104 s; Density of Ice = 0.917 x 103 kg/m3 Re = equatorial radius of earth = 6.374 x 106 m; Density of air (STP) = 1.29 kg/m3 Me = mass of earth = 5.976 x 1024 kg; Specific Heat of Water =1.0kcal/(kg K) Mm= mass of moon = 7.350 x 1022 kg Latent Heat of fusion for water =80 kcal/kg Mean radius of moon’s orbit around earth = 3.844 x 108 m Specific Heat of Aluminum = 0.215 cal/g C Ms = mass of sun = 1.989 x 1030 kg Specific Heat of Copper = 0.0924 cal/g C Moon’s orbital period = 27.3 days Volsphere = (4/3)πr3 Avg Earth to Sun distance = 1.5x1011 m NA = 6.023 x 1023 kB = 1.38 x 10-23 J/K; R = 8.31 J/(mol*K); 1 cal = 4.185 J Weight = mg xf = xi + vxit + 0.5axt2 θ(t) = θ o + ωot + (1/2) αt2 PV = nRT = N kB T vxf = vxi + axt ω(t) = ω o + αt Q = C T = m cv T vxf

2 = vxi2 + 2ax(xf-xi) ω2 - ω o2 = 2 α (θ - θ o) Eint = Q + W

vave = Δx/Δt; aave = Δv/ Δt ω ave = Δθ /t; αave = Δω / t W = - p dV ΣFi = Fnet = m a Στ i = I α KEave = (3/2) kB T = (1/2) m vrms

2 v =ω r acentripetal = v2/r acentripetal = ω 2 r

p = m v L = Iω Kinetic Energy = (1/2) m v2 KErot = (1/2) Iω 2 Ug = -G m1 m2 / r ΔUg = PE = mgh PE (spring) = (1/2) k x2 W = Work = F o r = F Δx cos (θ ) Work = τ Δθ friction, f = µN FG = (G M1 M2 )/ ( R2 ) is Newton’s Law of Universal Gravitation v = λ f vstring = (T/µ)1/2 Tpendulum = 2 π (l/g)1/2 Tspring = (2π ) (m/k)1/2 ω= 2 π f f = (1/T) SHM: x(t) = A sin (ωt); Wave: y(x) = A sin (2 π x/λ ) Isphere = (2/5)MR2; Ithin cylinder = MR2