physics 231 - michigan state university · 2012. 12. 4. · extra credit quiz force due to...

MSU Physics 231 Fall 2012 1

Physics 231

Review Session: Chapters 1-14

Wade Fisher

Dec 5-7 2012


Final Exam Seating Chart

Physics 231 section 2

ROW Students ROW

C Aben Through Berichon C

D Berlin Through Christopoulos D

E Christy Through Fleming E

F Floyd Through Heimburger F

G Hein Through Joyce G

H Kadzban Through Larcom H

I Lauwers Through Marier I

J Marte Through Morrison J

K Muench Through Peleman K

L Peltier Through Roof L

M Roper Through Sturgis M

N Swanson Through Walter N

O Walters Through Zigman O


Review Problems Chapters 1-5


From Last Year’s Exam A pitcher throws a ball horizontally with a speed of 39 m/s to a catcher 18.5 m away. When the ball is caught, the height has decreased by X m. Find X.

Different ways to transform this problem: 1) Give drop distance and ask for initial speed. 2) Give drop distance and ask for distance to catcher.


Same Problem Reworded A pitcher throws a ball with a speed of 39 m/s to a catcher some distance away. When caught, the ball’s height has decreased by 0.57 m. How far away is the catcher? g = 9.81 m/s2

y = -½gt2 = -0.57 m t = sqrt(20.57/g) = sqrt(0.116) = 0.34 sec x = vot = 39m/s (0.34 s) = 13.3 m

MSU Physics 231 Fall 2012 6 PHY 231 6

EXTRA CREDIT QUIZ Force due to airblower. A small block is put on a

frictionless slope. Attached on the block is an air-blower which keeps the block from slipping down the slope. Which of the following is true:

a) The force by the airblower on the block equals the total gravitational force on the block

b) The force by the airblower on the block is smaller than the total gravitational force on the block

c) The force by the airblower on the block is larger than the total gravitational force on the block

F = mgsin

Fp = mg cos

Fg=mg

n = -Fp


0.5 kg

A

20o

Is there a value for the static friction of surface A for which these masses do not slide? If so, what is it?

0.5 kg mass: F=ma (only vertical) T-mg=ma T-0.5g=0.5a

1 kg n Ffriction

Fg

T

T

Fg

Example

1 kg mass: F=ma (parallel to the slope) -Fg//-T+Ffriction=ma -mgsin(q)-T+smgcos(q)=ma -3.35-T+9.2s=a

No sliding: a=0, so T=0.5g=4.9 (from 0.5kg mass equation) -3.35-4.9+9.2s=0 s=0.9


Question

A boat is trying to cross a 1-km wide river in the shortest way (straight across). Its maximum speed (in still water) is 10 km/h. The river is flowing with 5 km/h. 1) At what angle does the captain have to steer the boat the go straight across? 2) How long does it take for the boat to cross the river? 3) If it doesn’t matter at what point the boat reaches the other side, at what angle should the captain steer to cross in the fastest way?


A boat is trying to cross a 1-km wide river in the shortest way (straight across). Its maximum speed (in still water) is 10 km/h. The river is flowing with 5 km/h. 1) At what angle does the captain have to steer

the boat the go straight across?

2) How long does it take the boat to cross the river?

sin = opposite/hypotenuse = 5/10 = 0.5 = sin-1(0.5) = 30o

tan = opposite/adjacent tan30o = 0.577 = 5/velocityhor velocityhor = 8.66 km/h Time = (1 km)/(8.66 km/h) = 0.115 h = 6.9 min

Counter balance flow=5km/h

Flow=5km/h

Answer


Answer

0o :the horizontal component of the velocity is then maximum.

A boat is trying to cross a 1-km wide river in the shortest way (straight across). Its maximum speed (in still water) is 10 km/h. The river is flowing with 5 km/h. 3) If it doesn’t matter at what point the boat reaches the other side, at what angle should the captain steer to cross in the fastest way?

Counter balance flow=5km/h

Flow=5km/h

MSU Physics 231 Fall 2012 11 PHY 231 11

Clicker Quiz!

5 kg 10 kg

5 kg

3 cases with different mass and size are standing on a floor. which of the following is true: a) The normal forces acting on the crates is the same b) The normal force acting on crate b is largest

because its mass is largest c) The normal force on crate c is largest because its

size is largest d) The gravitational force acting on each of the

crates is the same e) All of the above

a b c


Question

A ball of 1 kg rolls up a ramp, with initial velocity of 6 m/s. It reaches a maximum height of 1 m (i.e., the velocity is 0 m/s at that point). How much work is done by friction?

a) 0.0 J b) 8.2 J c) 9.8 J d) 18 J e) 27.8 J

Initial: ME = ½mv2 (kinetic only) = ½x1x62 = 18 J Final: ME = mgh (potential only) = 1x9.8x1 = 9.8 J Wnc = 18-9.8 = 8.2 J

kinetic energy: ½mv2 potential energy: mgh g=9.81 m/s2

MEi – MEf = Wnc (KEi+PEi) - (KEf+PEf) = Wnc


Conservation of mechanical energy What is the speed of m1 and m2 when

they pass each other?

At time of release: PE1 = m1gh1 =5.00*9.81*4.00 =196.2 J PE2 = m2gh2 =3.00*9.81*0.00 =0.00 J KE1 = ½m1v2 =0.5*5.00*(0.)2 =0.00 J KE2 = ½m1v2 =0.5*3.00*(0.)2 =0.00 J

Total =196. J

196.2 = 156.8 + 4.0v2 v = 3.13 m/s

M2

3 kg

5 kg

M1

4.0 m

At time of passing: PE1 = m1gh1 = 5.00*9.81*2.00 =98.0 J PE2 = m2gh2 = 3.00*9.81*2.00 =58.8 J KE1 = ½m1v2 = 0.5*5.00*(v)2 =2.5v2 J KE2 = ½m1v2 = 0.5*3.00*(v)2 =1.5v2 J

Total =156.8+4.0v2 J

(PE1+PE2+KE1+KE2)=constant


Friction (non-conservative) The pulley is now not frictionless. The friction force equals 5 N. What is the speed of the objects when they pass?

Wnc = Ffrictionx = 5.00*2.00 = 10.0 J

39.2 J - KEpassing = 10 J KE = 29.2 J = ½(m1+m2)v2 = ½(8)v2 = 4v2 v=2.7 m/s

M2

3 kg

5 kg

M1

4.0 m

(PE+ KE)start-(PE+KE)passing = Wnc

PEstart-PEpassing-KEpassing = Wnc

PEstart- PEpassing = (m1gh + m2g(0)) – (m1gh/2 + m1gh/2) = (5*9.81*4 + 0)- (5*9.81*2 + 3*9.81*2) = (196.2) – (98.1 + 58.9) = 39.2 J

Without Friction: v = 3.13 m/s


And another example…

Calculate how far the ball goes.


Clicker Quiz!

What is Newton’s 2nd Law? a) An object in a state of motion tends to remain in

that motion unless an external force is applied to it.

b) The relationship between an object’s mass, it’s acceleration, and the applied force is F=ma.

c) For every action, there is an equal and opposite reaction.

d) Mechanical energy is always conserved in the absence of non-conservative forces.

e) Momentum is conserved in elastic collisions.


And another example…

Calculate how far the ball goes.

Vertical direction y(t) = y(0) + v0sin()t - ½gt2 11 = 37 + 10.1 sin(59)t - ½9.8t2 4.9t2 - 8.66t - 26=0 t = 3.35 (solve quadratic equation)

This time correspond to how long it takes for the ball to land on the second building.

Horizontal direction x(t) = x(0) + vocos()t Use time derived from vertical direction X(3.35)=0+10.1cos(59)3.35 X(3.35)=17.4 m


C

A mass is oscillating horizontally while attached to a spring with spring constant k. Which of the following is true? a) When the magnitude of the displacement is largest, the magnitude of the acceleration is also largest. b) When the displacement is positive, the acceleration is also positive c) When the displacement is zero, the acceleration is non-zero

Clicker Quiz!


From Exam 2 The motor for a toy Ferris wheel creates a torque of 15 Nm. The wheel requires 31 seconds to reach its design frequency of one revolution per 12 seconds. What is the moment of inertia of the wheel?

Ways to modify this problem: 1) Give moment of inertia, solve for torque. 2) Give moment of inertia, solve for time. 3) Give moment of inertia, solve for frequency.


Same Problem Reworded The motor for a toy Ferris wheel with moment of inertia 1181 kgm2 creates a torque of 15 Nm. The wheel has a design frequency of one revolution per 12 seconds. How long does it take to reach this frequency? = I = t = 2π/12 sec = 0.524 rad/sec = 15/1181 = 0.0127 rad/sec2

/ = t = 0.524/0.0127 = 41.2 sec


An Example

A mass of 1 kg is hung from a spring. The spring stretches by 0.5 m. Next, the spring is placed horizontally and fixed on one side to the wall. The same mass is attached and the spring stretched by 0.2 m and then released. What is the acceleration upon release?

1st step: find the spring constant k Fspring =-Fgravity or -kd =-mg k = mg/d =1*9.8/0.5=19.6 N/m

2nd step: find the acceleration upon release Newton’s second law: F=ma -kx-mg=ma a=-kx/m a=-19.6(-0.2)/1 = 3.92 m/s2


The gravitational force depends inversely on the

distance squared. So if you increase the distance by

a factor of 2, the force will decrease by a factor of 4.

Clicker Quiz! Earth and Moon

a) one quarter

b) one half

c) the same

d) two times

e) four times

If the distance to the Moon were

doubled, then the force of attraction

between Earth and the Moon would be:

2R

MmGF


Translational equilibrium (Hor.) Fx=ma=0 n-Tx=n-Tcos37o=0 so n=Tcos37o Translational equilibruim (vert.) Fy=ma=0 sn-w-w+Ty=0 sn-2w+Tsin37o=0 (use n=Tcos37o) sTcos370-2w+Tsin370=0 0.4T-2w+0.6T=0 T=2w Rotational equilibrium:

=0 xw+2w-4Tsin370=0 (4sin37o)=2.4 &T=2w w(x+2-4.8)=0 x=2.8 m

s=0.5 coef of friction between the wall and the 4.0 meter bar (weight w). What is the minimum x where you can hang a weight w for which the bar does not slide?

w

sn

n

w

T Ty

Tx

(x=0,y=0)

37o


An Example

d

h A block with mass of 200 g is placed over an opening. A spring is placed under the opening and compressed by a distance d=5 cm. Its spring constant is 250 N/m. The spring is released and launches the block. How high will it go relative to its rest position (h)? (the spring can be assumed massless)

ME=½mv2+½kx2+mgh must be conserved. Initial: v=0, x=-0.05m h=0 0.5x0.2x02 + 0.5x250x0.052 + 0.2*9.8*0 = 0.3125 Final: v=0 (at highest point mass does not move), x=0 (spring no longer compressed), h=??? 0.5x0.2x02 + 0.5x250x0.02 + 0.2*9.8*h = 1.96 x h Conservation of ME: 0.3125 = 1.96 x h h=0.16 m


Conical motion

What is the speed of the mass undergoing conical motion if the mass is 1 kg and =20o? The length of the string swinging the mass is 1 m.

mg

T Tcos

Tsin

Vertical direction: F=ma Tcos-mg=0 So T=mg/cos

Horizontal direction: F=mac Tsin=mac mgsin/cos=mgtan=mac ac=gtan=9.8*0.36=3.6 m/s2

ac=v2/r so v=(acr)=(acLsin)=1.11 m/s

L

r=Lsin


Example

A monocycle (bicycle with one wheel) has a wheel that has a diameter of 1 meter. The mass of the wheel is 5 kg (assume all mass is sitting at the outside of the wheel). The friction force from the road is 25 N. If the cycle is accelerating with 0.3 m/s2, what is the force applied on each of the paddles if the paddles are 30 cm from the center of the wheel?

25N

0.5m

0.3m F

=I =a/r so =0.3/0.5=0.6 rad/s

I=(miri2)=MR2=5(0.52)=1.25 kgm2

friction=-25*0.5=-12.5

paddles=F*0.3+F*0.3=0.6F

0.6F-12.5=1.25*0.6, so F=22.1 N


An Example

A

B star

Two planets are orbiting a star. The orbit of A has a radius of 108 km. The distance of closest approach of B to the star is 5x107 km and its maximum distance from the star is 109 km. If A has a rotational period of 1 year, what is the rotational period of B?

RA = 108 km RB = ½(rmin+rmax) = ½(5x107+109) = 5.25x108 km

Rmin = distance of closest approach. = perihelion Rmax = maximum distance = aphelion

R3/T2 = constant RA3/TA2= RB3/TB2 so TB2=(RB3/RA3)TA2

So TB=(5.253 x (1 yr)2) = 12 years


Ballistic Pendulum

Mblock=1 kg

Mbullet=0.1 kg Vbullet=20 m/s

h

How high will the block go?

There are 2 stages: • The collision • The Swing of the block

The collision The bullet gets stuck in the block (perfect inelastic collision). Use conservation of momentum. m1v1i+m2v2i=vf(m1+m2) so: 0.1*20+1*0=vf(0.1+1) vf=1.8 m/s The swing of the block Use conservation of Mechanical energy. (mgh+½mv2)start of swing= (mgh+½mv2)at highest point 0+½1.1(1.8)2=1.1*9.81*h so h=0.17 m

Why can’t we use Conservation of ME right from the start??


An Example An architect wants to design a 5m high circular pillar with a radius of 0.5 m that holds a bronze statue that weighs 1.0E+04 kg. He chooses concrete for the material of the pillar (Y=1.0E+10 Pa). How much does the pillar compress?

5m

2

0

0

2

0 /

)/(

/

/

pillar

statue

pillarstatue

RY

gLML

LL

RgM

LL

AFY

R = 0.5m L0 = 5m Y = 1.0E+10 Pa M = 1.0E+04 kg

L = 6.2E-05 m


Another one An air mattress 2m long 0.5m wide and 0.08m thick and has a mass of 2.0 kg. A) How deep will it sink in water? B) How much weight can you put on top of the mattress before it sinks? water=1.0E+03 kg/m3

A) h= objectVobject/(waterA) h=Mobject/(1.0E+03*2*0.5)=2.0/1.0E+03=2.0E-03m=2mm

B) if the objects sinks the mattress is just completely submerged: h=thickness of mattress. 0.08=(Mweight+2.0)/(1.0E+03*2*0.5) So Mweight=78 kg


example

A police car using its siren (frequency 1200Hz) is driving west towards you over Grand River with a velocity of 25m/s. You are driving east over grand river, also with 25m/s. a)What is the frequency of the sound from the siren that you hear? b) What would happen if you were also driving west (behind the ambulance)? vsound=343 m/s

Hz

f

vv

vvff

source

observer

138916.11200

25343

253431200

a) b)

Hz

f

vv

vvff

source

observer

1200.11200

)25(343

253431200


An example ?? N

1 kg of water inside thin hollow sphere

A) ?? N

7 kg iron sphere of the same dimension as in A)

B)

Two weights of equal size and shape, but different mass are submerged in water. What are the weights read out?

B=waterVdisplacedg w=sphereVsphereg A) B= waterVsphereg w=waterVsphereg so B=w and 0 N is read out! B) B= waterVsphereg=Mwater sphereg w= ironVsphereg=Miron sphereg=7Mwater sphereg

T=w-B=6*1*9.8 =58.8 N


And another! An airbubble has a volume of 1.50

cm3at 950 m depth (T=7oC). What is

its volume when it reaches the

surface

(water=1.0x103 kg/m3)?

P950m=P0+watergh

=1.013x105+1.0x103x950x9.81

=9.42x106 Pa

293

10013.1

280

1050.11042.9

566

2

22

1

11

surfaceV

T

VP

T

VP

Vsurface=1.46x10-4 m3=146 cm3


A helium balloon in an air-

filled glass jar floats to the top.

If the air is replaced with

helium, what will happen to

the helium balloon?

a) It still floats at the top because it has positive buoyancy

b) It stays in the middle because it has neutral buoyancy

c) It sinks to the bottom because it has negative buoyancy

d) The balloon shrinks in size due to the surrounding helium

d) The balloon grows in size due to the lack of surrounding air

The balloon floats initially because the displaced air weighs more than the

balloon, so the buoyant force provides a net upward force. When the

balloon is in the lighter helium gas (instead of air), the displaced helium

gas does not provide enough of an upward buoyant force to support the

weight of the balloon.

A Helium Balloon


Heating your Home A house has a floor area of 190 m2 and 2.3-m ceilings. (A)What energy (in kWh) is required to raise the air

temperature by from 20°C to 21°C at constant pressure? (specific heat of air cP=30 J/mol°C and 1 mol of air = 24 L)

(B)At $0.16/kWh, what’s the cost to heat this air by 1°C? (assume a 80% heating efficiency)

A) V = (190 m2)(2.3m) = 437 m3 = 4.37E5 Liters 1 mol = 30 L, n = (4.37E5)/(3E1) = 1.82E4 mol Q = n cPT = (1.82E4 mol)(30 J/mol °C)(1 °C) = 5.46E5 Joule 1 kWh = 1000 (J/s)x(3600s) = 3.6E6 J Thus, the heat required is (5.46E5)/(3.6E6) = 0.152 kWh B) The cost is $0.16/kWh at 80% efficiency. Cost = (0.152 kWh)($0.16/kWh)/(0.80) = $0.03


Testing an Unknown Substance A block of unknown substance with a mass of 8 kg, initially at T=280K is thermally connect to a block of copper (5 kg) that is at T=320 K (ccopper=0.093 cal/g

0C). After the system has reached thermal equilibrium the temperature T equals 290K. What is the specific heat of the unknown material in cal/goC?

????

copper

Qcold+Qhot=0 Q=cmT Qcold=-Qhot

munknowncunknown(Tfinal-Tunknown)=-mcopperccopper (Tfinal-Tcopper)

cunknown=-mcopperccopper(Tfinal-Tcopper) munknown (Tfinal-Tunknown)

cunkown=-50000.093(290-320) = 0.174 cal/goC 8000(290-280)


Examples for this chapter

One mole of an ideal gas initially at 00C undergoes an expansion at constant pressure of one atmosphere to four times its original volume. a) What is the new temperature? b) What is the work done on the gas?

a) PV/T=constant so if V x4 then T x4 273K*4=1092 K b) W=-PV use PV=nRT before expansion: PV=1*8.31*273=2269 J after expansion: PV=1*8.31*1092=9075 J W=-PV=-(PV)=-[(PV)f-(PV)i]=-(9075-2269)=-6806 J -6806 J of work is done on the gas.


Question An ideal monatomic gas is

taken around the cycle ABCDA

as shown below.

What is the amount of energy

removed by heat during one

cycle?

a. 0

b. P1V1

c. 2P1V1

d. 4P1V1

e. Impossible to determine

The path is counter clockwise, so this is a fridge. Net work (P1V1) was performed on the gas using heat from the gas, which was thus removed


Another Example A gas goes from initial state I to final state F, given the parameters in the figure. What is the work done on the gas and the net energy transfer by heat to the gas for: a) path IBF b) path IF c) path IAF (Ui=91 J Uf=182 J)

a) work done on gas: -area under graph: note the ‘-’ sign before area W=-(0.8-0.3)10-3*2.0*105=-100 J U=Uf-Ui=182-91=91 U=W+Q 91=-100+Q so Q=191 J b) W=-[(0.8-0.3)10-3*1.5*105 + ½(0.8-0.3)10-3*0.5*105]=-87.5 J U=W+Q 91=-87.5+Q so Q=178.5 J

c) W=-[(0.8-0.3)10-3*1.5*105]=-75 J U=W+Q 91=-75+Q so Q=166 J

The volume expands, positive work is done by the gas. So negative work is done on the gas. W is defined as Work on the gas.


Although the water is indeed hot, it releases only 1 cal/g of heat as it

cools. The steam, however, first has to undergo a phase change into

water and that process releases 540 cal/g, which is a very large amount of

heat. That immense release of heat is what makes steam burns so

dangerous.

Hot Water

Which will cause more severe burns

to your skin: 100°C water or 100°C

steam?

a) water

b) steam

c) both the same

d) it depends...


Aluminum is the only material that has a larger b value than the

steel ring, so that means that the aluminum rod will expand more

than the steel ring. Thus, only in that case does the rod have a

chance of reaching the top of the steel ring.

Thermal Expansion

Coefficient of volume expansion b (1/°C )

Glass Hg Quartz Air

Al Steel

a) aluminum

b) steel

c) glass

d) aluminum and steel

e) all three

A steel ring stands on edge with a rod of

some material inside. As this system is

heated, for which of the following rod

materials will the rod eventually touch the

top of the ring?

physics 231 - michigan state university · 2012. 12. 4. · extra credit quiz force due to...

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