Artificial Gravity - Advanced Physics - Solved Exam, Exams of Advanced Physics

This is the Solved Exam of Advanced Physics which includes Free Body Diagrams, Accelerate Upwards, Density of Oil, Freshness of Battery, Spherical Shell, Net Charge, Distribution of Charges etc. Key important points are: Artificial Gravity, Rotating Cylinders, Gravitational Fields, Angular Velocity, Forces Acting on Astronaut, Rotating Space Station, Unextended Length, Two Identical Masses

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93/11(a) Semester 1, 2009 Page 1 of 1
THE UNIVERSITY OF SYDNEY
PHYS 1901
PHYSICS 1A (ADVANCED)
Solutions
JUNE 2009
Time allowed: THREE Hours
MARKS FOR QUESTIONS ARE AS INDICATED
TOTAL: 90 MARKS
INSTRUCTIONS
All questions are to be answered.
Use a separate answer book for section A and section B.
All answers should include explanations in terms of physical principles.
DATA
Density of fresh water =
33
1.000 10 kg.m
Free fall acceleration at earth's surface g =
2
9.80 m.s
Gravitational constant G =
11 2 2
6.67 10 N.m .kg

Speed of light in a vacuum c =
18
3.00 10 m.s
Speed of sound in air v =
1
344 m.s
Avogadro constant NA =
123
6.023 10 mol
Universal gas constant R =
11
8.314 J.mol .K

Boltzmann constant k =
123
1.380 10 J.K
Stefan-Boltzmann constant σ =
8 2 4
5.67 10 W.m .K
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93/11(a) Semester 1, 2009 Page 1 of 1

THE UNIVERSITY OF SYDNEY

PHYS 1901

PHYSICS 1A (ADVANCED)

Solutions

JUNE 2009

Time allowed: THREE Hours

MARKS FOR QUESTIONS ARE AS INDICATED

TOTAL: 90 MARKS

INSTRUCTIONS

**- All questions are to be answered.

  • Use a separate answer book for section A and section B.
  • All answers should include explanations in terms of physical principles.**

DATA

Density of fresh water  =

3 3 1.000 10 kg.m

 

Free fall acceleration at earth's surface g =

2 9.80 m.s

Gravitational constant G =

11 2 2 6.67 10 N.m .kg

  

Speed of light in a vacuum c =

8 1 3.00 10 m.s

 

Speed of sound in air v =

1 344 m.s

Avogadro constant N A

=

23 1 6.023 10 mol

 

Universal gas constant R =

1 1 8.314 J.mol .K

 

Boltzmann constant k =

23 1 1.380 10 J.K

  

Stefan-Boltzmann constant σ =

8 2 4 5.67 10 W.m .K

   

ADV_Q

It has been proposed that future space stations should be rotating cylinders, providing the

astronauts with artificial gravity. Consider such a cylinder of radius R rotating with an angular

velocity of in deep space, far away from any gravitational fields.

(a) Identify the forces acting on an astronaut who is rotating with the space station. By

comparing to an astronaut standing on the surface of the Earth, briefly describe how the

astronaut in the rotating space station experiences “weight”.

(b) If an astronaut in a rotating space station of radius R is to experience the same weight as

she would on Earth, show that the space station must be rotated at an angular velocity of

g

R

Ensure that you justify your answer.

(c) Suppose an astronaut rotating with the space station releases a ball from her hand.

Qualitatively describe the motion of the ball as observed by an external observer (not

rotating with the space station).

(5 marks)

ADV_Q02=REG_Q

An astronaut constructs an accelerometer for his rocket using two identical springs of

unextended length L and spring constant k , and two identical masses of mass m. He hangs the

springs and masses in the rocket as shown below.

Assume the rocket remains within the gravitational field of the Earth and that the acceleration

due to gravity is g at all times.

(a) After launch, the rocket accelerates vertically upwards with an acceleration a. Show that

the length of the accelerometer (the total length of the springs, assuming the masses have

negligible length) is:

m g a L k

(b) Eventually the rocket runs low on fuel and the motors are shut down. What is the total

length of the accelerometer after shutdown of the motors? Explain your result.

(5 marks)

Solution

(a)

Use Newton’s second law and take upwards to be positive:

Hookes law states that

F   k x

where x is the extension of the spring beyond its natural length and k is the spring constant.

For Mass A:

F 1 (^)  F 2  m gma

For Mass B:

F 2  m gma

(1 mark)

F 2  m g   a

Using the expression for F 2 we get

F 1 (^)  2 m g   a .

Hookes law gives the extensions for each spring:

 

 

1 1

1

F k x m g a

m g a x k

and

 

 

2 2

2

F k x m g a

m g a x k

(1 mark)

Ignoring the signs which just indicate direction then the total length is given by

REG_Q03=ADV_Q

Two solid spherical balls (A and B) are released at the same time from the top of the inclined

ramp. Both have the same mass M , but Ball A has twice the radius of Ball B. They both roll

down the ramp without slipping. The dimensions of the ramp are much larger than the radius of

either ball.

(a) Which ball reaches the bottom of the ramp first? Justify your answer.

Hint: Consider the potential, rotational, and translational kinetic energies.

In a second experiment, a solid spherical ball of radius R and a solid cylinder with a circular

cross section of radius R are released at the same time from the top of an inclined ramp. Both

have mass M. The dimensions of the ramp are much larger than the radius R of the ball or

cylinder.

Which object reaches the bottom of the ramp first if:

(b) The ramp is frictionless. Justify your answer.

(c) There is a frictional force between each object and the ramp so that they roll without

slipping. Justify your answer.

Moment of Inertia Data

Moment of Inertia for Solid Sphere

I  M R

Moment of Inertia for Solid Cylinder

I  M R

(5 marks)

Solution

(a)

The potential energy lost as an object comes down the ramp is equal to the sum of the

kinetic energy of translation and the kinetic energy of rotation at the bottom. If the vertical

height of the ramp is h , the final velocity of the object (of mass M and moment of inertia I ) at the

bottom of the ramp is v , and the rotational angular velocity of the object is then we have:

M g h  M v  I .

If the object rolls without slipping then

v  r 

where r is the radius of the object.

Apply this to:

Object 1: solid sphere of radius R rolling without slipping

2 2 2

2 2 2

v M g h M v MR R

g h v v v

Object 2: solid sphere of radius 2R rolling without slipping

Same answer as for Object 1 because the answer does not depend on radius.

Hence the two objects will have the same velocity (at all times going down the ramp as h is

arbitrary). They will reach the bottom of the ramp at the same time.

(2 marks)

(b)

Apply energy conservation equation to:

Object 1: solid sphere of radius R on ramp with no friction

Object slides so

2

2

M g h M v

g h v

Object 2: solid cylinder of radius R on ramp without friction

Object slides and answer is the same as for Object1 because the answer does not depend on the

moment of inertia of the object.

Hence the two objects will have the same velocity (at all times going down the ramp as h is

arbitrary). They will reach the bottom of the ramp at the same time.

(1 mark)

(c)

Apply energy conservation equation to:

Object 1: solid sphere of radius R on ramp rolling without slipping

ADV_Q

(a) Why does the water temperature at your favourite beach only vary by a few degrees

between peak summertime and mid-winter, while the daytime air temperature can vary

by as much as 30 degrees between seasons?

(b) Whilst lying on the beach on a hot day, it is common to feel a breeze blowing towards

you from the sea. Explain briefly why this occurs.

(c) In what fundamental way does a Stirling engine differ from an internal combustion

engine, such as that used in cars?

(d) Suppose you place an ideal Stirling engine on a block of ice. If the surrounding air

temperature is22.0 C

 , calculate the maximum efficiency of the engine.

(5 marks)

Solution

(a)

The large heat capacity of water means that a relatively large amount of heat needs to be

absorbed or lost to produce even a small temperature change. So the temperature of the water

will vary to a lesser extent than the air temperature.

(1 mark)

(b)

During daytime, land is hotter than the sea so hot air rises up from land creating a region of

lower pressure over the land. Air then moves from the region of higher pressure over the sea to

that of lower pressure over the land. This is the sea breeze, often experienced later in the day.

(1½ marks)

(c)

The heat sink in a Stirling engine is external to the working fluid, so it is an external combustion

engine.

(1 mark)

(d)

Use Carnot efficiency:

c

h

T

e T

(1½ marks)

ADV_Q

Harry stands near a bird’s nest. The bird tries to scare him away by flying in circles around his

head while emitting a sound wave with a constant frequency. The bird keeps a constant distance

from Harry.

Sally, standing at a large distance from Harry, also hears the bird (see the above diagram).

(a) At which position(s) of the bird (A,B,C,D) does Sally hear the highest frequency?

Briefly explain your answer.

(b) At which position(s) of the bird (A,B,C,D) does Harry hear the highest frequency?

Briefly explain your answer.

(c) At which position(s) of the bird (A,B,C,D) do Harry and Sally hear the same frequency?

Briefly explain your answer.

(5 marks)

Solution

(a)

At position B as the bird is approaching Sally with the greatest speed. The doppler shift( towards

higher frequency) of the sound emitted by the bird is largest at this position.

(2 marks)

(b)

All positions (A,B,C,D) are the same. The bird has no component of velocity towards Harry and

this is no doppler shift of the sound heard by him.

(1½ marks)

(c)

Positions A and C. At these positions there is no velocity towards Sally and likewise no velocity

towards Harry. There is no doppler shift and both Harry and Sally hear the same frequency.

(1½ marks)

ADV_Q

A proton of mass m moves in one dimension in a potential energy function given by

2

U x ( ) x x

where and  are positive constants. The proton is released from rest at x 0

In the following, ensure that you justify your answers.

(a) Show that the potential can be written as

2

0 0 2 0

x x U x x x x

 ^   

(b) Sketch a graph of the potential as a function of x and identify the point x 0.

(c) Give a qualitative description of the motion, identifying the minimum and maximum

values of x reached during the motion.

(d) What is the force on the proton as a function of x?

(e) Let the proton now be released from rest at 1

x

. Give a qualitative description of

the motion; how does this differ from the motion described in (c)?

(10 marks)

Solution

(a)

In the potential energy function

2

U x ( ) x x

put 0

x

2

2 2

2 2

2 2

2 2 0 0 2 2

2 0 0 2 2 0

U x x x

x x

x x

x x

x x

x x x

  ^ 

 ^ 

(2 marks)

(b)

0

U x (  x )  0

(2 marks)

(c)

Starts from rest and moves outwards as force accelerates proton to higher speed (force is

outwards). Past the minimum, the force is negative and so the proton decelerates but gets to

infinity with ever decreasing speed.

(2 marks)

(d)

Force is the negative gradient of the potential.

2 0 0 2 2 0

x x U x x x x

so

2 0 0 2 3 2 0

dU x x F x dx x x x

or, in terms of the potential function originally given in the question

2

U x ( ) x x

   

3 2 3 2

dU F x dx x x x x

 ^ ^  ^ 

(2 marks)

(e)

ADV_Q

On a warm summer’s day at the South Pole, a worker is pushing boxes up a rough plank inclined

at an angle  above the horizontal. The plank is partially covered in ice, with more ice near the

bottom of the plank than near the top. He pushes the boxes at the bottom and each slides up the

plank to a height that depends on how hard he pushed.

The coefficient of friction increases with distance x up the plank such that:

  x

where  is a positive constant. The bottom of the plank is at x = 0. Assume that the coefficients

of static and kinetic friction are equal, so that  s (^)   k .

(a) Sketch a diagram showing all the forces acting on a box while on the plank.

(b) Show that as a box of mass m slides from x = 0 to x = A , the work done by friction on the

box is

2

cos 2

A

W m g

(c) If the box has an initial velocity of v 0 as it starts up the plank at x = 0, show that the

maximum distance along the plank that the mass travels, B , is the solution to the equation

2 2

 g B cos   2 g B sin   v 0  0

Derive this equation, but do not solve it.

(d) Again consider the mass sliding up the slope with an initial velocity v 0. Show that when

the box first comes to rest, it remains at rest if

2 2 0

3 sin

cos

g v

 

(10 marks)

Solution

(a)

(1 mark)

(b)

Take the x direction upwards along the plank and the y direction upwards and perpendicular to

the plank.

Decomposing forces in the y direction:

cos 0

cos

N m g

N m g

(1 mark)

Kinetic friction force is

Ff   N  mg cos.

The question tells us that varies along the plank as follows:

  x

Hence kinetic friction force is given by

Ff   xmg cos

Work done is given by

0

cos

A W  (^)  F dxf  (^)   mgx dx

(1 mark)

Therefore

2

cos 2

A

W mg

(1 mark)

(c)

Conservation of energy at point B gives

2 2 0

cos sin 2 2

B

mv mg mg h mg B

(1 mark)

This can be rearranged as

ADV_Q

It is a hot day and the air temperature in your room is 30.0 C so you turn on the air conditioner,

which brings the temperature down to 22.0 C. The room is sealed and contains 2500 moles of

air.

(a) Calculate the change in entropy in the room. Use

1 1 cair 29.11 J.mol .K

  .

(b) Using the refrigerator statement, explain how the room can continue to cool below the

surrounding outside temperature (30C) without violating the 2

nd law of thermodynamics.

(c) Suppose once the temperature is constant in your room that you open a door leading to an

adjacent room of identical volume that you’ve kept in a vacuum for the purpose of

conducting “secret physics experiments”. Calculate the entropy change as the N

molecules of air expand freely to occupy the volume of both rooms. Assume there is no

change in the air temperature.

(10 marks)

Solution

(a)

Use

2

1

dQ S

T

 

(1 mark)

2 2 2 3 1

1 1 1

ln (2500)(29.11)ln 1.947 10 J.K 303

dQ dT T S nc nc T T T

        

(2 marks)

(b)

The refrigerator statement says that energy cannot flow from a cooler body to a hotter body

without work being done. As the room cools to a temperature below the surrounding

temperature, it can continue to cool only because work is being done by the air conditioning unit

to continually remove heat from the room to outside.

(3 marks)

(c)

Use

Sk ln

w 2

w 1

where w 1 is the number of microscopic states of the system when it occupies volume 1 and w 2 is

the number of microscopic states of system when it occupies volume 2.

(1 mark)

So

2 2 1 ln2^ ln

N N ww  SkN k

(1 mark)

and

23 27 Nn NA  (2500)(6.02 10 ) 1.51 10  molecules.

So

4 1 S 1.44 10 J.K

   .

(2 marks)

Can also use

2

1

ln ln(2) ln(2)

V

S n R n R N k V

which corresponds to an equivalent reversible process – isothermal expansion, with QW , so

2

1

V

V

dQ W dV S n R T T V

 

.