ASTR 101 Lecture Notes - Lecture 4: Tidal Locking, Orbital Period, Hyperbola

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tealdinosaur910

22 Nov 2017

School

Queen's University

Department

Physics, Engineering Physics and Astronomy

Course

ASTR 101

Professor

David Alan Hanes

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Document Summary

Acceleration: rate of change of velocity (m/s2) On earth, g 10m/s2: speed increases 10 m/s with each second of falling. Galileo showed that g is the same for all objects, regardless of their mass. All falling objects accelerate at the same rate (not counting the friction of air resistance). A net force changes momentum of a spinning or orbiting object known as angular momentum. Mass - a measure of the amount of matter in an object. Weight - the force that a scale exerts upon an object. Weightlessness is due to a constant state of free-fall. Angular momentum: describes objects that are spinning or moving in circles. A special description of the force, a torque, is needed to change an object"s angular momentum. Realized the same physical laws that operate on earth also operate in the heavens. Much more; experiments with light, first reflecting. Objects continue at constant velocity because of conservation of momentum.

Related Questions

a.) A hockey puck is observed to be sliding along a flatfrictionless surface at a speed of 4 cm/min. There is no net forceacting on the puck. Assuming it doesn't smash into anything, howfast will the puck be going 17 min later?
cm/min

b.) An object is accelerating. Which one of the followingstatements is true?
The object must be speeding up.
Gravity must be causing the object to accelerate.
The object must be changing directions.
The object must be slowing down.
There is a net force acting on the object.

c.) A 100 kg block weighs 220 lb on Earth. If the block weretransported to a planetary body that had a smaller free fallacceleration (as compared to Earth), which one of the followingwould be true?
The block would have the same weight and a smaller mass.
The block would have the same mass and a bigger weight.
The block would have the same mass and the same weight.
The block would have the same weight and a bigger mass.
The block would have the same mass and a smaller weight.
d.) What does it mean (in physics) to say that a quantity isconserved?
It means that the quantity is not being used.
It means the total amount is zero.
It means the total amount never changes.
It means it's being saved up.

e.) At a certain point in time, the total energy of a closed systemis observed to be 176 J. Which one of the following statements istrue?
Three hours later the energy of the system is 352 J.
Four hours later the energy of the system is 88 J.
All of the energy must be kinetic if the objects are moving.
Twelve hours later the energy of the system is 176 J.
Some of the energy must be potential.

g.) Which of the following statements are impossible? Choose allthat apply.
The parked truck had 8770 J of KE.
The scooter had 0 J of KE and 0 kgÂ·m/s of momentum.
A ball moving in a circle had 0 N of net force acting on it.
A box had an acceleration of 0 m/s2 while a net force of 311 Nacted on it.
Jack exerted 18 N of force due east on a cart. As a direct result,the cart exerted 18 N of force due west upon Jack.
A toothpick exerts a smaller gravitational force upon the Moon thanthe Moon exerts upon it.

a.) Two balls make up a closed system. Ball A has 701 J of energyand ball B has 78 J. After they collide, ball A has 665 J. How muchenergy does ball B have after they collide?
J

b.) An asteroid and a comet collide head on. Prior to thecollision, the asteroid has a momentum of 700,000 kgÂ·m/s up and thecomet has a momentum of 800,000 kgÂ·m/s down. After the collisionthe asteroid has a momentum of 750,000 kgÂ·m/s down. After thecollision, what is the comet's momentum?
kgÂ·m/s
North
South
East
West
Up
Down
Left
Right
No direction because the comet has stopped

c.) How much does a 28 kg object weigh on Moon? Please feel free toreference the Solar System Data table located at the beginning ofUnit II.
______N

d.) A 660 N net force acts on a 42 kg object. Determine theobject's acceleration.
_____m/s2

e.) Calculate the gravitational force between a 657 kg object(resting on Earth) and the Moon. The mass of the Moon is 7.347Ã1022kg and the distance between the object & the Moon is 3.844Ã108m.
____N

1.

While helping a friend move, you push a 11.5 kg dolly with a forceof 11.0 N. Assume there are no other forces acting on the dolly. Ifthe dolly starts at rest, how far does it move in 3.50 s?
1 m

2.

At a lumberjack competition an axe with a mass of 1.25 kg is thrownat a target with a speed of 41 m/s. When it hits the target, itpenetrates to a depth of 0.055 m.
(a) What was the average force exerted by the target on theaxe?
1 N

(b) If the mass of the axe is halved, and the force exerted by thetarget on the axe remains the same, by what multiplicative factordoes the penetration depth change?
2
It doubles. It increases by a factor of the square root of two. Itdoes not change. It decreases by a factor of the square root oftwo. It halves.
Your work in question(s) will also be submitted or saved.
3

Your 1500 kg car pulls a 540 kg trailer away from a stop sign withan acceleration of 2.00 m/s2. (Note the answers should be given inkN.)
(a) What is the net force exerted by the car on the trailer?
1 kN

(b) What force does the trailer exert on the car?
2 kN

(c) What is the net force acting on the car?
3 kN

4.

Two boxes sit next to, and touching, each other on a frictionlesshorizontal surface. The lighter box has a mass of 5.31 kg and theheavier box has a mass of 7.29 kg.
(a) Find the contact force between these boxes when a horizontalforce of 6.1 N is applied to the light one.
1 N

(b) If the 6.1 N force is applied to the heavy box instead, is thecontact force between the boxes the same as, greater than, or lessthan the contact force in part (a)?
2

(c) Verify your answer to part (b) by calculating the contact forcein this case.
3 N

5.

A child sits in a wagon and together they have a mass of 25 kg. Thewagon is pulled with a constant speed by a handle that is at anangle of 22Â° above the horizontal. If the normal force exerted onthe wagon is 210 N, what is the force F applied to thehandle?
1 N

6.

An object acted on by three forces moves with constant velocity.One force acting on the object is in the positive x direction andhas a magnitude of 6.7 N; a second force has a magnitude of 4.5 Nand points in the negative y direction. Find the direction andmagnitude of the third force acting on the object.
1 N
2Â° (counterclockwise from the +x axis)

7.

A chain is used to lift a 1200 kg car vertically upward with anacceleration of 0.90 m/s2. What is the tension in the chain?
1 N

8.

A 22.5 kg box (m1) rests on a table.

(a) What is the weight of the box?
1 N
What is the normal force acting on it?
2 N (upward)
(b) A 12.5 kg box (m2) is placed directly on top of the 22.5 kgbox.
Determine the normal force that the table exerts on the 22.5 kgbox.
3 N (upward)
Determine the normal force that the 22.5 kg box exerts on the 12.5kg box.
4 N (upward)

9.

Your friend has a weight of 127 lb when standing on a scale. Youtake the scale to an elevator where you notice your friend's weighton the scale reads 108 lb. What are the direction and magnitude ofthe elevator's acceleration?
Direction 1
Magnitude 2 m/s2

10.

When you lift a package with a force of 82 N, the packageaccelerates upward at a rate of a. If, instead, you lift with aforce of 93 N, the package now has an acceleration of 8 multipliedby a.
(a) Find the weight of the package.
1 N

(b) Find the acceleration a.
2 m/s2

11.

One 3.1 kg paint bucket is hanging by a massless cord from another3.1 kg paint bucket, also hanging by a massless cord, as shownbelow.

(a) If the buckets are at rest, what is the tension in eachcord?
(lower cord) 1 N
(upper cord) 2 N

(b) If the two buckets are pulled upward with an acceleration of1.30 m/s2 by the upper cord, calculate the tension in eachcord.
(lower cord) 3 N
(upper cord)5

12)
Your friend has a pair of fuzzy dice with a mass of 1.32 g hangingfrom his rear-view mirror.
(a) When he accelerates from a stoplight, the dice deflect backwardtoward the rear of the car hanging at an angle of 6.18Â° relative tothe vertical. Find the tension in the string holding thedice.
N

(b) At what angle to the vertical will the tension in the string betwice the weight of the dice?
Â°

A boy in a wheelchair (total mass 50.5 kg) has a speed of 1.40 m/s at the crest of a slope 2.40 m high and 12.4 m long. At the bottom of the slope, his speed is 6.50 m/s. Assume air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N. Find the work he did in pushing forward on his wheels during the downhill ride.

2 Answers

Relevance

electron1
Lv 7

4 years ago

There are two forces that cause the wheelchair to accelerate. These two forces are the component of the wheelchair’s weight that is parallel to the hill and the force the boy exerts. The net force on the wheelchair is equal to the sum of these two forces minus the friction force. Let’s use the following equation to determine the acceleration.

vf2 = vi^2 + 2 * a * d

6.5^2 = 1.4^2 + 2 * a * 12.4

24.8 * a = 40.29

a = 40.29 ÷ 24.8

This is approximately 1.62 m/s^2.

Net force = 50.5 * 40.29 ÷ 24.8 = 2034.645 ÷ 24.8

This is approximately 82 N.

Force parallel = 50.5 * 9.8 * 2.40 ÷ 12.4 = 1187.76 ÷ 12.4

This is approximately 95.8 N. This force and the force that the boy exerts are the forces that cause the wheelchair to accelerate.

Net force = F + (1187.76 ÷ 12.4) – 41

Set the two net forces equal to each other.

F + (1187.76 ÷ 12.4) – 41 = 2034.645 ÷ 24.8

F = 41 + (2034.645 ÷ 24.8) – (1187.76 ÷ 12.4)

F is approximately 27.3 N

Work = [41 + (2034.645 ÷ 24.8) – (1187.76 ÷ 12.4)] * 12.4 = 337.9625 J
JupiterSky
Lv 6

4 years ago

His change in kinetic energy is due to gravitational- and his own work,

while the friction does negative work.

∆K = ∆U +∆W + ∆Wf

½m(Vf²−Vi²) = mgh + ∆W + Ff*d

∆W = ½m(Vf²−Vi²) − mgh − Ff*d

∆W = ½*50.5(6.5²−1.4²) − (50.5*9.8*2.4) − (−41*12.4)

∆W = 508 J

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Next Question

---

asked in Science & MathematicsPhysics · 8 years ago

What is the spring constant? please help me!!!!?

A block of mass 487 g is pushed against the
spring (located on the left-hand side of the
track) and compresses the spring a distance
4.7 cm from its equilibrium position (as shown
in the ﬁgure below). The block starts from
rest, is accelerated by the compressed spring,
and slides across a frictionless horizontal track
(as shown in the ﬁgure below). It leaves the
track horizontally, ﬂies through the air, and
subsequently strikes the ground.
acceleration = 9.81
 
A) What is the spring constant?
B)What is the speed v of the block when it leaves
the track?
C) What is the total speed of the block when it
hits the ground?
...Show more

Update:

Height Above ground = 1.9 m
Lands = 4.59m away

What is the spring constant? please help me!!!!?

3 Answers

Relevance

Jánošík
Lv 7

8 years ago

Favorite Answer

I'm sure there is some crucial information in "the figure below" !!!

Edit:

OK... now we got something to work with ... ;-)

Time needed to fall a vertical distance of 1.9 m is:

t = √( 2h / g )

t = √[ 2×(1.9 m) / (9.81 m/s²) ]

t = 0.6224 s

In that time, the block must travel a horizontal distance of 4.59 m, so the horizontal velocity is:

r = v×t

(4.59 m) = v×(0.6224 s)

v = 7.37 m/s < - - - - - - - - - - - - - - - - - answer B

That is also the speed of the block as it leaves the spring.

Its kinetic energy at that time is:

KE = m×v²/2

KE = (0.487 kg)×(7.3749 m/s)²/2

KE = 13.244 J

That is also the elastic potential energy stored in the spring at maximum compression. So the spring constant is:

KE = PEe = k×d²/2

(13.244 J) = k×(0.047 m)²/2

k = 12 N/m < - - - - - - - - - - - - - - - - - - answer A

The gravitational potential energy of the block on the table is:

PEg = m×g×h

PEg = (0.487 kg)×(9.81 m/s²)×(1.9)

PEg = 9.077 J

The total energy of the block on the table is:

TE = PEg + PEe

TE = (9.077 J) + (13.244 J)

TE = 22.32 J

When hitting the ground, all that energy is converted to kinetic energy, so

TE = KE = m×v²/2

(22.32 J) = (0.487 kg)×v²/2

v = 9.57 m/s < - - - - - - - - - - - - - - - - - - answer C

...Show more
Technobuff
Lv 7

8 years ago

There must be a height above ground and a distance from the end of the track on the "as shown in the figure below".??
STALKER
Lv 4

8 years ago

use f=kl

where k is the spring constant and l is extension from the natural length.

k=f/l

since you have mass in grams. use w=mg

w=0.487 x 9.8 = 4.77N

Now use the equation,

k=4.77/4.7

k=1.012

Note: my answer could be wrong since I do not know the springs original length from the diagram shown. But, the method of working out spring constant is same.

...Show more

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Next Question

ひいらぎ

Lv 5

asked in Science & MathematicsPhysics · 1 decade ago

Physics Momentum, Rolling?

Pat builds a track for his model car out of wood. The track is 10.00 cm wide and 3.00 m long, and is solid. The runway is cut such that it forms a parabola by the equation y=(x - 3)^2/9. Locate the horizontal position of the center of gravity of this track.
 
A washing machine load in the "spin" cycle, (much like a centrifuge used to separate fluids), can be modeled as a cylinder which has zero density toward the center, and starting from some inner radius where the clothes (or fluids) begin, increases in density as you move radially outward. The washing machine has density of clothes (in SI units) of p=14r+5, a height of 0.51 m, and only has clothes present from inner radius 0.2m out to 0.7m from the center.
-What is the moment of Inertia of the wash load?
-What is the power of motor that drives the clothes from rest up to 4 rev/sec, in 14 seconds if the basin has moment of Inertia of 4 kg*.m2
...Show more

Physics Momentum, Rolling?

1 Answer

Relevance

Anonymous

1 decade ago

Favorite Answer

I can't visualize the first problem.

The second: Moment of inertia is I = m r^2. In a shell of thickness dr there is a mass of p * 2 * pi * r * h * dr. The moment of inertia of the shell is 2 pi h p r^3 dr. Now integrate this from r = 0.2m to r = 0.7m.

Kinetic energy is 1/2 m v^2. In rotations the moment of inertia is like the mass and the angular velocity is like the velocity, so rotational energy is 1/2 I omega^2. Omega is in rad/s. The total inertia of the system is the sum of the inertia of the clothes and the basin. Divide the energy by the time needed to get the power.

...Show more

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Next Question

asked in Science & MathematicsPhysics · 1 decade ago

Physics:You and your friends push a 75-kg...Help!?

You and your friends push a 75-kg greased pig up an aluminum slide at the county fair, starting from the low end of the slide. The coefficient of kinetic friction between the pig and the slide is 0.070.
 
(a) All of you pushing together (parallel to the incline) manage to accelerate the pig from rest at the constant rate of 4.9 m/s2 over a distance of 1.8 m, at which point you release the pig. The pig continues up the slide, reaching a maximum vertical height above its release point of 50 cm. What is the angle of inclination of the slide? __degree
 
(b) At the maximum height the pig turns around and begins to slip down the slide, how fast is it moving when it arrives at the low end of the slide? __m/s
...Show more

Physics:You and your friends push a 75-kg...Help!?

1 Answer

Relevance

Anonymous

1 decade ago

Favorite Answer

♣the distance the pig was pushed up is

s=0.5*a*t^2, where s=1.8m, a=4.9m/s^2, t is time of push-up;

speed of the pig gained after s=1.8m is v=a*t; thus v^2=2as;

♠ kinetic energy of the pig at the moment it’s released after push-up is

E=0.5*m*v^2, which is enough for the pig to move by itself L distance more along the ramp, where L=h/sin(b), h=50cm=0.5m, b is angle in question;

♦pig’s potential energy at vertex (when stopped) is E1=mgh;

the energy E2=E-E1 is gone on friction, that is E2=f*L is work of friction, where

f=μ*w1, w1=mg*cos(b) is component of pig’s weight normal to the aluminum ramp; or;

f= μ*mg*cos(b);

♦Thus E2=E-E1; or; f*L = 0.5*m*v^2 –mgh; or;

(μ*mg*cos(b)) * (h/sin(b)) =0.5*m* 2as –mgh; or;

μ*gh*cos(b)/sin(b) =as –gh, hence cot(b) =(as –gh)/(μ*gh), hence

(a); b=atan(μ*gh/(as-gh))= atan(0.07*9.8*0.5/(4.9*1.8 -9.8*0.5)) =5°;

♥ pig’s potential energy above the low end of the ramp is

W= mg*(s*sin(b) +h); pig’s ramp position is p=s +L = s + h/sin(b);

Pig’s final kinetic energy is W1=W-W2, where W2=f*p is work of friction;

and W1=0.5*m*v^2; thus

0.5*m*v^2 = mg*(s*sin(b) +h) – μ*mg*cos(b)*(s + h/sin(b));

v^2 =2g*(s*sin(b) +h)*(1 –μ*cot(b)) =2.575;

(b); v =1.605 m/s;

...Show more

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ASTR 101 Lecture Notes - Lecture 4: Tidal Locking, Orbital Period, Hyperbola

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