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CCS Physics Chapter 7: Circular Motion, Rotary Motion, Laws of Universal Gravitation, Kepler's Laws, Satellites, and Simple Machines
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Terms in this set (145)
Which type of motion occurs when the motion of the object is outside of the center of rotation?
circular motion
What is the definition of circular motion?
motion of the object is outside the center of rotation
What is an example of circular motion?
the moon
The moon is an example of what kind of motion?
circular motion
What are the two aspect of circular motion that we need to know how to calculate?
centripetal acceleration and centripetal force
Is centripetal acceleration a scaler or vector quantity?
vector
Is centripetal force a scaler or vector quantity?
vector
Where is the direction of centripetal force and acceleration always pointing to?
the center
What is the formula for finding centripetal acceleration?
ac= v2/r
What does T stand for?
period
A runner going around a track goes at a speed of 8.8 m/s around the curve. If the radius of the curve is 25 meters, what is the centripetal acceleration?
ac= v2/r
V= 8.8m/s
r= 25m
ac= 8.8^2/25= 3.1 m/s2
ANSWER= 3.1 m/s2 to the center
A car is going around a curve with a radius of 65 meters. If it takes the car 35 seconds to go all the way around, what is the centripetal acceleration of the car?
Use the formula: ac= 4n2r/T2
r= 65m
T= 35s
ac= 4n2(65)/35^2= 2.1 m/s2
ANSWER: 2.1 m/s2 to the center
What exists in non-intertial frameworks but does not exist in an inertial frameworks?
centripetal force
What is the formula we use to find centripetal force?
Fc= mv2/r (v= 2nr/T)
There's a 13g stopper attached to a string that is 0.93m long. It takes 1.18 seconds to make one revolution around a horizontal circle. What is the centripetal force exerted?
First, use the formula V= 2nr/T
r= 0.93m
T= 1.18s
V= 2n(0.93)/1.18= 4.95m/s
Now that you have V, you can plug the information into the equation Fc= mv2/r
M= 13g (you have to convert it to kg and you get 0.013kg)
r= 0.93m
V= 4.95m/s
Fc= (0.013)(4.95)^2/0.93= 0.34 N
ANSWER: 0.34 N
What would be the centripetal force of a person on earth who weighs 97kg? Would this force be enough to throw them off the planet? (Earth's radius= 6.38x10^3km. Earth's period is 24 hours)
First gather the information:
r= 6.38x10^3km (must be changed to m, we get 6.38x10^6m)
m= 97kg
T= 24hrs (must be changed to seconds, 24x60x60= 86400 seconds)
We then use the formula V=2nr/T
V= 2n(6.38x10^6)/86400= 464m/s
Now that we know the velocity, we can plug the information into the equation Fc= mv2/r
Fc= (97)(464)^2/6.38x10^6= 3.3 N
ANSWER: 3.3 N. No, this is not enough force to throw them off the planet (why? idk)
Find the centripetal force of a 0.45kg object going in a circle with a radius of 1.25m. It goes through 20 revolutions in 15.2 seconds.
First, we must find the T.
To do this, simply divide 15.2 seconds by 20 revolutions and get 0.76rev/sec
m= 0.45kg
r= 1.25m
T= 0.76rev/sec
Plug information into V= 2nr/T
V= 2n(1.25)/0.76= 10.3m/s
V= 10.3m/s
Now, plug it into the equation Fc= mv2/r
Fc= (0.45)(10.3)^2/1.25= 38.2 N
ANSWER: 38.2 N
What is the centripetal force of a car with a mass of 1100kg going 29.2m/s around a curve with a radius of 46.5m?
We don't need to use the velocity equation since it has already been given to us so we just plug our information into the equation Fc= mv2/r
m= 1100kg
v= 29.2m/s
r= 46.5m
Fc= (1100)(29.2)^2/46.5 = 20170 N
ANSWER: 20170 N
A 900kg car navigates a curve with a center of rotation of 55m. If the car is traveling 18.5m/s, what is the centripetal force?
Fc = mv^2/r
m= 900kg
v= 18.5m/s
r= 55m
Fc= (900)(18.5)^2 / 55 = 5600 N
ANSWER: 5600 N
A 0.05kg stopper makes 20 cycles in 12.5 seconds. If the length of the string is 0.85 meters, what is the centripetal force?
V = 2nr/T
Fc= mv^2/r
T= 12.5 / 20 = 0.625
V= 2n(0.85) / 0.625 = 8.55m/s
Fc= (0.05)(8.55)^2 / 0.85 = 4.3 N
ANSWER: 4.3 N
Which type of motion occurs when the motion of the object is inside the center of rotation?
rotary motion
When does rotary motion occur?
when the motion of the object is inside the center of rotation
What are the two aspects of rotary motion that we need to know how to calculate?
angular velocity and acceleration
What is the difference between rotary motion and circular motion?
circular motion occurs when the motion of the object is outside the center of rotation and rotary motion occurs when the motion of the object is inside the center of rotation
What is an example of rotary motion?
the earth spinning on its axis
The Earth spinning on its axis is an example of what type of motion?
rotary motion
What is the formula for angular velocity?
w=@/t
Describe what each variable is in the equation: w=@/t
w= omega (which is angular velocity)
@= angular displacement (change in feta)
t= time
What does the term omega refer to?
angular velocity
What term is used to refer to angular velocity?
omega
A sphere rotates 5.7 revolutions in 1.6 seconds. What is the angular velocity?
w = @/t
@ = 5.7rev x 2nrad = 35.8 rad
w = 35.8rad / 1.6 sec = 22 rad/sec
ANSWER: 22 rad/sec
A flywheel turns 2.50 revolutions in 0.600 seconds. What is the angular velocity?
w = @/t
@ = 2.50rev x 2nrad = 15.7rad
w = 15.7 / 0.600sec = 26.2rad/sec
ANSWER= 26.2 rad/sec
A carwheel turns 35.2 revolutions in 6.50 seconds. What is the angular velocity?
w = @/t
@ = 35.2rev x 2nrad = 221.2
w= 221.2 / 6.50sec = 34.0 rad/sec
ANSWER: 34.0 rad/sec
An object spins 3.2 revolutions in 9.5 seconds. What is omega?
w = @/t
@ = 3.2rev x 2nrad = 20.1 rad
w = 20.1 / 9.5sec = 2.1rad/sec
ANSWER: 2.1 rad/sec
Convert 520 degrees to radians.
520 degrees x 2nrad / 360 degrees = 9.08 rads
ANSWER: 9.08 radians
What is the formula we use when finding angular acceleration?
& = w/t (wf-wi/t)
What do each of the variables stand for in the equation & = w/t?
&= alpha (angular acceleration)
w= change in omega (angular velocity)
t= time
A bike starting from rest attains an angular velocity of 45.2rad/sec in 7.60 seconds. What is the angular acceleration?
Use the formula &=w/t (wf-wi/t)
w= 45.2 - 0
(it is 0 since it is starting at rest)
t= 7.60sec
Plug in the information and get:
& = 45.2-0/7.60 = 5.95rad/sec^2
ANSWER: 5.95 rad/sec^2
A sphere rotating 2.5rev/sec accelerates to 5.3rev/sec in 0.91 seconds. What is the angular acceleration?
wi = 2.5rev x 2nrad = 15.7 rad/sec
wf = 5.3rev x 2nrad = 33.3 rad/sec
&= wf - wi / t
& = 33.3 - 15.7 / 0.91 = 19 rad/sec^2
ANSWER: 19 rad/sec^2
What is the label for alpha?
rad/sec^2
A pulley starting at rest attains an angular velocity of 3.7rev/sec in 14 seconds. What is the angular acceleration?
Use the formula &=w/t (wf-wi/t)
wi= 0rad/sec
wf= 3.7rev/sec (must change to rad/sec- 3.7rev x 2nrad / 1sec x 1rev = 23.25rad/sec)
t= 14sec
Plug in information:
&= 23.25-0 / 14 = 1.7rad/sec^2
ANSWER: 1.7 rad/sec^2
What is the resistance of a rotating object to changes in its angular velocity?
rotational inertia
What is the definition of rotational inertia?
resistance of a rotating object to changes in its angular velocity
What is the amount of force applied tangent through a certain distance from the center called?
torque
What is a torque?
the amount of force applied tangent through a certain distance from the center
What is the label for a torque?
nm
What is the formula used when finding a torque?
Torque = Fr (or)
Torque = I& (Inertia x alpha)
When comparing a thin ring to a disk, which object has the most inertia and why?
the thin ring has the most inertia because all of the mass is distributed on the outside. the mass of the disk is distributed evenly
What are two things that increase rotational inertia?
how the mass is distributed and the amount of mass
What kind of distribution of mass would increase inertia?
mass distributed further away from the rotation
When there is less inertia, does the object go faster or slower?
faster
When there is more inertia, does the object go faster or slower?
slower
Finish the statement:
The less inertia, the __________
faster it goes
Finish the statement:
The more inertia, the __________
slower it goes
A disk with a mass of 0.85kg and with a radius of 0.220m starts at rest. If the disk is hit with a torque of 2.75mn, after 3.5 seconds, what is the final angular velocity?
First you have to find the angular acceleration.
To do this, use the torque equation Torque = I&.
In order to find & (alpha), we must rearrange the equation to look like & = Torque/ I.
We know the Torque = 2.75nm
We need to find I. To do this, use the equation for inertia of a disk ( I = 1/2mr^2)
Plug in information and get:
I = 1/2(0.85)(0.220)^2 = 0.0206kgm^2
Now that we have the inertia, we can find alpha.
& = Torque / I
& = 2.75/0.0206 = 133 rad/sec^2
&= 133 rad/sec^2
Now that we have the angular acceleration, we must find the angular velocity. To do this, we look at our equation sheet and find that the best equation to use in this situation is Wf= Wi + &t
Wi= 0
&= 133 rad/sec^2
t= 3.5 sec
Wf = 0 + (133)(3.5)= 466rad/sec
ANSWER: 466 rad/sec
There is a 1.7kg disk with a radius of 0.35m. If a torque of 15mn is applied, what would the angular acceleration be?
First, we must decipher which equation to use based off of the shape of the object. Since it is a disk, we know the equation we will use will be I = 1/2mr^2.
m= 1.7kg
r= 0.35m
We plug in the info:
I = 1/2(1.7)(0.35)^2= 0.10kgm^2
We then use the Torque equation to find alpha.
Torque = I&
We rearrange it and get & = Torque / I
Torque= 15mn
I= 0.10kgm^2
& = 15/0.10 = 150 rad/sec^2
ANSWER: 150 rad/sec^2
If a torque of 5.6mn is applied to a thin ring that is 0.85kg (with a radius of 0.22m) that is at rest, how long will it take to reach a final angular velocity of 47rad/sec^2?
First determine which equation will be used based on the shape of the object. Since it is a thin ring, we know we will use I =mr^2
m= 0.85kg
r= 0.22m
I = (0.85)(0.22)^2 = 0.041kgm^2
Now, we must find alpha, we do this by using the torque equation and rearranging it to look like:
& = Torque / I
We plug in the information:
& = 5.6 / 0.041 = 140rad/sec^2
Now that we know this information, we must find an equation that will allow us to find the time.
Wf= wi + &t
We rearrange it to get:
t = wf/&
t = 47 / 140 = 0.34s
ANSWER: 0.34 seconds
What is the equation that goes along with Newton's Law of Universal Gravitation?
Fg= G (mi x m2/d^2)
What do each of the variables represent in the formula Fg = G (mi-m2/d^2)?
Fg= force of gravity
G= 6.67x10^-11 nm^2/kg^2 (universal standard)
mi= mass of one object
m2= mass of the other objecy
d= distance between the centers of gravity of both objects
If there is a 50kg person standing on the earth, what is the force of gravity according to Newton's Law of Universal Gravitation?
(The mass of the earth is 5.98x10^24kg. The radius of the earth is 6.378x10^6m)
Use the formula Fg=G(mi x m2/d^2)
G= 6.67x10^-11 nm^2/kg^2
mi= 5.98x10^24kg
m2= 50kg
d= 6.378x10^6m
Plug in the information:
Fg= 6.67x10^-11 (5.98x10^24 x 50 / 6.378x10^6 ^2) = 490 N
ANSWER: 490 N
What would be the force of gravity if you (a 90kg person) were 100km above the earth?
(The mass of the earth is 5.98x10^24kg. The radius of the earth is 6.378x10^6m. G= 6.67x10^-11)
Fg=G(mi x m2/d^2)
Convert 100km to m and get: 1x10^5m.
Add the 1x10^5m to the radius of the earth (6.378x10^6) to get the distance.
Plug in the information:
G= 6.67x10^-11
mi= 5.98x10^24kg
m2= 90kg
d= 6.378x10^6 + 1x10^5m
Fg= (6.67x10^-11)(90 x 5.98x10^24 / (6.378x10^6 + 1x10^5) ^2 = 855 N
ANSWER: 855 N
What is the angular acceleration of a clock's second hand?
0
How much gravitational force exists between the earth and the moon? (Earth's mass is 5.98x10^24. Moon's mass is 7.36x10^22.
Earth's radius is 6.378x10^6m. Moon's radius is 1.74x10^6m. Earth's apogee is 406,700km. Earth's perigee is 356,400km.)
We use the formula Fg= G (mi x m2/d^2)
First, we need to determine what the distance would be. To do this, we first add Earth's apogee and perigee and divide them by 2. (406,700 + 356,400 / 2 = 381550km) We need to convert the km to m so we get 3.8155x10^8m.
We then add that to the earth and moon's radii and get 3.81550x10^8 + 1.74x10^6 + 6.378x10^6 = 3.89668x10^8m
DISTANCE= 3.89668x10^8m
Now that we have all the distance, we can plug information in.
G= 6.67x10^-11 nm^2/kg^2
mi= 5.98x10^24kg
m2= 7.36x10^22kg
d= 3.89668x10^8m
We then plug it into the equation:
Fg = 6.67x10^-11 (5.98x10^24 x 7.36x10^22 / 3.89668x10^8 ^2) = 1.933x10^20 N
ANSWER: 1.933x10^20 N
What would be the force of gravity if you were a 55kg person standing on Mars. (Mars mass is 6.42x10^23kg. Mars radius is 3.40x10^6m. G= 6.67x10^-11nm^2/kg^2.)
We use the equation Fw= G(mi x m2 /d^2)
G= 6.67x10^-11 nm^2/kg^2
mi= 6.42x10^23 kg
m2= 55 kg
d= 3.40x10^6 m
Plug in the information:
Fw= (6.67x10^-11)(6.42x10^23 x 55 / 3.40x10^6 ^2)= 204 N
ANSWER: 204 N
What is "motion that results from the application of a torque that tends to displace the axis of rotation of a rotating object"?
precession
What occurs when there is an uneven torque?
precession
What method is used when determining the direction of angular velocity?
right-hand rule
What is a big example of precession?
the earth's rotating process
How long does it take the earth to go through on full precession?
26,000 years
Name Kepler's three laws of planetary motion
1. Planets orbit the sun in ellipses with the sun at one focus
2. A line drawn from a planet to the Sun sweeps out equal areas in equal intervals of time.
3. The square of the orbital period of a planet is directly proportional to the cube of its average distance.
What does Kepler's first law state?
Planets orbit the Sun in ellipses with the Sun at one focus.
Which of Kepler's laws states that planets orbit the sun in ellipses with the sun at one focus?
first law
What does Kepler's second law state?
A line drawn from a planet to the Sun sweeps out equal areas in equal intervals of time.
Which of Kepler's laws states that a line drawn from a planet to the sun sweeps out equal areas in equal intervals of time?
second law
According to Kepler's second law, when are planets moving the fastest?
when they are near the sun
When do planets move slower according to Kepler's second law?
when they are further from the sun
What does Kepler's third law state?
The square of the orbital period of a planet is directly proportional to the cube of its average distance.
What equation goes along with Kepler's third law?
Ti^2 / T2^2 = ri^3 / r2^3
Find the period of Mercury.
Mercury's distance from the sun= 5.79x10^10m
Earth's distance from the sun= 1.50x10^11m
Earth's period= 365.25 days
We use the equation Ti^2 / T2^2 = ri^3 / r2^3
We must rearrange it to solve for t2.
We get T2= SQUARE ROOT of Ti^2 x r2^3 / ri^3
Ti= 365.25 days
ri= 1.50x10^11 m
r2= 5.79x10^10 m
T2= ?
Plugin the information:
T2= SQUARE ROOT of 365.25^2 x 5.79x10^10 ^3 / 1.50x10^11 ^3 = 87.6 days
ANSWER: 87.6 days
Find the period of Jupiter.
Jupiter's distance from the sun= 776.3x10^8 KM
Earth's distance from the sun= 149.8x10^8 KM
Earth's period= 1 year
We use the equation Ti^2 / T2^2 = ri^3 / r2^3
We must rearrange it to solve for t2.
We get T2= SQUARE ROOT of Ti^2 x r2^3 / ri^3
Ti= 1 year
ri= 149.8x10^8 KM
r2= 776.3x10^8 KM
T2= ?
Plugin the information:
T2= SQUARE ROOT of 1^2 x 776.3x10^8 ^3 / 149.8x10^8 ^3 = 11.8 years
ANSWER: 11.8 years
Find the radius of Pluto's orbit.
Pluto's period = 248 years
Earth's period = 1 year
Earth's radius = 1.50x10^11m
We use the equation T1^2 / T2^2 = r1^3 / r2^3
We must rearrange the equation to solve for r2.
We get r2 =. CUBED ROOT of T2^2 x r1^3 / T1^2
We plug in information:
T2= 248 years
r1= 1.50x10^11 m
T1= 1 year
r2 = CUBED ROOT of 248^2 x 1.50x10^11 ^3 / 1^2 = 5.92x10^12m
ANSWER: 5.92x10^12 meters
What equation do we use to find the period of a satellite?
T = 2n SQUARE ROOT of r^3 / Gm
What equation do we use to find the velocity of a satellite?
V = SQUARE ROOT of Gm / r
A satellite needs to achieve an orbit of 250km above the earth's surface. How long would it take and how fast would it need to go?
Earth's radius is 6.38x10^6m
Earth's mass is 5.97x10^24kg
G= 6.67x10^-11nm^2/kg^2
First we find the period by using the equation
T = 2n SQUARE ROOT of r^3 / Gm
Then, we need to convert 250 KM to M. We get 250,000m or 2.50x10^5m.
To find the radius, we take our 2.50x10^5m and add to it the Earth's radius which is 6.38x10^6m.
We then plug in the information:
T = 2n SQUARE ROOT of (2.50x10^5 + 6.38x10^6)^3 / (6.67x10^-11)(5.97x10^24) = 5400 seconds
Then, we use the equation to find velocity.
V = SQUARE ROOT of Gm / r
We plug in our information:
V = SQUARE ROOT of (6.67x10^-11)(5.97x10^24) / (2.50x10^5 + 6.38x10^6) = 7750m/s
ANSWER:
T= 5400 seconds
V= 7750 m/s
What is the period and velocity of the International Space Station that is orbiting 250 miles above the earth's surface?
Mass of the Earth is 5.97x10^24kg
Radius of the Earth is 6.38x10^6m
G= 6.67x10^-11nm^2/kg^2
First, we find the period by using the equation
T = 2n SQUARE ROOT of r^3 / Gm
We need to convert 250 miles to m (250mi x 1609m / 1mi = 4.02x10^5m)
Next, we must add the 4.02x10^5m to the earth's radius of 6.38x10^6m. We now have the radius.
Plugin the information
T = 2n SQUARE ROOT of (4.02x10^5 + 6.38x10^6)^3 / (6.67x10^-11)(5.97x10^24) = 5600 seconds
Now, we use the equation for velocity.
V = SQUARE ROOT of Gm / r
We plugin the information:
V = SQUARE ROOT of (6.67x10^-11)(5.97x10^24) / 4.02x10^5 + 6.38x10^6 = 7660 m/s
ANSWER:
T= 5600 seconds
V= 7660 m/s
How fast would the International Space Station have to go to achieve an orbit of 300 miles?
Earth's mass= 5.97x10^24kg
Earth's radius= 6.38x10^6m
G= 6.67x10^-11nm^2/kg^2
First find the radius.
Convert 300 miles to meters and get 482700m.
Add the 482700m to Earth's radius of 6.38x10^6.
482700 + 6.38x10^6 = 6862700m
Now plug in the information to find the velocity.
V = SQUARE ROOT of Gm / r
G= 6.67x10^-11
m= 5.97x10^24
r= 6862700m
V = SQUARE ROOT of 6.67x10^-11 x 5.97x10^24 / 6862700 = 7617 m/s
ANSWER: 7617 m/s
A satellite needs to obtain an orbit of 150 miles above the surface of Mars. What velocity needs to be obtained? How long would it take?
Mar's radius= 3398km
Mar's mass= 6.4x10^23kg
G= 6.67x10^-11nm^2/kg^2
First, find the actual radius.
To do this, we take the 150mi (convert to meters- 2.41x10^5m) and add it to the radius of Mars (which we also must convert to become 3.398x10^6m)
r = 2.41x10^5 + 3.398x10^6 = 3.639x10^6m
Now that we have the radius, we can plug in the other information to find the period-
T = 2n SQUARE ROOT of r^3 / Gm
r= 3.639x10^6m
m= 6.4x10^23kg
T = 2n SQUARE ROOT of 3.639x10^6 ^3 / 6.67x10^-11 x 6.4x10^23 = 6675 seconds
PERIOD= 6675 seconds
Now, find the velocity- V = SQUARE ROOT of Gm / r
G= 6.67x10^-11nm^2/kg^2
m= 6.4x10^23kg
r= 3.639x10^6m
V = SQUARE ROOT of 6.67x10^-11 x 6.4x10^23 /
3.639x10^6 = 3425m/s
VELOCITY= 3425 m/s
What would be the IMA of a ramp with a height of 5m and a distance of 24m?
IMA = da / h
da= 24m
h= 5m
IMA = 24 / 5 = 4.8
ANSWER: 4.8
What would be the AMA of a ramp with a 35N object being pushed up by a force of 15N?
AMA = Fw / Fa
Fw= 35N
Fa= 15N
AMA= 35 / 15 = 2.3
AMA= 2.3
What would be the AMA of a ramp with a 325N object being pushed up by a force of 20N?
AMA = Fw / Fa
Fw= 325N
Fa= 20N
AMA = 325 / 20 = 16.3
AMA= 16.3
Determine the efficiency of a ramp with a height of 2m, a distance of 12m with a 783N object being pushed up with a force of 150N.
E = AMA / IMA x 100%
AMA = Fw / Fa
IMA = da / h
First we need to find the AMA.
Fw= 783N
Fa= 150N
AMA= 783 / 150 = 5.2
Now, find the IMA.
da= 12m
h= 2m
IMA= 12 / 2 = 6
Now that we have both AMA and IMA, we can find the efficiency.
E = AMA/IMA x 100
AMA = 5.2
IMA = 6
E = 5.2 / 6 x 100 = 86.7%
ANSWER: 86.7%
Find the IMA of a lever with a resistance distance of 12m and an effort distance of 37m.
IMA = de / dr
de= 37m
dr= 12m
IMA = 37 / 12 = 3.08
ANSWER: 3.08
Find the IMA of a lever with a resistance distance of 30m and an effort distance of 24m.
IMA = de / dr
de= 24m
dr= 30m
IMA = 24 / 30 = 0.8
ANSWER: 0.8
Find the IMA of a lever with a resistance distance of 7m and an effort distance of 15m.
IMA = de / dr
de= 15m
dr= 7m
IMA = 15 / 7 = 2.14
ANSWER: 2.14
Find the AMA of a lever with a resistance force of 327N and an effort force of 45N.
AMA = Fr / Fe
Fr= 327N
Fe= 45N
AMA = 327 / 45 = 7.27
ANSWER= 7.27
Find the AMA of a lever with a resistance force of 597N and an effort force of 75N.
AMA = Fr / Fe
Fr= 597N
Fe= 75N
AMA = 597 / 75 = 7.96
ANSWER= 7.96
Determine the efficiency for a lever with a resistance distance of 6.3m, an effort distance of 5.8m, a resistance force of 45N, and an effort force of 57.7N.
E = AMA / IMA x 100%
IMA = De / Dr
AMA = Fr / Fe
First, find the IMA.
De= 5.8m
Dr= 6.3m
IMA = 5.8 / 6.3 = 0.92
Now, find the AMA.
Fr= 45N
Fe= 57.7N
AMA = 45 / 57.7 = 0.78
Finally, find the efficiency.
AMA= 0.78
IMA= 0.92
E = 0.78 / 0.92 x 100 = 84.8%
ANSWER: 84.8%
What is the IMA of a pulley system with a 10N weight connected to a supporting rope with one side hanging over the pulley?
1
What is the IMA of a pulley system with a 10N weight connected to two supporting ropes with the second rope hooking under the pulley?
2
What is the IMA of a pulley system with 6 ropes, one of them being over the top of the final pulley?
5
If a pulley system is lifting 100N, how much weight does each rope handle if there are 5 supporting ropes?
Fw / IMA = N
Fw= 100N
IMA= 5
100 / 5 = 20 N
ANSWER= 20 N
If a pulley system is lifting 300N, how much force does each rope handle if there are 2 supporting ropes?
Fw / IMA = N
Fw= 300N
IMA= 2
300 / 2 = 150N
ANSWER= 150 N
What is the IMA of a pulley system with 5 ropes, one of them being under the final pulley and being lifted up?
5
Determine the efficiency of a pulley system lifting 100N with 5 supporting ropes. The force used to pull the system is 22N.
E = AMA / IMA x 100%
AMA = Fw / Fa
IMA = Supporting ropes
Find AMA.
Fw= 100N
Fa= 22N
AMA = 100 / 22 = 4.55
Find IMA.
IMA = 5
Find efficiency.
AMA= 4.55
IMA= 5
E = 4.55 / 5 x 100 = 91%
ANSWER: 91%
Determine the efficiency of a pulley system lifting 100N with 5 supporting ropes if the force used to pull the system is 27N.
E = AMA / IMA x 100%
AMA = Fw / Fa
IMA = Supporting ropes
Find AMA.
Fw= 100N
Fa= 27N
AMA= 100 / 27 = 3.7
Find IMA.
IMA = 5
Find efficiency.
AMA= 3.7
IMA= 5
E = 3.7 / 5 x 100 = 74%
ANSWER= 74%
What is the IMA of a wheel with a diameter of 12cm and an axle with a radius of 1.5cm?
IMA = Rw / Ra
Rw= 1/2 of 12cm = 6cm
Ra= 1.5cm
IMA= 6 / 1.5 = 4
ANSWER: 4
What is the IMA of a wheel with a radius of 13cm and an axle with a radius of 3cm?
IMA = Rw / Ra
Rw= 13cm
Ra= 3cm
IMA = 13 / 3 = 4.3
ANSWER= 4.3
What are the six simple machines?
lever, inclined plane, pulley, wedge, screw, wheel and axle
All simple machines are variations of which two main machines?
levers and inclined planes
Which machines are related to the inclined plane?
wedge and screw
Which machines are related to the lever?
pulley and wheel and axle
Why do we use simple machines?
Make work easier
Do simple machines increase the amount of work we put in?
No. You cannot get more work out than you put in
Why can't you get more work out than you put in?
friction
How do simple machines make work easier?
lessens the required force (multiplies input force)
changes the direction of the force
increases the speed through which a force acts
Which lever-class changes the direction of force?
first-class
Which lever-class has the effort, fulcrum, and then resistance?
first-class
What is an example of a first-class lever?
seesaw
Describe a first-class lever.
effort, fulcrum, resistance
changes the direction of force
ex: teeter-totter
Which lever-class lessens the required force?
second-class
Which lever-class has the fulcrum, resistance, and then effort?
second-class
What is an example of a second-class lever?
wheelbarrow
Describe second-class levers.
fulcrum, resistance, effort
lessens the required force
ex: wheelbarrow
Which lever-class increased the speed?
third-class
Which lever-class has the resistance, effort, and then fulcrum?
third-class
What is an example of a third-class lever?
biceps, golf-club, baseball bat, broom
Describe third-class levers.
resistance, effort, fulcrum
increases the speed
ex: biceps, baseball bat
What is sacrificed when a machine multiplies the input force?
distance
What is the formula for work output?
w=fd
Wo=Fw x h
What is the formula for work input?
WI= Fa x da
What is mechanical advantage?
how many times a simple machine multiplies the force
What is "how many times a simple machine multiplies the force" called?
mechanical advantage
Which type of mechanical advantage works with forces?
actual mechanical advantage
Which type of mechanical advantage works with distance?
ideal mechanical advantage
Which type of mechanical advantage involves friction?
actual MA
Which type of mechanical advantage does NOT involve friction?
ideal MA
What equation do we use to find AMA?
AMA = Fw / Fa
What equation do we use to find IMA?
IMA = da / h
Describe AMA.
works with forces
involves friction
AMA =. Fw / Fa
Describe IMA.
works with distance
frictionless
IMA = da / h
What is the equation we use to find the efficiency of a simple machine?
AMA / IMA x 100%
Wo / Wi x 100%
What do we call a machine that has more work output than input?
perpetual motion machine
True or False: Simple machines can be 100% efficient.
false
There is an inclined plane that is 1.5m high and 5.5m long. An 800N box is pushed up by a force of 250N. What is the ideal and actual mechanical advantage and efficiency?
E = AMA / IMA x 100%
AMA = Fw / Fa
IMA = d / h
First find IMA.
d= 5.5m
h= 1.5m
IMA= 5.5 / 1.5 = 3.7
IMA= 3.7
Now find AMA.
Fw= 800N
Fa= 250N
AMA= 800 / 250 = 3.2
AMA= 3.2
Now, efficiency.
AMA= 3.2
IMA= 3.7
E = 3.2 / 3.7 x 100 = 86.5%
ANSWER= 86.5%
There is an incline plane that is 1.25m heigh and 3.65m long. An 810N box is pushed up the plane by a force of 324N. What is the work output, work input, ideal mechanical advantage, actual mechanical advantage, and efficiency?
First find the work input- Wi = Fa x d
Fa= 324 N
d= 3.65m
Wi = (324)(3.65) = 1180nm
WORK OUTPUT: 1180 nm
Now, find the work output- Wo = Fw x h
Fw= 810N
h= 1.25m
Wo = (810)(1.25) = 1010nm
WORK OUTPUT: 1010 nm
Now, find the ideal mechanical advantage-
IMA = d / h
d= 3.65m
h= 1.25m
IMA = 3.65 / 1.25 = 2.92
IMA= 2.92
Find the actual mechanical advantage-
AMA = Fw / Fa
Fw= 810N
Fa= 324N
AMA = 810 / 324 = 2.5
AMA = 2.5
Next, find the efficiency- E = AMA / IMA x 100%
AMA= 2.5
IMA= 2.92
E = 2.5 / 2.92 x 100% = 85.6%
EFFICIENCY = 85.6%
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