Recognize the need for a centripetal force when rounding a curve
Predict the effect of an inadequate centripetal force
Relate speed to centripetal force
What Do You See/What Do You Think November 14th
Screen_shot_2011-11-15_at_9.07.13_PM.png
Screen_shot_2011-11-15_at_9.07.13_PM.png
I see a car driving on top of a winding cliff. The car is going fast so its turns are sharp resulting in the car almost falling off.Why is the sign indicating to slow down?It says to slow down because driving on a sharp turn to quickly could result in losing control of the vehicle and falling off the cliff.How is the amount you should slow down determined?The amount is determined by the size of your car and how sharp the turn is.
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InvestigationNovember 15th
Part 1: Can a car go faster around a wide turn or sharp turn?
Hypothesis: Answer the question posed in the title of Part 1 and give a reason for this hypothesis
Procedure:
1. Choose two radii to place your circular mass on, one towards the center of your Lazy Susan and one towards the edge. The radius towards the center will simulate the car on a sharp turn while the radius towards the edge will simulate the car on a wide turn.
Sharp Turn: 4.5 cm
Wide Turn: 24.5 cm
2. Spin the Lazy Susan with the mass at the inner radius for 10 revolutions then spin the Lazy Susan with the mass at the outer radius for 10 revolutions, recording the time each takes to make 10 revolutions.
OKAY!
3. Calculate the time it took the Lazy Susan to make one revolution for each radii.
Sharp Turn: 1.781 s
Wide Turn: 3.049 s
4. Use the formula C=2πr to solve for the circumference of the path that the weight traveled for each radii. Next, Divide the Circumference for each path by the time for the weight to make one revolution for the inner and outer radius. This will tell you the speed of the weight while it traveled around the Lazy Susan.
Screen_shot_2011-11-15_at_8.33.57_AM.png
Screen_shot_2011-11-15_at_8.33.57_AM.png
Sharp Turn:
v = (2)(3.14)(4.5)/1.781
v = 28.26 cm/1.781s
v = 15.88 cm/s
Wide Turn:
v = (2)(3.14)(14.5)/3.049
v = 91.06 cm/3.049s
v = 29.87 cm/s
5. Create a table to organize your data for Part 1.
Radius
Max Speed
Sharp
4.5 cm
15.88 cm/s
Wide
14.5 cm
29.87 cm/s
6. Can you achieve a larger maximum speed on a wide turn (large radius) or a sharp turn (small radius)? How does your data show this?
You can achieve a larger maximum speed on a larger radius, or wide turn. Our data shows this because the wide turn has a larger maximum speed by 13.98 cm/s.
Part 2: Can a Car Go Faster Turning on an Icy Surface or Dry Surface?
Hypothesis: Answer the Question posed in the Title of Part 2 and give a reason for this hypothesis.
Procedure:
1. Spin the Lazy Susan with the mass on the Wood Surface for 10 revolutions then spin the Lazy Susan with the mass at the same radius on the sandpaper for 10 revolutions, recording the time each takes to make 10 revolutions.
Wood Surface: 3.13 s
Sandpaper Surface: 1.58 s
Radius: 14.5 cm
2. Calculate the time it took the Lazy Susan to make one revolution for each surface
Wood Surface:
v = (2)(3.14)(14.5)/3.13
v = 91.06 cm/3.13s
v = 29.1 cm/s
Sandpaper Surface:
v = (2)(3.14)(14.5)/1.58
v = 91.06 cm/1.58s
v = 57.63 cm/s
3. Use the formula C=2πr to solve for the circumference of the path that the weight traveled for each radii. Next, Divide the Circumference for each path by the time for the weight to make one revolution for the inner and outer radius. This will tell you the speed of the weight while it traveled around the Lazy Susan.
Screen_shot_2011-11-16_at_8.51.42_AM.png
Screen_shot_2011-11-16_at_8.51.42_AM.png
4. Create a table similar to below to organize your data for Part 2
Radius
Max. Speed
Wood
14.5 s
29.1 cm/s
Sandpaper
14.5 s
57.63 cm/s
5. Can you achieve a larger maximum speed on sandpaper (normal asphalt) or on the wood surface (slippery road conditions)? How does your data show this?
You can go faster on normal asphalt than on slippery road conditions. The difference between the two maximum speeds is 28.53.
Part 3: Can a more massive vehicle go faster turning or a less massive vehicle go faster turning?
Hypothesis: Answer the Question posed in the Title of Part 3 and give a reason for this hypothesis.
Procedure:
1. Spin the Lazy Susan with the lighter mass on the Wood Surface for 10 revolutions, then spin the Lazy Susan with the heavier mass at the same radius on the sandpaper for 10 revolutions, recording the time each takes to make 10 revolutions.
Lighter Mass: 1.64 s
Heavier Mass: 1.84 s
Radius: 4.5 cm
2. Calculate the time it took the Lazy Susan to make one revolution for each mass.
Lighter Mass:
v = (2)(3.14)(4.5)/1.64
v = 28.26 cm/1.64s
v = 17.23 cm/s
Heavier Mass:
v = (2)(3.14)(4.5)/1.84
v = 28.26 cm/1.84s
v = 15.35 cm/s
3. Create a table similar to below to organize your data for Part 2
Radius
Max. Speed
Lighter
4.5 cm
17.23 cm/s
Heavier
4.5 cm
15.35 cm/s
4. Can you achieve a larger maximum speed before skidding out in a heavier vehicle or a lighter vehicle? How does your data show this?
The mass is irrelevant.
Homework:November 15th
Physics Words
Force: a push or a pull Centripetal force: a force directed toward the center to keep an object in a circular path Centripetal acceleration: a change in the direction of the velocity with respect to time
Checking Up Questions 1. What is the direction of the force that keeps an object moving in a circle? The force is always toward the center of the circle. 2. What is the name of a force that keeps an object moving in a circle? The force is call centripetal force. 3. Name the force that keeps an automobile moving in a circular path on a road. The force that keeps an automobile moving is friction between the tires and the road. 4. Explain how the velocity of an object can change even if the speed is not changing. A car can move in a circle at a constant speed, but the velocity changes because it is changing direction. 5. Describe three situations in which acceleration can take place. Acceleration can take place when an object speeds up, slows down, or changes direction. 6. What is the force that keeps the Earth moving in a circle around the Sun? The force that keeps Earth moving around the Sun is gravity.
Centripetal Acceleration: a change in velocity caused not by a change in speed. but just by a change in direction.
Physics To GoNovember 16th
1. C=2πr C=2π(6400) C=40212.39/ 24 V=1675.52km/h 1675.52 km x 1000m = 1675516.082m/3600s V=465.42m/s
2. C=2π(1.5x10^8) / 24h = 39269908.17km/h-
3. C=2π(15) C=94.25 / 60 V= 1.57cm/sec
4. a. The automobile can go faster if the curve is tighter. b. The automobile should go slower if the road surface becomes slippery. c. The automobile should go slower around the curve because the road is slippery and there won't be as much friction to hold the car now.
5. An example is a race car racing on the track. The track is curved to allow the drivers to make sure they don't slide. Gravity is pulling them to always stay in the curved path,
7. The driver turns the steering wheel so the wheels will turn on a curved road. However, it is the road that actually moves the entire automobile and forces it to continue following the wheels. The road has a force (gravity, centripetal force) that is making the car follow the road.
8. a=v^2/r a= 270^2 / 1000 a= 72.9m/s^2
9. The fist explanation is not correct because if you turn to the left, the force will make your body go to the left too. The second explanation is correct because your body wanted to go straight but the force/friction made your body go to the left as well. Also, the passenger door had a centripetal force on it because that was what made it continue in the circular path.
10. Centripetal force acts on the race car to go in a circular path when it is on the track.
11. When you are traveling on a wider turn (gentle turn), you can gain more maximum speed. However, on sharp turns, you need to go slower. Drivers need to realize this and they need to know to slow down on these types of curves to prevent sliding off the road and getting into a dangerous accident.
12. Curve bends right: You would end up into the lane of oncoming traffic because your car would slide on the road and go straight into the other lane. Curve bends left: You would end up into the ditch because, as said before, your car would continue to slide and accelerate off the road.
Active Physics Plus:November 16th
Centripetal Force (1-4)
Screen_shot_2011-11-17_at_8.26.27_AM.png
Centripetal Acceleration:(5-7)
Screen_shot_2011-11-17_at_8.28.25_AM.png
Force: measured in Newtons (N) F=mv^2/r (kg)(m^2/s^2) x (1/m)= N
1). f=13,720 N m= 2000 kg r= 10m v= ? 13720=2000v^2/10 68.6=v^2 V= 8.28 m/s
2). f=6860 N m=2000 kg r= 10m v= ? 6860=2000v^2/10 34.3=v^2 V=5.86m/s
3). f= 20,580 N m=3000 kg r= ? v= 10 m/s 20580=(3000)(10)^2/ r 20580=300000/r r=14.58m
4). v=5m/s m=2200 kg f= 6000 N r=? 6000=2200(5^2)/r r=9.17m
SECTION SEVEN
Learning Outcomes:
What Do You See/What Do You Think
November 14th
InvestigationNovember 15th
Part 1: Can a car go faster around a wide turn or sharp turn?
Hypothesis: Answer the question posed in the title of Part 1 and give a reason for this hypothesis
Procedure:
1. Choose two radii to place your circular mass on, one towards the center of your Lazy Susan and one towards the edge. The radius towards the center will simulate the car on a sharp turn while the radius towards the edge will simulate the car on a wide turn.
Sharp Turn: 4.5 cm
Wide Turn: 24.5 cm
2. Spin the Lazy Susan with the mass at the inner radius for 10 revolutions then spin the Lazy Susan with the mass at the outer radius for 10 revolutions, recording the time each takes to make 10 revolutions.
OKAY!
3. Calculate the time it took the Lazy Susan to make one revolution for each radii.
Sharp Turn: 1.781 s
Wide Turn: 3.049 s
4. Use the formula C=2πr to solve for the circumference of the path that the weight traveled for each radii. Next, Divide the Circumference for each path by the time for the weight to make one revolution for the inner and outer radius. This will tell you the speed of the weight while it traveled around the Lazy Susan.
Sharp Turn:
v = (2)(3.14)(4.5)/1.781
v = 28.26 cm/1.781s
v = 15.88 cm/s
Wide Turn:
v = (2)(3.14)(14.5)/3.049
v = 91.06 cm/3.049s
v = 29.87 cm/s
5. Create a table to organize your data for Part 1.
You can achieve a larger maximum speed on a larger radius, or wide turn. Our data shows this because the wide turn has a larger maximum speed by 13.98 cm/s.
Part 2: Can a Car Go Faster Turning on an Icy Surface or Dry Surface?
Hypothesis: Answer the Question posed in the Title of Part 2 and give a reason for this hypothesis.
Procedure:
1. Spin the Lazy Susan with the mass on the Wood Surface for 10 revolutions then spin the Lazy Susan with the mass at the same radius on the sandpaper for 10 revolutions, recording the time each takes to make 10 revolutions.
Wood Surface: 3.13 s
Sandpaper Surface: 1.58 s
Radius: 14.5 cm
2. Calculate the time it took the Lazy Susan to make one revolution for each surface
Wood Surface:
v = (2)(3.14)(14.5)/3.13
v = 91.06 cm/3.13s
v = 29.1 cm/s
Sandpaper Surface:
v = (2)(3.14)(14.5)/1.58
v = 91.06 cm/1.58s
v = 57.63 cm/s
3. Use the formula C=2πr to solve for the circumference of the path that the weight traveled for each radii. Next, Divide the Circumference for each path by the time for the weight to make one revolution for the inner and outer radius. This will tell you the speed of the weight while it traveled around the Lazy Susan.
4. Create a table similar to below to organize your data for Part 2
5. Can you achieve a larger maximum speed on sandpaper (normal asphalt) or on the wood surface (slippery road conditions)? How does your data show this?
You can go faster on normal asphalt than on slippery road conditions. The difference between the two maximum speeds is 28.53.
Part 3: Can a more massive vehicle go faster turning or a less massive vehicle go faster turning?
Hypothesis: Answer the Question posed in the Title of Part 3 and give a reason for this hypothesis.
Procedure:
1. Spin the Lazy Susan with the lighter mass on the Wood Surface for 10 revolutions, then spin the Lazy Susan with the heavier mass at the same radius on the sandpaper for 10 revolutions, recording the time each takes to make 10 revolutions.
Lighter Mass: 1.64 s
Heavier Mass: 1.84 s
Radius: 4.5 cm
2. Calculate the time it took the Lazy Susan to make one revolution for each mass.
Lighter Mass:
v = (2)(3.14)(4.5)/1.64
v = 28.26 cm/1.64s
v = 17.23 cm/s
Heavier Mass:
v = (2)(3.14)(4.5)/1.84
v = 28.26 cm/1.84s
v = 15.35 cm/s
3. Create a table similar to below to organize your data for Part 2
4. Can you achieve a larger maximum speed before skidding out in a heavier vehicle or a lighter vehicle? How does your data show this?
The mass is irrelevant.
Homework:November 15th
Physics Words
Force: a push or a pull
Centripetal force: a force directed toward the center to keep an object in a circular path
Centripetal acceleration: a change in the direction of the velocity with respect to time
Checking Up Questions
1. What is the direction of the force that keeps an object moving in a circle?
The force is always toward the center of the circle.
2. What is the name of a force that keeps an object moving in a circle?
The force is call centripetal force.
3. Name the force that keeps an automobile moving in a circular path on a road.
The force that keeps an automobile moving is friction between the tires and the road.
4. Explain how the velocity of an object can change even if the speed is not changing.
A car can move in a circle at a constant speed, but the velocity changes because it is changing direction.
5. Describe three situations in which acceleration can take place.
Acceleration can take place when an object speeds up, slows down, or changes direction.
6. What is the force that keeps the Earth moving in a circle around the Sun?
The force that keeps Earth moving around the Sun is gravity.
Centripetal Acceleration: a change in velocity caused not by a change in speed. but just by a change in direction.
Physics To GoNovember 16th
1. C=2πr
C=2π(6400)
C=40212.39/ 24
V=1675.52km/h
1675.52 km x 1000m = 1675516.082m/3600s
V=465.42m/s
2. C=2π(1.5x10^8) / 24h
= 39269908.17km/h-
3. C=2π(15)
C=94.25 / 60
V= 1.57cm/sec
4. a. The automobile can go faster if the curve is tighter.
b. The automobile should go slower if the road surface becomes slippery.
c. The automobile should go slower around the curve because the road is slippery and there won't be as much friction to hold the car now.
5. An example is a race car racing on the track. The track is curved to allow the drivers to make sure they don't slide. Gravity is pulling them to always stay in the curved path,
7. The driver turns the steering wheel so the wheels will turn on a curved road. However, it is the road that actually moves the entire automobile and forces it to continue following the wheels. The road has a force (gravity, centripetal force) that is making the car follow the road.
8. a=v^2/r
a= 270^2 / 1000
a= 72.9m/s^2
9. The fist explanation is not correct because if you turn to the left, the force will make your body go to the left too. The second explanation is correct because your body wanted to go straight but the force/friction made your body go to the left as well. Also, the passenger door had a centripetal force on it because that was what made it continue in the circular path.
10. Centripetal force acts on the race car to go in a circular path when it is on the track.
11. When you are traveling on a wider turn (gentle turn), you can gain more maximum speed. However, on sharp turns, you need to go slower. Drivers need to realize this and they need to know to slow down on these types of curves to prevent sliding off the road and getting into a dangerous accident.
12. Curve bends right: You would end up into the lane of oncoming traffic because your car would slide on the road and go straight into the other lane. Curve bends left: You would end up into the ditch because, as said before, your car would continue to slide and accelerate off the road.
Active Physics Plus: November 16th
Centripetal Force (1-4)
Centripetal Acceleration:(5-7)
Force: measured in Newtons (N)
F=mv^2/r
(kg)(m^2/s^2) x (1/m)= N
1).
f=13,720 N
m= 2000 kg
r= 10m
v= ?
13720=2000v^2/10
68.6=v^2
V= 8.28 m/s
2).
f=6860 N
m=2000 kg
r= 10m
v= ?
6860=2000v^2/10
34.3=v^2
V=5.86m/s
3).
f= 20,580 N
m=3000 kg
r= ?
v= 10 m/s
20580=(3000)(10)^2/ r
20580=300000/r
r=14.58m
4).
v=5m/s
m=2200 kg
f= 6000 N
r=?
6000=2200(5^2)/r
r=9.17m
5).
a=?
v=10m/s
r=12
a=v^2/r
a=10^2/12
a= 8.33m/s^2
6).
a= ?
v= 20m/s
r= 12
a=v^2/r
a=20^2/12
a= 33.3m/s^2
7).
r= 24
v=10m/s
a=?
a=10^2/24
a= 4.17m/s^2