Download Unit 4 Objectives: Circular Motion Standard: SP1. Students will

Unit 4 Objectives: Circular Motion
Standard: SP1. Students will analyze the relationships between force, mass, gravity, and the motion of objects.
Category One: Measure and calculate centripetal force.
1. What’s the difference between tangential speed and rotational speed?tangential speed is the circularl
distance covered per unit of time and Rotational speed is the number of times something will rotate
per minute.
2. The circular distance covered per unit of time is called tangential speed.
3. What is the number of rotations per unit of time called? RPM
4. What is the unit that rotational speed is commonly expressed in? RPM
5. What happens to the rotational speed of an object as it moves away from the axis of rotation toward
the outer edge of a rotating system? Rotational speed stays constant no matter where you are located
in a rotating system.
6. Is the tangential speed greater or smaller on the outer edge of a rotating object than it is closer to the
axis? Explain. Greater because the object must move faster in order to cover a greater distance in the
same amount of time.
7. What happens to the tangential speed of an object as it moves away from the axis of rotation toward
the outer edge of a rotating system? The tangential speed becomes greater as it moves toward the
outer edge of a rotating system.
8. A 30-m tree is in the process of being cut down by a lumberjack. A squirrel is in her nest about
halfway up the tree. Another squirrel is at the very top of the tree. When the tree hits the
ground, what will be different about the motion of the squirrels? What will be the
same? The
squirrel at the very top will be moving faster than the one half-way up will be moving slower. They will
both have the same rotational speed.
9. Does the Earth rotate or revolve? Explain. It rotates on an internal axis- which is what something does
when it rotates. It also revolves around the sun- the sun acts like an external axis that it goes around.
10. What formula is used to determine the velocity of an object that is moving in a circle?
V = 2πr/t
11. What is the velocity of a merry-go-round that has a radius of 4 meters and takes 4 seconds to complete
one rotation? V =2(3.14)(4)/4 = 6.28 m/s
12. Contrast centripetal and centrifugal force. Which one is considered a “fictitious” force and why?
Centripetal
Centrifugal
A center seeking force
Definition?
A center fleeing force
Can calculate this force w/
Calculation?
Cannot calculate – it’s not a
equation
real force
Causes the centrifugal force
Cause/Effect?
Is the effect of the centripetal
force
13. When riding in a car around a curve, identify the centripetal and centrifugal forces. Centripetal ForceFor a level curve, the centripetal force will be supplied by the friction force between the tires and roadway.
Centrifugal Force – is what you feel (as if being pushed outward) when going around the curve.
14. An object attached to the end of a rope is swung around in a circular motion. How would shortening
the rope affect the motion (velocity) of the object?
 The object’s velocity would be slower b/c it wouldn’t need to go as fast to get around the circle
b/c the radius is smaller.
 For example – Use the 1’s rule to show this v = 2πr/t If you assume that the radius of the
object is 1 m to begin with then the new radius will be 0.5 m.
 Keep all of the other information in the equation constant by filling in their values with Solve
for v with a full length rope. v=2(π)(1)/1 = 6.28 m/s
 This is the speed of the object before shortening the rope.
 Solve for velocity with a shortened rope (half the length). V = 2π(.5)/1 = 3.14. See, the shorter
rope has a smaller speed!
How would increasing the mass of the object affect the object’s circular motion (velocity)?
 Use the 1’s rule again keeping everything constant except mass this time.
 Fc = mv2/r Let’s say that the mass is originally 1 kg. Use the number 1 for all of the other values.
 Solve for velocity- 1 = (1)(v)2/1 V = 1 m/s
 Now double the mass 1 =(2)(v)2/1
 1 = 2v2
 ½ =v2
 Sq. root ½ = v
 0.707 m/s = v
How would increasing the force applied to the swing affect the object’s motion (velocity)?
 Use this equation to do the 1’s rule. Fc = mv2/r
 If the force is originally 1 N, then then the equation would look like this using the 1’s rule :
 1 = 1(v)2/1 1 m/s = v Now increase the force by 2.
 Fc = mv2/r 2 = 1(v)2/1
 2 = v2 Now take the square root of 2 and that is V. V= 1.41 m/s
 So if you increase the force then the object’s velocity increased from 1 m/s to 1.41 m/s.
15. A 3-kg ball attached to a 1.5m rope is spun in a circle. It took 2 seconds for the ball to complete the
circle. What is the tangential speed of the ball? What is the ball’s acceleration? What is the
centripetal force of the string on the ball?
 V = 2πr/t v = 2π(1.5)/2 = 4.71 m/s
Category Two: Determine the conditions required to maintain a body in a state of static equilibrium
16. Write the formula for calculating Torque. T = Fd
Write the formula for Newton’s 2nd law of motion. F = ma
What is similar about each formula? What is different about each formula?
Torque
Unit is the N x m
Causes Rotation
Force is always perpendicular to
the motion
Similarities
Both are Vector Quantities
Both involve force
Force
Unit is the Newton
Causes Acceleration
Force can be in any direction to
the motion
17. When describing torque, what is a fulcrum? The center or the support base underneath something
18. What is the torque on a board when a 300-N man stands 3 meters from the fulcrum? T = Fd
T = 300(3) = 900 Nm
19. Describe the net torque of a system that is in static equilibrium. Net torque is always zero when a
system is in static equilibrium.
20. A 200 N weight is place on a meter stick 40 cm from the fulcrum. How far from the fulcrum should a
250-N weight be placed to create static equilibrium? F x d = F x d (200)(40) = 250 (d) 8000 = 250d
d= 8000/250 d = 32 cm from the pivot.
21. A 600N child sits on a seesaw 1 meter from the fulcrum. A 400N child sits on the other side of the
seesaw 2 meters from the fulcrum. What is the net torque? F x d = F x d (600)(1) = 400(2)
600 = 800
To get the net torque subtract the value on the right from the left. 800-600 = 200 Nm
22. Jim and Joan sit on a seesaw. If Jim is twice as massive as Joan, where should they each sit to create
equilibrium? If they were at equilibrium and Joan leaned forward, what would happen?
F x d = F x d JimBIG (distance small) = Joansmall (distance big) Jim should sit twice as close to the
pivot as Joan since he is twice her size.
Vocabulary:
1. Tangential speed
2. Rotational speed
3. Rotation
4. Revolution
5. Centripetal force
6. Centrifugal force
7. Centripetal acceleration
8. Radius
9. Torque
10. Fulcrum
11. Equilibrium
12. Lever arm