1. Circular Motion; Gravitation
Chapter 7
Circular Motion; Gravitation
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2. Centripetal Acceleration
Centripetal means “Center Seeking” and the centripetal force on an object moving in a circle always acts towards the center of the circle.
By definition: a = V2/R where V = linear or tangential speed, and R is the radius of the circle.
Try p. 236 Practice A
2 & 4
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3. Herriman High Honors Physics
Centripetal Force
Centripetal Acceleration results from a Centripetal Force since accelerations are always caused by outside forces
Since F = ma and centripetal acceleration is ac = V2/R then it follows that Centripetal Force, Fc = mV2/R
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4. Herriman High Honors Physics
Centripetal Force
According to Newton’s 1st Law. An object in motion tends to stay in motion. It has been shown that this is straight line motion.
Things want to move in straight lines, so when something is moving in a circle there must be a force causing it to stay in the circle and constantly changing the direction of its motion.
Examples
When car is rounding a curve, friction between the tires and the road which is providing (not opposing) the centripetal force.
A satellite in orbit is held their by gravity (i.e. gravity is providing the centripetal force.
The object actually wants to travel in a straight line, tangent to the circular path, but the provided centripetal force keeps it in a circle.
Try p. 238
Practice B
1, & 3
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5. A Ball Revolving in a Vertical Circle
At the top of the circle
the tension on the string is in the same direction as the weight
together they equal the Net Force which provides the centripetal force
Fta + mg = mV2/R
Minimum velocity to stay in a circle, minimum tension
At the bottom of the circle
the tension on the string is in the opposite direction as the weight
Fta - mg = mV2/R
Maximum Velocity, maximum tension
Fta mg
Ftb mg
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6. Special Case Rounding a CurveFN
Fc = Ff mg
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7. Special Case Rounding a Banked Curve
Fn sin θ θ
Fn cos θ Fn
Fc= Ff Fw θ
The red arrows are the Vertical and Horizontal components of FN. The vertical component is FN cos  and the horizontal component is FN sin .
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8. Universal Gravitation
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9. Universal Gravitation
So F = Gm1m2/R2 and F = ma
Attraction between earth and another object
mpg = Gmpme/R2
Since mp cancels on both sides you get:
g = Gme/R2
So this tells us that gravity caused by any object is the ratio of the mass to the distance between the centers of the objects multiplied by the universal gravitation constant
Try p. 242
Practice C
1, & 3
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10. Special Case: Satellites
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11. Herriman High Honors Physics
Kepler’s Laws
Kepler studied planetary motion. He found:
All planetary bodies move in elliptical orbits
Planetary objects sweep equal areas in equal amounts of time
If two satellites revolve around the same object they are related by the equation:
(t1/t2)2 = (r1/r2)3
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12. Herriman High Honors Physics
Kepler’s Laws
Try p. 251
Practice D
1 & 2
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13. Torque Torque = Fr = force x radius
Torque is measured in newton•meters which means that it has the same units as work in a linear system.
Torque is positive if it causes counterclockwise rotation and negative if it causes clockwise rotation.
Try p. 258
Practice E
1 & 3
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14. Herriman High Honors Physics
Simple Machines
The six simple machines are the screw, the wedge, the inclined plane, the lever, the pulley, and the wheel & axel.
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15. Herriman High Honors Physics
Simple Machines
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16. Herriman High Honors Physics
Rotational Motion
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17. Herriman High Honors Physics
Important Variable
θ (theta) – angular displacement – radians
ω (omega) – angular velocity – radians/sec
α (alpha) = angular acceleration – radians/sec2
t – still time and still in seconds
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18. Measuring Angular Displacement
θ = L/r
360° = 2π radians
= 1 revolution
v= ωr =
Linear velocity =
angular velocity x radius
a = αr
Linear acceleration =
angular acceleration x radius θ r L
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19. Herriman High Honors Physics
Sample Problems
If an exploding fireworks shell makes a 10° angle in the sky and you know it is 2000 meters above your head, how many meters wide is the arc of the explosion?
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20. Herriman High Honors Physics
Solution
θ = L/r
θ = 10° x π rad/180° = 0.17 rad
0.17 rad = L/2000 meters
L = 2000 meters x 0.17 rad = 349 meters
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21. Herriman High Honors Physics
Sample Problems
Convert the following measures from Radians to degrees:
3.14 rad
150 rad
24 rad
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22. Herriman High Honors Physics
Solution
3.14 rad x 180°/π rad = 180°
150 rad x 180°/π rad = 8594°
24 rad x 180°/π rad = 1375°
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23. Herriman High Honors Physics
Sample Problems
What is the linear speed of a child seated 1.2 meters from the center of a merry-go-round if the ride makes one revolution in 4 seconds?
What is the child’s acceleration?
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24. Herriman High Honors Physics
Solution
a) v = ωr and ω=2π rad/4 sec = 1.6 rad/s
since r = 1.2 m then
v = 1.58 rad/s x 1.2 m = 1.9 m/s
b) Since the linear velocity is not changing there is no linear or tangential acceleration, but the child is moving in a circle so there is centripetal or radial acceleration which you will recall fits the equation:
a c = v2/r and since v = ωr ac = ω2r
so a c = (1.9 m/s)2/1.2 m = 3.0 m/s2 or
ac = ω2r = (1.6 rad/s)2(1.2 m) = 3.0 m/s2
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25. Angular Kinematic Equations
Linear Angular
V = v0+at ω=ω0+ αt
x = v0t+½at2 θ=ω0t+½ αt2
V2 = v02+2ax ω2=ω02+2αθ
Vavg = (v0+vf)/2 ωavg=(ω0+ωf)/2
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26. Herriman High Honors Physics
Sample Problem
A angular velocity of wheel changes from 10 rad/s to 30 rad/s in 5 seconds.
What is its angular acceleration?
What is its angular displacement while it is accelerating?
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27. Herriman High Honors Physics
Solution
α = (ωf – ω0)/t =
=(30 rad/s – 10 rad/s)/5 sec = 4 rad/s2
θ = ω0t = ½αt2
= 10 rad/s + ½(4 rad/s2)(5 sec)2
= 100 rad
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28. Rotational Inertia
Remember that linear inertia is a property of mass. The rotational inertia of an object depends upon its shape and its mass.
The general equation for rotational inertia is _ MR2
The specific equations for each shape are given on page 905 in the text
Torque causes rotation. We previously defined T = Fr but can also be defined as T= Iα or rotational inertia x angular acceleration
Hence Fr = Iα
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29. Herriman High Honors Physics
Sample Problem
A force of 10 newtons is applied to the edge of a bicycle wheel (a thin ring mass 1 kg and radius of 0.5 meters). What is the resulting angular acceleration of the wheel?
If the wheel was at rest when the force was applied and the force is applied for 0.4 seconds what is the angular velocity of the wheel immediately after it is applied?
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30. Herriman High Honors Physics
Solution
Since Fr = Iα then α = Fr/I and since the wheel is a thin ring:
I = mr2 = (1 kg)(0.5 m)2 = 0.25 kg m2
So α = (10 N)(0.5 m)/0.25 kg m2
= 20 rad/s2
ωf = ω0 + αt
= 0 rad/s + (20 rad/s2)(0.4 sec)
= 8 rad/s
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