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Chapter 7 and 8 notes
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Chapter 7 Rotational Motion and The Law of Gravitation
1. Rotational Motion – motion of a body that spins about an axis
a. Axis of rotation – line about which rotation occurs
i. Perpendicular to motion
ii. Through center of motion
b. Linear equations will not work b/c direction of rotational motion is constantly changing.
2. Analyzing circular motion
a. describe through angles of motion
b. all points on rotating body move through same angle
c. set up a fixed reference line
i. r = distance from center
ii. Θ = angle from reference line
iii. s = arc length – distance moved along the circumference of circle.
3. Measuring Angles
a. Use radians – angle whose arc length (s) is equal to the radius (r)
Θ=s
r
i. Pure number therefore, use (rad) as label
ii. 1 revolution (s) = 2πr
iii. One complete revolution = 360o
Θ = s = 2πr = 2 π rad
r
r
b. To convert angle in degrees to radians
i. Θ (rad) = π
Θ (deg.)
180o
c. Practice converting the following angles to radians.
i. 25o
ii. 35o
iii. 128o
iv. 270o
4. Angular Displacement – angle through which a point, line, or body is rotated in a specific direction and
about a specific axis.
a. ∆Θ (rad) = ∆s
r
i. (+) ∆s = counterclockwise rotation
ii. (-) ∆s = clockwise rotation
b. Practice Problems
i. While riding a carousel that is rotating clockwise, a child travels through an arc length of
11.5 m. If the child’s angular displacement is 165o, what is the radius of the carousel?
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ii. Earth has an equatorial radius of approximately 6380 km and rotates 360o every 24 h.
1. What is the angular displacement (in degrees) or a person standing at the equator
for 1.0 h?
5. Angular Speed – rate that a body rotates around an axis
a. ω = greek letter omega
b. ωavg = ∆Θ (rad)
∆t
c. SI = rad / s
d. 1 revolution = 2π rad
6. Angular Acceleration – rate of change of ω
a. α = Greek letter alpha
b. αavg = ω2 – ω1 = ∆ω
or
(a/d) → tangential acceleration (m/s2) / distance (m)
t2 – t1
∆t
c. SI = rad/s2
d. All points on a rigid body have the same ω and α
i. Pg 251 has a table that compares rotational and linear kinematic equations
e. Practice Problems
i. A car’s tire rotates at an initial angular speed of 21.5 rad/s. The driver accelerates, and
after 3.5 s the tire’s angular speed is 28.0 rad/s. What is the tire’s average angular
acceleration during the 3.5 s time interval?
ii. A top that is spinning at 15 rev/s spinning for 55 s before coming to a stop. What is the
average angular acceleration of the top while it is slowing?
7. Newton’s Law of Gravitation
a. Gravitational Force – attractive force between two objects
i. Increased mass = increased Fg
ii. Increased distance between = decreased Fg
iii. Fg = G (m1)(m2)
r2
iv. G =Constant of Universal Gravitation = 6.67 x 10-11 N • m2
kg2
b. Calculate Fg of an object on the surface of a spherical object
i. Fg = G (ME)(m2)
2
(RE)2
*Remember Fg = (m) (g)
c. Practice Problems
i. Find the distance between a 0.300 kg billiard ball and a 0.400 kg billiard ball if the
magnitude of the Fg is 8.92 x 10-11 N.
ii. Find the Fg exerted on the moon (mass = 7.36 x 1022 kg) by Earth (mass = 5.98 x 1024 kg)
when the distance between them is 3.84 x 108 m.
Chapter 8 Rotational Equilibrium and Dynamics
8. Torque – the ability of a force to rotate an object around some axis
a. Net torque produces rotation
b. Occurs around an axis of rotation – usually a hinge.
i. Imaginary line passing through a hinge
9. Torque depends on force and lever arm
a. Ease of rotation depends on:
i. How much force is applied
ii. Where the force is applied
1. Farther from the axis of rotation the easier the rotation
2. More torque produced
b. Lever arm – perpendicular distance from axis of rotation to a line drawn along the direction of the
force
i. Distance of lever arm = d
ii. Shorter lever arm = smaller torque
10. Torque depends on the angle between force and the lever arm
a. Changing the angle will change the ease that the object will move with
b. Torque = () = Greek letter Tau
c. SI unit = N • m
d. Torque is (+) or (-) depending on direction of the rotation
i. (+) When rotation is counterclockwise
ii. (-) When rotation is clockwise
e. When more than 1 force causes rotation then net torque is the sum of all torques involved.
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Section 8-2 Center of Mass
11. Center of Mass – point at which all of the mass of the body can be considered to be concentrated when
analyzing transitional motion.
a. Regular shaped objects (i.e. sphere, cube) center of mass is at the geometric center of the object.
i. Different for oddly shaped objects
ii. Average position of an object’s mass
b. Center gravity – an average position at which the gravitational force of the object acts.
i. In this book Center of Mass and Center of Gravity are equivalent.
12. Moment of Inertia
a. Mini Lab: Moment of inertia of a rod
i. Pg 285
ii. Calculate “I” for rod in each position
1. Why is it easier to rotate the rod around some axis and not others?
2. The moment of inertia changes as hand position changes.
b. Moment of Inertia –measure of the resistance of an object to changes in rotational motion
i. (I)
ii. Indicates how much an object’s rotation will change with a force
iii. Depends on:
1. Object’s mass
2. Distribution of mass around the axis of rotation
iv. Farther the mass is from the axis of rotation
1. More difficult to rotate
2. Greater moment of inertia
v. Closer mass is to axis of rotation
1. Less difficult to rotate
2. Smaller moment of inertia
13. Calculate Moment of Inertia
a. When net torque acts on an object, the resulting change in rotational motion depends on the
object’s moment of inertia
b. Table 8-1 helps you calculate the moment of inertia for common shapes
c. SI units
i. Kg • m2
d. Only the distance of a mass from the axis of rotation is important in determining the moment of
inertia for a shape
14. Rotational equilibrium
a. If 2 forces acting on an object are equal in magnitude and opposite in direction, the object will
rotate in place.
b. Translational equilibrium – net force on an object is equal to zero
c. Rotational equilibrium – net torque on an object is equal to zero
15. Rotational Dynamics
a. Newton’s 2nd law for Rotating Objects
i. net = Iα
1. I = moment of inertia
2. α = angular acceleration
3. (+) or (-) depending on direction of 
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b. Translational vs. Rotational
i. F = ma
ii. net = Iα
c. Practice Problems
i. A student tosses a dart using only the rotation of her forearm to accelerate the dart. The
forearm rotates in a vertical plane about an axis at the elbow joint. The forearm and dart
have a combined moment of inertia of 0.075 kg • m2 about the axis, and the length of the
forearm is 0.26 m. if the dart has a tangential acceleration of 45 m/s2 just before its is
released, what is the net torque on the arm and dart?
16. Momentum
a. Angular momentum (L)- product of a rotating objects moment of inertia and angular speed around
the same axis
i. L = Iω
1. I = moment of inertia
2. ω = angular speed
ii. Kg • m2/s → derived from (Kg • m2) (rad/sec) (Rad have no value)
b. L may be conserved
i. Law of conservation of L – When τnet = 0 and ∆L = 0
1. A skater spinning with arms out spins slower than when he/she brings arms in
a. More mass is brought closer to the body therefore decreasing “I” therefore,
increasing “ω” to compensate for decreased “I”, “L” is conserved
c. Practice Problem
i. A 65 kg student is spinning on a merry-go-round that has a mass of 5.25 x 102 kg and a
radius of 2.00 m. She walks from the edge of the merry-go-round toward the center. If the
angular speed of the merry-go-round is initially 0.20 rad/s, what is its angular speed when
the student reaches a point 0.50 m from the center?
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Section 8-4 Simple Machines
17. Machine – any device that transmits or modifies force usually by changing the force applied to an object
a. Made of a combination of simple machines
18. Simple Machine – change the direction or magnitude of an input force
a. 6 types
i. lever
ii. pulley
iii. inclined plane
iv. wheel and axle
v. wedge
vi. screw
b. Simple Machine Hand-out – Shows each simple machine and how to calculate the ideal
mechanical advantage for each.
c. Characterized by comparing how large the input force is relative to the output force = Mechanical
Advantage
MA = Output force = Fout
Input force = Fin
d. Consider a hammer:
i. Input force is applied to one end of the handle
ii. Output force is exerted by the handle in the head of the nail
1. Rotational equilibrium is maintained therefore, input torque must equal output
torque.
in = out
Fin din = Fout dout
e. Mechanical advantage may then be calculated by:
MA = Fout = din
Fin = dout
f.
Longer the input lever as compared to the shorter output lever will result in greater MA
19. Machines can alter the force and the distance moved
a. A machine can increase (or decrease) the force acting on an object at the expense (or gain) or the
distance moved
i. Small distance = large force
ii. Large distance = small force
b. Work done on an object is always constant
20. Efficiency is a measure of how well a machine works
a. Measure of how much input energy is lost compared with how much energy is used to perform
work on an object.
i. Friction responsible for much energy loss
b. If frictionless – ME is conserved
i. Ideal machine
ii. Input work = output work
iii. Efficiency is 1 or 100%
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