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Chapter 8
Rotational Motion
Rotation- circular
movement of an object
around a central point or axis
Revolution- a single turn
around the center point of a
circular path.
Specific points on a rigid
object will move in a circular
path
Measuring a fraction of a revolution
• Degrees-360°=1 revolution

Arbitrary number possibly chosen because it
approximates the days in a year, is divisible by 24
different divisors including all numbers from 1-10
except 7
• Grad- 1/400 of a revolution; 400 grad=1 revolution
 Sometimes used in surveying and seems to be derived
from the fact that is equal to 1/100 of a right angle
• Radian- is a ratio equal to the arc length divided by
the radius

Credited to Roger Cotes an English mathematician
who worked closely with Isaac Newton; 1714
For a full revolution:
2 r

 2 rad
r
Arc length s
 (in radians) 

Radius
r
2 rad  360
Converting radians to degrees
ϴdegrees= ϴradians x
180°
𝜋
Converting degrees to radians
ϴradians = ϴdegrees x
𝜋
180°
Adjacent Synchronous Satellites
Synchronous satellites are put into
an orbit whose radius is 4.23×107m.
If the angular separation of the two
satellites is 2.00 degrees, find the
arc length that separates them.
Arc length s
 (in radians) 

Radius
r
  rad 
  0.0349 rad
2.00 deg
 180 deg 
7
s  r  4.23 10 m0.0349 rad 
s  1.48 106 m (920 miles)
DEFINITION OF ANGULAR DISPLACEMENT
 The angle through which the object rotates is
called angular displacement.
 By convention, the angular displacement is
positive if it is counterclockwise and negative if it
is clockwise.
 SI Unit of
Angular Displacement: radian (rad)
Rotational Dynamics
• Changes in rotational movements; starting an
object rotating or speeding up or slowing down a
rotation
• Forces which change angular velocity change the
rotation of an object.
• Opening and closing a door requires a force which
causes the door to rotate on an axis of rotation; in
this case the axis of rotation is an imaginary line
vertical to the hinges.
When opening a door,
the force is exerted on
the doorknob at a right
angle to the axis away
from the hinges
Which of the following diagrams
will need more force to open the
door the same distance. (same
angular displacement)
The change in angular
velocity (rotation) is
determined by 3 things
• Magnitude of the
force
• Distance from the axis
to where the force is
applied
• The direction of the
force.
Torque- a measure of how effectively a force
causes a rotation; The magnitude of the torque
is the product of the force and the lever arm;
units N·m
• Lever arm- the perpendicular distance
from the axis to the point where the force
is exerted.
• If the force is perpendicular to the radius
of rotation the lever arm is simply the
distance from the axis
• For the doorknob it is the distance from
the hinges
𝝉 = 𝑭𝒓
If the force is applied at an angle the lever arm is
shortened and is calculated by multiplying the
distance (r) times sin of θ
  Fr sin 
Force vs. Torque
• Force causes things to
accelerate.
• If you want something
to accelerate, apply a
force.
• F = ma
• Units: Newtons (N)
• Torque causes things to
rotate.
• If you want something
to rotate, apply a
torque.
• Torque = perpendicular
force x lever arm
• Units: mN or Nm
HOMEWORK pg. 203 11-15
Torque and Balancing
• If a teeter-totter is
balanced, that means that
the torque on both sides
is equal but in the
opposite direction.
• A common mistake is that
students think the force is
the same on both sides
but that is not true. THE
TORQUE MUST BE THE
EQUAL!
𝜏1=𝜏2
𝐹𝑔1𝑟1 = 𝐹𝑔2𝑟2
Torque Practice Problem
F =?
40 kg
• A 40 kg child sits 1.5 meters
away from the center of a
teeter-totter. How much force
does his dad need to exert 2
meters from the center (on the
other side) to prevent the child
from moving?
Example; Kerriann with a mass of 56 kg and Aysha with a mass of 43 Kg
want to balance on a 1.75 meter seesaw. Where does the pivot point
(fulcrum ) need to be?
Define Kariann’s distance in terms of
the length of the seesaw and Aysha’s
distance.
RK= 1.75m- rA
Where there is no rotation the sum of
the torques is equal to zero.
Known:
Mk= 56kg
M A= 43 kg
Rk + RA = 1.75m
Kariann:
FgK=mKg
FgK=(56 kg)
(9.8 m/s2)
FgK=550 N
Unknown:
Rk= ?
RA=?
Aysha:
FgA=mAg
FgA=(43 kg)
(9.8m/s2)
FgA=420 N
Fgkrk=FgArA
Fgk (1.75m - rA) = FgArA
Solve for rA
550N (1.75- rA )= 420N rA
962.5-550 rA = 420rA
962.5= 970 rA
rA= .99 m from Aysha
Moment of inertia- the resistance of an object to
rotation.
𝐼 = 𝑚𝑟
2
The farther the mass is from the center of rotation
the greater the moment of inertia
The distribution of mass matters here—these two
objects have the same mass, but the one on the
left has a greater moment inertia, because so
much of its mass is far from the axis of rotation.
Moment of inertia involving several objects being rotated.
𝐼=
𝑚1𝑟12 + 𝑚2𝑟22+ . . . . . .
A baton that is .65 m has 2 objects (ball and tip) on its
ends that both have a mass of .30 kg. Find the moment of
inertia of the baton if it is rotated at the midpoint
between the objects. NEGLECT THE MASS OF THE ROD
CONNECTING THE OBJECTS.
M=.30 kg
Length of baton= .65 m
I= m1r12 + m2r22
I= (.30 kg)(.325m)2 + (.30 kg)(.325m)2
I= .063 kg ∙m2
What if the baton was rotated around one end?
I= (.30 kg)(.65 m)2
I= .13 kg ∙m2
Newton’s 2nd Law for rotational motion
𝐹
a= 𝑚
α=
Replace a with angular acceleration α
F with net torque τnet
m with moment of inertia I
τnet
𝐼
EXAMPLE: a solid steel wheel has a mass of 15 kg and a
diameter of .44 m You want to rotate at 8.0 rev/s.
What torque must be applied?
If you apply the torque by wrapping a strap around the
outside how much force needs to be applied?
HOMEWORK; Pg. 208 21, 22, 24
Pg. 210 25-29