Transcript Physics Toolkit - Effingham County Schools

Rotational Motion
Describing and Dynamics
Rotational Motion

Describing Rotational Motion

Fractions of revolution measured in grads, degrees, or radians
Grad =
revolution
Degree =
revolution
Radian =
revolution
Rotational Motion

Angular Displacement

Theta, θ, represents angle of revolution

Counterclockwise rotation positive, clockwise negative

Change in angle = angular displacement
d = rθ

Displacement (d) = rotation through angle,
θ, at distance, r, from center
Rotational Motion

Angular Velocity

Velocity = displacement divided by time

Angular velocity is angular displacement divided by time required to
make displacement

Angular velocity is represented by the Greek letter omega, ω
Rotational Motion

Angular Velocity

If velocity changes over time, average velocity not equal to
instantaneous velocity at any given instant

Angular velocity = average angular velocity over a time interval, t

Instantaneous angular velocity = slope of graph of angular position
versus time

Measured in rad/s

So, for Earth, ωE = (2π rad)/[(24.0 h)(3600 s/h)] = 7.27×10─5 rad/s
Rotational Motion

Angular Velocity

Counterclockwise rotation also results in positive angular velocity

If angular velocity is ω, then linear velocity of point at distance, r,
from axis of rotation given by
v = rω

Speed at which object on Earth’s equator moves due to Earth’s rotation
is v = r ω = (6.38×106 m) (7.27×10─5 rad/s) = 464 m/s
Rotational Motion

Angular Acceleration

Change in angular velocity divided by time required to make that
change

Measured in rad/s2

If △𝑣 positive, then ω also positive

Angular acceleration also average angular acceleration over time
interval Δt
Rotational Motion

Angular Acceleration

Find instantaneous angular acceleration by finding slope of graph of
angular velocity as function of time

Linear acceleration of point at r from axis of object with angular
acceleration, α, given by
a = r
Rotational Motion

Angular Acceleration

A summary of linear and angular relationships
Rotational Motion

Angular Frequency

Number of complete revolutions made by object in 1s called angular
frequency

Angular frequency, f, is given by the equation:
Rotational Motion

Describing Rotational Motion


Jupiter, the largest planet in our solar system, rotates around its own
axis in 9.84 h. The diameter of Jupiter is 1.43 x 108 m.

What is the angular speed of Jupiter’s rotation in rad/s?

What is the linear speed of a point on Jupiter’s equator, due to
Jupiter’s rotation?
A computer disk drive optimizes the data transfer rate by rotating the
disk at a constant angular speed of 34.1 rad/s while being read. When
the computer is searching for a file, the disk spins for 0.892 s.

What is the angular displacement of the disk during this time?

Through how many revolutions does the disk turn during this time?
Rotational Motion

Describing Rotational Motion

A cyclist wants to complete 10.0 laps around a circular track 1.0 km in
diameter in exactly 1.0 h. At what linear velocity must this cyclist
ride?

A 75.0 g mass is attached to a 1.0 m length of string and whirled
around in the air at a rate of 4.0 rev/s when the string breaks.

What is the breaking force of the string?

What was the linear velocity of the mass as soon as the string
broke?
Rotational Motion

Rotational Dynamics

Change in angular velocity depends on magnitude of force, distance
from axis to point where force exerted, and direction of force

To open door, you exert force. Doorknob near outer edge of door.
Exert force on doorknob at right angles to door, away from hinges

To get most effect from least force,
exert force as far from axis of
rotation as possible
Rotational Motion

Rotational Dynamics

Magnitude of force, distance from axis to point where force exerted,
and direction of force determine change in angular velocity

Change in angular velocity depends on lever arm, perpendicular
distance from axis of rotation to point where force exerted
Rotational Motion

Rotational Dynamics

For door, distance from the hinges to point where force exerted

If force perpendicular to radius of rotation then lever arm is distance
from axis, r. If force not exerted perpendicular to radius, lever arm
reduced
Rotational Motion

Rotational Dynamics

Lever arm, L, calculated by equation, L = r sin θ,

θ = angle between force and radius from axis of rotation to point
where force applied

Torque measures how effectively force causes
rotation. Measured in newton-meters (N·m)
Rotational Motion

Rotational Dynamics
Rotational Motion

Rotational Dynamics

In order for a bolt to be tightened, a torque of 45.0 N•m is needed. You
use a 0.341 m long wrench, and you exert a maximum force of 189 N.
What is the smallest angle, with respect to the wrench, at which you
can exert this force and still tighten the bolt?

Chloe, whose mass is 56 kg, sits 1.2 m from the center of a seesaw.
Josh, whose mass is 84 kg, wants to balance Chloe. Where on the
seesaw should Josh sit?
Rotational Motion

The Moment of Inertia

Equal to mass of object times square
of object’s distance from axis of rotation

Resistance to changes in rotational
motion

Represented by symbol I and has units
of mass times square of distance
Rotational Motion

Newton’s Second Law for Rotational Motion

Angular acceleration directly proportional to net torque and inversely
proportional to moment of inertia

Changes in torque, or moment of inertia, affect rate of rotation
Rotational Motion

Newton’s Second Law for Rotational Motion

A fisherman starts his outboard motor by pulling on a rope wrapped
around the outer rim of a flywheel. The flywheel is a solid cylinder
with a mass of 9.5 kg and a diameter of 15 cm. The flywheel starts
from rest and after 12 s, it rotates at 51 rad/s.

What torque does the fisherman apply to the flywheel (α= τ/I)?

How much force does the fisherman need to exert on the rope to apply
this torque?
Rotational Motion

The Center of Mass

The center of mass of an object is the point on the object that moves in
the same way that a point particle would move

The path of the center of mass of the object below is a straight line
Rotational Motion

The Center of Mass

To locate the center of mass of an object, suspend the object from any
point

When the object stops swinging, the center of mass is along the
vertical line drawn from the suspension point

Draw the line, and then suspend the object from another point. Again,
the center of mass must be below this point

Draw a second vertical line. The center of mass is at the point where
the two lines cross

A wrench, racket, and all other freely-rotating objects, rotate about an
axis that goes through their center of mass
Rotational Motion

The Center of Mass
Rotational Motion

The Center of Mass of a Human Body

The center of mass of a person varies with posture

For a person standing with his or her arms hanging straight down, the
center of mass is a few centimeters below the navel, midway between
the front and back of the person’s body
Rotational Motion

The Center of Mass of a Human Body

When the arms are raised, as in ballet, the center of mass rises by 6
to10 cm

By raising her arms and legs while in the air, as shown below, a ballet
dancer moves her center of mass closer to her head

The path of the center of
mass is a parabola, so the
dancer’s head stays at
almost the same height
for a surprisingly long
time
Rotational Motion

Center of Mass and Stability
Rotational Motion

Center of Mass and Stability

An object is said to be stable if an external force is required to tip it

The object is stable as long as the direction of the torque due to its
weight, τw tends to keep it upright. This occurs as long as the object’s
center of mass lies above its base

To tip the object over, you must rotate its center of mass around the
axis of rotation until it is no longer above the base of the object

To rotate the object, you must lift its center of mass. The broader the
base, the more stable the object is
Rotational Motion

Center of Mass and Stability

If the center of mass is outside the base of an object, it is unstable and
will roll over without additional torque

If the center of mass is above the base of the object, it is stable

If the base of the object is very narrow and the center of mass is high,
then the object is stable, but the slightest force will cause it to tip over
Rotational Motion

Conditions for Equilibrium

An object is said to be in static equilibrium if both its velocity and
angular velocity are zero or constant

First, it must be in translational equilibrium; that is, the net force
exerted on the object must be zero

Second, it must be in rotational equilibrium; that is, the net torque
exerted on the object must be zero
Rotational Motion

Rotating Frames of Reference

Newton’s laws are valid only in inertial or nonaccelerated frames

Newton’s laws would not apply in rotating frames of reference, as they
are accelerated frames

Motion in a rotating reference frame is important to us because Earth
rotates

The effects of the rotation of Earth are too small to be noticed in the
classroom or lab, but they are significant influences on the motion of
the atmosphere and therefore on climate and weather
Rotational Motion

Centrifugal “Force”

An observer on a rotating frame, sees an object attached to a spring on
the platform

He thinks that some force toward the outside of the platform is pulling
on the object

Centrifugal “force” is an apparent force that seems to be acting on an
object when that object is kept on a rotating platform
Rotational Motion

Centrifugal “Force”

As the platform rotates, an observer on the ground sees things
differently

This observer sees the object moving in a circle

The object accelerates toward the center because of the force of the
spring

The acceleration is centripetal acceleration and is given by
Rotational Motion

Centrifugal “Force”

It also can be written in terms of angular velocity, as:

Centripetal acceleration is proportional to the distance from the axis of
rotation and depends on the square of the angular velocity
Rotational Motion

The Coriolis “Force”

A person standing at the center of a rotating disk throws a ball toward
the edge of the disk. An observer standing outside the disk sees the
ball travel in a straight line at a constant speed toward the edge of the
disk
Rotational Motion

The Coriolis “Force”

An observer stationed on the disk and rotating with it sees the ball
follow a curved path at a constant speed

A force seems to be acting to deflect the ball
Rotational Motion

The Coriolis “Force”

An apparent force that seems to cause deflection to an object in a
horizontal motion when the observer is in a rotating frame of reference
is known as the Coriolis “force”

It seems to exist because we observe a deflection in horizontal motion
when we are in a rotating frame of reference
Rotational Motion

The Coriolis “Force”

An observer on Earth, sees the Coriolis “force” cause a projectile fired
due north to deflect to the right of the intended target

The direction of winds around high- and low-pressure areas results
from the Coriolis “force.” Winds flow from areas of high to low
pressure
Rotational Motion

The Coriolis “Force”

Due to the Coriolis “force” in the northern hemisphere, winds from the
south blow east of low-pressure areas

Winds from the north, however, end up west of low-pressure areas

Therefore, winds rotate counterclockwise around low-pressure areas in
the northern hemisphere

In the southern hemisphere however, winds rotate clockwise around
low-pressure areas