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## PowerPoint Slideshow about ' ROTATIONAL MOTION' - deborah-meyers

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AND EQUILIBRIUM

Angular Quantities of Rotational Motion

Rotational Kinematics

Torque

Center of Gravity

Moment of Inertia

Rotational Kinetic Energy

Angular Momentum

Rotational Equilibrium

ANGULAR SPEED AND ANGULAR ACCELERATION:

w

= angular speed

rad/s

deg/s

rev/s

a

2

2

2

= angular acceleration

rad/s

deg/s

rev/s

D

q

q

w

=

D

t

rotating

drum

D

w

a

=

D

t

axis

p

1 rev = 2

rad = 360 deg

p

rad = 180 deg

Convert 246

to radians.

o

Convert 16.4 rev to degrees.

Uniform angular acceleration:

q

= instantaneous angular position

w

= instantaneous angular speed

Through what angle does a wheel rotate if it begins from rest

and accelerates at 3

o

/s

2

for 15 seconds? Also calculate its final

angular speed.

Through what angle does a wheel rotate if it begins from rest

and accelerates at 3

o

/s

2

for 15 seconds? Also calculate its final

angular speed.

o

pivot

57

F = 16N

EXAMPLE:

Calculate the torque produced by

the force acting on the rod as shown.

o

pivot

57

F = 16N

line of action

EXAMPLE:

Calculate the torque produced by

the force acting on the rod as shown.

1 - Draw Line of Action

1.25m

o

pivot

57

F = 16N

line of action

EXAMPLE:

Calculate the torque produced by

the force acting on the rod as shown.

1 - Draw Line of Action

2 - Draw Moment Arm

1.25m

o

pivot

57

F = 16N

line of action

EXAMPLE:

Calculate the torque produced by

the force acting on the rod as shown.

1 - Draw Line of Action

2 - Draw Moment Arm

3 - Calculate Moment Arm

1.25m

o

pivot

57

F = 16N

line of action

EXAMPLE:

Calculate the torque produced by

the force acting on the rod as shown.

1 - Draw Line of Action

2 - Draw Moment Arm

3 - Calculate Moment Arm

4 - Calculate Torque

t = F·d = 16N·1.05m = 16.8Nm

1.25m

o

pivot

57

Rotation will be counter-clockwise

F = 16N

line of action

EXAMPLE:

Calculate the torque produced by

the force acting on the rod as shown.

1 - Draw Line of Action

2 - Draw Moment Arm

3 - Calculate Moment Arm

4 - Calculate Torque

5 - See Rotation

t = F·d = 16N·1.05m = 16.8Nm

When g is considered constant cg is usually referred to as center of mass.

EXAMPLE: center of mass.

y

Suppose the three masses in the diagram

=

m

4 kg

at (0, 3m)

are fixed together by massless rods and are

2

free to rotate in the xy plane about their

r

2

center of gravity. The coordinates of their

center of gravity are (1m, 1m) as calculated

r

r

m

= 3 kg at (4m, 0)

3

1

3

in the previous example.

x

Calculate the moment of inertia of these

m

= 5 kg at (0, 0)

masses about this axis.

1

ANSWER: center of mass.

y

Suppose the three masses in the diagram

=

m

4 kg

at (0, 3m)

are fixed together by massless rods and are

2

free to rotate in the xy plane about their

r

2

center of gravity. The coordinates of their

center of gravity are (1m, 1m) as calculated

r

r

m

= 3 kg at (4m, 0)

3

1

3

in the previous example.

x

Calculate the moment of inertia of these

m

= 5 kg at (0, 0)

masses about this axis.

1

EXAMPLE: center of mass. A solid ball rolls down a 40o incline from a height of 4m without slipping. The radius of the ball is 20cm and it begins from rest. What is the linear speed of the ball at the bottom of the incline?

Basic Equation center of mass.

Values and expressions for initial and final quantities

Continued on next slide

ANSWER: A solid ball rolls down a 40o incline from a height of 4m without slipping. The radius of the ball is 20cm and it begins from rest. What is the linear speed of the ball at the bottom of the incline?

Continues from previous slide center of mass.

Substitutions made to produce working equation

Simplify and solve for v

The speed of the ball does not depend on its mass nor on its radius. It even does not depend on the angle of the incline, just the height from which it starts.

Do the same calculation with a cylinder and determine which will reach the bottom of the incline first, the sphere or the cylinder if released together.

EXAMPLE: center of mass. A merry-go-round on a playground has a radius or 1.5 m and a mass of 225 kg. One child is sitting on its outer edge as it rotates 1 rev/s. If the mass of the child is 50 kg, what will the new rotation rate be when the child crawls half way to the center of the merry-go-round?

ANSWER: center of mass. A merry-go-round on a playground has a radius or 1.5 m and a mass of 225 kg. One child is sitting on its outer edge as it rotates 1 rev/s. If the mass of the child is 50 kg, what will the new rotation rate be when the child crawls half way to the center of the merry-go-round?

An object in equilibrium has no linear and no angular accelerations. That does not mean it isn’t moving, just not accelerating. When an object is stationary it is said to be in static equilibrium. Many of the equilibrium problems here will be static equilibrium problems.

EXAMPLE: accelerations. That does not mean it isn’t moving, just not accelerating. When an object is stationary it is said to be in static equilibrium. Many of the equilibrium problems here will be static equilibrium problems.

cable

l

A uniform rod of length

= 4m

m

and mass

= 75kg is hinged to

a wall at the left and supported

wall

rod

at the right by a cable. The

o

45

l

cable is attached to the rod

/4

from its right edge. The rod

makes a 30

o

angle with the

o

30

horizontal and the cable makes

hinge

o

a 45

angle with the rod.

Calculate the tension in the

cable and the force exerted on

the rod by the hinge.

Solution on next several slides

STEP 1: accelerations. That does not mean it isn’t moving, just not accelerating. When an object is stationary it is said to be in static equilibrium. Many of the equilibrium problems here will be static equilibrium problems.

cable

wall

T

rod

o

45

o

30

hinge

R

W

Add Force Vectors: T = Tension W = Weight R = Reaction

STEP 1: accelerations. That does not mean it isn’t moving, just not accelerating. When an object is stationary it is said to be in static equilibrium. Many of the equilibrium problems here will be static equilibrium problems.

165O

210O

T

R

W

cable

wall

rod

o

45

pivot

o

30

hinge

Add Force Vectors: T = Tension W = Weight R = Reaction

Add Needed Angles

STEP 1: accelerations. That does not mean it isn’t moving, just not accelerating. When an object is stationary it is said to be in static equilibrium. Many of the equilibrium problems here will be static equilibrium problems.

165O

210O

T

R

W

Add Force Vectors: T = Tension W = Weight R = Reaction

Add Needed Angles

Add Pivot and Moment Arms: dT = Moment Arm For Tension

dW = Moment Arm For Weight

On the next slide these force equations will be used to solve for Rx and Ry.

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