Chapter 12. Rotation of a Rigid Body

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Chapter 12. Rotation of a Rigid BodyChapter 12. Rotation of a Rigid Body

Not all motion can be described as that of a particle. Rotation requiresthe idea of an extended object. This diver is moving toward the water along a parabolic trajectory, and she’s rotating rapidly around her center of mass.

Chapter Goal: To understand the physics of rotating objects.


Student objectives

• To extend the particle model to the rigid-body model.

• To understand the equilibrium of an extended object.


TorqueConsider the common experience of pushing open a door. Shown is a top view of a door hinged on the left. Four pushing forces are shown, all of equal strength. Which of these will be most effective at opening the door?

The ability of a force to cause a rotation depends on three factors:1. the magnitude F of the force.2. the distance r from the point of application to the pivot.3. the angle at which the force is applied.


Torque

Let’s define a new quantity called torque τ (Greek tau) as


Signs, magnitudes of torque


Two perspectives of torque

τ = r Ft where Ft is the tangential (i.e. perpendicular) component of force

Ft = (F sin φ)

Note: If you use the acute angle, put sign in manually



Line of action: extension of the force vector

moment arm aka lever arm (d): perpendicular distance between the pivot point and the line of action.



|τ| =dF where d is the moment arm.

d = r sinφ


EXAMPLE 12.9 Applying a torque

QUESTION:


EXAMPLE 12.9 Applying a torque


EQUILIBRIUM OF A RIGID BODY

A rigid body is in equilibrium if:

0 0yF0 xF

This analysis will most likely involve multiple equations with more than one unknown.


Reasoning Strategy1. Select the object to which the equations for equilibrium are to be applied.

2. This is the most important step. Draw an extended free-body diagram that shows all of the external forces acting on the object You will no longer be able to use a particle to represent the object.

3. Choose a convenient set of x, y axes and resolve all forces into componentsalong these axes. The weight force acts at the center of gravity. Assume the center of

gravity is at the mid point of the object, unless told otherwise.

4. Apply Newton’s 1st Law (since ma=0 for equilibrium) in component form.

5. Select a pivot point where one or more of the unknown forces will have a torque of zero. Set the sum of the torques about this axis equal to zero. The pivot you pick does not have to be an actual axis of rotation for this object!

6. Solve the equations for the desired unknown quantities.


Example 3 A Diving BoardA woman whose weight is 530 N is poised at the right end of a diving board with length 3.90 m. The board has negligible weight and is supported by a fulcrum 1.40 m away from the left end. Find the forces that the bolt and the fulcrum exert on the board. knownW - normal force of woman on board, numerically equal to her weight, Lw = 3.90mL2 = 1.4 m

FindF1 force of bolt on boardF2 force of fulcrum on board


A diving board

1. Select the object (the board).2. Draw a free-body diagram

(diagram b). Note the direction of forces. The woman pushes down on the board. The fulcrum pushes up (a normal force) With the woman on the board, the diving board would rotate in a clockwise direction about the fulcrum. What prevents this? The downward force of the bolt.


A diving board

3. Choose a convenient set of x, y axes and resolve all forces into components along these axes (All forces are in the y direction –object weight negligible – so true horizonal/ vertical axes are appropriate).

y

x


A diving board4. Apply Newton’s Law along

the y axis:

y

x

0yF

F2 - F1 – W = 0F2 = F1 + WBoth F1 and F2 are unknownsUh-ohknownW - normal force of woman on board, numerically equal to her weight, Lw = 3.90mL2 = 1.4 m

FindF1 force of bolt on boardF2 force of fulcrum on board


A diving board5. Select a pivot point where

one or more of the unknown forces will have a torque of zero. The pivot point is not necessarily the point about which the object is most likely to pivot. In this case, we can choose as the pivot point the position of the fulcrum, or the position of the bolt, since they both have one unknown force with a torque of zero about that point (no lever arm).

y

x


A diving board5 (cont’d) In this problem, we choose

the bolt to be the pivot point. Set the sum of the torques about this axis equal to zero. Negative sign is if the force causes the board to rotate cw

y

x

0022 WWF

N 1480

m 1.40

m 90.3N 5302 F

about the bolt


Diving Board

021 WFFFy

0N 530N 14801 F

N 9501 F

obvious that F1 must point down for board to be in equilibrium, so it gets a negative sign in Newton’s 2nd Law


EXAMPLE 12.17 Will the ladder slip?

QUESTION:


EXAMPLE 12.17 Will the ladder slip?


The uniform beam has a weight of 1220 N. It is attached to a vertical wall at one end and is supported by a cable at the other end. A 1960-N block hangs from the far end of the beam. Find:

a. The magnitude of the tension in the cable.

b. The magnitude of the horizontal

c. and vertical components of the force that the wall exerts on the left end of the beam.


The uniform beam has a weight of 1220 N. It is attached to a vertical wall at one end and is supported by a cable at the other end. A 1960-N block hangs from the far end of the beam. Find:

a. The magnitude of the tension in the cable. 2260N

b. The magnitude of the horizontal

c. and vertical components of the force that the wall exerts on the left end of the beam. Both = 1450N.


Guess my weight

• A rigid, 100 N board is placed across 2 bathroom scales separated by a distance of 2.00m. A person of that height lies on the board. The scale under his head reads 475 N and the scale under his feet reads 365 N. Find: a. his weight.

• b. the position of the center of mass relative to his head.


Guess my weight

• A rigid, 100-N board is placed across 2 bathroom scales separated by a distance of 2.00 m. A person of that height lies on the board. The scale under his head reads 475 N and the scale under his feet reads 365 N. Find:

• a. his weight 740 N

• b. his position of center of mass relative to his head 0.85 m


A student holds a meter stick horizontally. The number, mass and locations of the hanging objects vary. Rank the situations on the basis of how hard it would be for the student to keep the meter stick from rotating. Show ties. Explain your reasoning:


Rotation About the Center of Mass

An unconstrained object (i.e., one not on an axle or a pivot) on which there is no net force rotates about a point called the center of mass. The center of mass remains motionless while every other point in the object undergoes circular motion around it.


Consider a a particle of mass mi that is located at position (xi ,yi ),


The position of the center of mass is found by the following:

This works for a system which consists of discrete particles. For a continuous solid…


Rotation About the Center of Mass

The center of mass is the mass-weighted center of the object.


EOC #16

Find the coordinates of the center of mass.


EOC #16

Find the coordinates of the center of mass.

xcm = 6 cm

ycm = 4 cm


EOC # 54

• Find the position of the center of mass for this uniform plate with a cutout section.


EOC # 54

• Find the position of the center of mass for this uniform plate with a cutout section.

• xcm = -.25 cm

• ycm = .125 cm

Chapter 12. Rotation of a Rigid Body

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