Two-Dimensional Rotational Dynamics8.01W09D2

Young and Freedman 1.10 (Vector Product),

10.1-10.2, 10.4, 11.1-11.3

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Next Reading Assignment Young and Freedman

1.10 (Vector Product) 10.1-10.2, 10.5-10.6

11.1-11.3

Rigid Bodies

- Rigid body An extended object in which the

distance between any two points in the object is

constant in time. Examples sphere, disk - Effect of external forces (the solid arrows

represent forces)

Main Idea Fixed Axis Rotation of Rigid Body

- Torque produces angular acceleration about center

of mass - is the moment of inertial about the center

of mass - is the angular acceleration about center of

mass

Recall Fixed Axis Rotation Kinematics

- Angle variable
- Angular velocity
- Angular acceleration
- Mass element
- Radius of orbit
- Moment of inertia
- Parallel Axis Theorem

Torque as a Vector

- Force exerted at a point P on a rigid

body. - Vector from a point S to the point P.

Torque about point S due to the force exerted at

point P

Summary Cross Product

- Magnitude equal to the area of the parallelogram

defined by the two vectors - Direction determined by
- the Right-Hand-Rule

Torque Magnitude and Direction

- Magnitude of torque about a point S due to force

acting at point P

Direction of torque Perpendicular to the

plane formed by and .Determined

by the Right-Hand-rule.

Properties of Cross Products

Cross Product of Unit Vectors

- Unit vectors in Cartesian coordinates

Components of Cross Product

Concept Question Torque

- Consider two vectors with x gt 0 and
- with Fx gt 0 and Fz gt 0 . The cross product
- points in the
- x-direction
- -x-direction
- y-direction
- -y-direction
- z-direction
- -z-direction
- None of the above directions

Concept Question Torque Answer

- Answer 4. We calculate the cross product noting

that in a right handed choice of unit vectors,

and - Since and , the direction of the cross

product is in the negative y-direction.

Concept Question Magnitude of Torque

- In the figure, a force of magnitude F is applied

to one end of a lever of length L. What is the

magnitude of the torque about the point S? - FL sin?
- FL cos?
- FL tan?
- None of the above

Concept Question Magnitude of Torque Answer

- Answer 2

Torque due to Uniform Gravitational Force

- The total torque on a rigid body due to the

gravitational force can be determined by placing

all the gravitational force at the center-of-mass

of the object.

Recall Rotational Kinematics

- Individual element of mass
- Radius of orbit
- Tangential velocity
- Tangential acceleration
- Radial Acceleration

Dynamics Newtons Second Law and Torque about S

- Tangential force on mass element produces torque
- Newtons Second Law
- Torque about S
- z-component of torque about S

Torque, Moment of Inertia and Angular Acceleration

- Component of the total torque about an axis

passing through S is the sum over all elements - Recall Moment of Inertia about and axis passing

through S - Summary

Fixed Axis Rotational Dynamics

Worked Example Moment of Inertia Wheel

- An object of mass m is attached to a string

which is wound around a disc of radius Rd. The

object is released and takes a time t to fall a

distance s. What is the moment of inertia of the

disc?

Analysis Measuring Moment of Inertia

- Free body force diagrams and force equations
- Rotational equation
- Constraint
- Solve for moment of inertia
- Time to travel distance s

Demo Moment of Inertia Wheel

- Measuring the moment of inertia.
- Radius of disc
- Mass of disc
- Mass of weight holder
- Theoretical result

Concept Question Chrome Inertial Wheel

- A fixed torque is applied to the shaft of the

chrome inertial wheel. If the four weights on the

arms are slid out, the component of the angular

acceleration along the shaft direction will - increase.
- decrease.
- remain the same.

Concept Question Chrome Inertial Wheel Answer

- Answer 2. Torque about the central axis is

proportional to the component of the angular

acceleration along that axis. The proportionality

constant is the moment of inertia about that

axis. By pushing the weights out, the moment of

inertia has increased. If the applied torque is

constant then the component of the angular

acceleration must decrease.

Problem Solving StrategyTwo Dimensional Rotation

- Step 1 Draw free body force diagrams for each

object and indicate the point of application of

each force - Step 2 Select point to compute torque about

(generally select center of mass) - Step 3 Choose coordinate system. Indicate

positive direction for increasing rotational

angle. - Step 4 Apply Newtons Second Law and Torque Law

to obtain equations - Step 5 Look for constraint condition between

rotational acceleration and any linear

accelerations. - Step 6 Design algebraic strategy to find

quantities of interest

Rotor Moment of Inertia

Table Problem Moment of Inertia Wheel

- A steel washer is mounted on a cylindrical

rotor . The inner radius of the washer is R. A

massless string, with an object of mass m

attached to the other end, is wrapped around the

side of the rotor and passes over a massless

pulley. Assume that there is a constant

frictional torque about the axis of the rotor.

The object is released and falls. As the mass

falls, the rotor undergoes an angular

acceleration of magnitude a1. After the string

detaches from the rotor, the rotor coasts to a

stop with an angular acceleration of magnitude

a2. Let g denote the gravitational constant. - What is the moment of inertia of the rotor

assembly (including the washer) about the

rotation axis?

Torque and Static Equilibrium

Conditions for Static Equilibrium

- Translational equilibrium the sum of the forces

acting on the rigid body is zero. - (2) Rotational Equilibrium the vector sum of the

torques about any point S in a rigid body is zero.

Concept Question Tipping

- A box, with its center-of-mass off-center as

indicated by the dot, is placed on an inclined

plane. In which of the four orientations shown,

if any, does the box tip over?

Concept Question Tipping Answer

- Answer 3. In order to tip over, the box must

pivot about its bottom left corner. Only in case

3 does the torque about this pivot (due to

gravity) rotate the box in such a way that it

tips over.

Problem Solving Strategy Static Equilibrium

- Force
- Identify System and draw all forces and where

they act on Free Body Force Diagram - Write down equations for static equilibrium of

the forces sum of forces is zero - Torque
- Choose point to analyze the torque about.
- Choose sign convention for torque
- Calculate torque about that point for each force.

(Note sign of torque.) - Write down equation corresponding to condition

for static equilibrium sum of torques is zero

Table Problem Standing on a Hill

A person is standing on a hill that is sloped at

an angle a with respect to the horizontal. The

persons legs are separated by a distance d, with

one foot uphill and one downhill. The center of

mass of the person is at a distance h above the

ground, perpendicular to the hillside, midway

between the persons feet. Assume that the

coefficient of static friction between the

persons feet and the hill is sufficiently large

that the person will not slip. a) What is the

magnitude of the normal force on each foot? b)

How far must the feet be apart so that the normal

force on the upper foot is just zero? (This is

the instant when the person starts to rotate and

fall over.)

Rotational Work

- Tangential force
- Displacement vector
- work for a small displacement

Rotational Work

- Newtons Second Law
- Tangential acceleration
- Work for small displacement
- Summation becomes integration for continuous body

Rotational Work

- Rotational work for a small displacement
- Torque about S
- Infinitesimal rotational work
- Integrate total work

Rotational Work-Kinetic Energy Theorem

- Infinitesimal rotational work
- Integrate rotational work
- Kinetic energy of rotation about S

Rotational Power

- Rotational power is the time rate of doing

rotational work - Product of the applied torque with the angular

velocity

Table Problem Change in Rotational Energy and

Work

- Suppose that a rotor of moment of inertia Ir

is slowing down during the interval t1, t2

according to - ?(t) ?(t1)-a t,
- where a ?(t1)/(t2 ). Use work energy

techniques to find the frictional torque acting

on the rotor.

Table Problem Solution Change in Rotational

Energy and Work

- Suppose that a rotor of moment of inertia Ir

is slowing down during the interval t1, t2

according to - ?(t) ?(t1)-a t,
- where a ?(t1)/(t2). Use work energy

techniques to find the frictional torque acting

on the rotor.