Physics - BRHS

•Rotational dynamics torque moment of inertia angular momentum

conservation of angular momentum

rotational kinetic energy

Last lecture:

1. Rotations angular displacement, velocity (rate of change of angular

displacement), acceleration (rate of change of angular velocity) motion with constant angular acceleration

2. Gravity laws

Review Problem: A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her?

An engineer wishes to design a curved exit ramp for a toll road in such a way that a car will not have to rely on friction to round the curve without skidding. She does so by banking the road in such a way that the force causing the centripetal acceleration will be supplied by the component of the normal force toward the center of the circular path. Find the angle at which the curve should be banked if a typical car rounds it at a 50.0-m radius and a speed of 13.4 m/s.

Rotational Equilibrium and

Rotational Dynamics

Consider force Consider force required to open required to open door. Is it easier to door. Is it easier to open the door by open the door by pushing/pulling pushing/pulling awayaway from hinge or from hinge or closeclose to hinge?to hinge?

closeclose to hinge to hinge

awayaway from from hingehinge

Farther Farther from from from from hinge, hinge, larger larger rotational rotational effect!effect!

Physics concept: Physics concept: torquetorque

Torque, , is the tendency of a force to rotate an object about some axis

is the torque d is the lever arm (or moment arm) F is the force

Fd

Door example:

The lever arm, d, is the shortest (perpendicular) distance from the axis of rotation to a line drawn along the the direction of the force

d = L sin Φ

It is not necessarily the distance between the axis of rotation and point where the force is applied

Torque is a vector quantity The direction is perpendicular

to the plane determined by the lever arm and the force

Direction and sign: If the turning tendency of the

force is counterclockwise, the torque will be positive

If the turning tendency is clockwise, the torque will be negative

Direction of torque: out of page

UnitsUnits

SISI Newton meter (Nm)Newton meter (Nm)

US CustomaryUS Customary Foot pound (ft lb)Foot pound (ft lb)

The force could also be resolved into its x- and y-components The x-component, F

cos Φ, produces 0 torque

The y-component, F sin Φ, produces a non-zero torque sinFL

F is the forceF is the force

L is the distance along the objectL is the distance along the object

Φ is the angle between force and objectΦ is the angle between force and object

LL

You are trying to open a door that is stuck by pulling on the doorknob in a direction perpendicular to the door. If you instead tie a rope to the doorknob and then pull with the same force, is the torque you exert increased? Will it be easier to open the door?

1. No2. Yes

You are trying to open a door that is stuck by pulling on the doorknob in a direction perpendicular to the door. If you instead tie a rope to the doorknob and then pull with the same force, is the torque you exert increased? Will it be easier to open the door?

1. No2. Yes

You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is most effective in loosening the nut? List in order of descending efficiency the following arrangements:

You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is most effective in loosening the nut? List in order of descending efficiency the following arrangements:

What if two or more different forces act on lever arm?

The net torque is the sum of all the torques produced by all the forces

Remember to account for the direction of the tendency for rotation Counterclockwise torques are positive Clockwise torques are negative

Given:

weights: w1= 500 Nw2 = 800 N

lever arms: d1=4 m d2=2 m

Find:

= ?

1. Draw all applicable forces

(500 )(4 ) ( )(800 )(2 )

2000 1600

400

N m N m

N m N m

N m

2. Consider CCW rotation to be positive

500 N 800 N

4 m 2 m

Rotation would be CCW

N

Determine the net torque:

Where would the 500 N person have to be relative to fulcrum for zero torque?

Given:

weights: w1= 500 Nw2 = 800 N

lever arms: d1=4 m = 0

Find:

d2 = ?

1. Draw all applicable forces and moment arms

2

2 2

(800 )(2 )

(500 )( )

800 2 [ ] 500 [ ] 0 3.2

RHS

LHS

N m

N d m

N m d N m d m

500 N 800 N

2 md2 m

According to our understanding of torque there would be no rotation and no motion!

N’ y

What does it say about acceleration and force?

Thus, according to 2nd Newton’s law F=0 and a=0! NN

NNNFi

1300'

0)800(')500(

First Condition of Equilibrium The net external force must be zero

This is a necessary, but not sufficient, condition to ensure that an object is in complete mechanical equilibrium

This is a statement of translational equilibrium

Second Condition of Equilibrium The net external torque must be zero

This is a statement of rotational equilibrium

0

0 and 0x y

F

F F

CCCCCCCCCCCCCC

0

So far we have chosen obvious axis of rotation

If the object is in equilibrium, it does not matter where you put the axis of rotation for calculating the net torque The location of the axis of rotation is

completely arbitrary Often the nature of the problem will

suggest a convenient location for the axis When solving a problem, you must specify

an axis of rotation Once you have chosen an axis, you must maintain

that choice consistently throughout the problem

The force of gravity acting on an object must be considered

In finding the torque produced by the force of gravity, all of the weight of the object can be considered to be concentrated at one point

1. The object is divided up into a large number of very small particles of weight (mg)

2. Each particle will have a set of coordinates indicating its location (x,y)

3. The torque produced by each particle about the axis of rotation is equal to its weight times its lever arm

4. We wish to locate the point of application of the single force , whose magnitude is equal to the weight of the object, and whose effect on the rotation is the same as all the individual particles.

This point is called the center of gravity of the object

The coordinates of the center of gravity can be found from the sum of the torques acting on the individual particles being set equal to the torque produced by the weight of the object

The center of gravity of a homogenous, symmetric body must lie on the axis of symmetry.

Often, the center of gravity of such an object is the geometric center of the object.

i

iicg

i

iicg m

ymyand

m

xmx

Given:

masses: m1= 5.00 kgm2 = 2.00 kgm3 = 4.00 kg

lever arms: d1=0.500 m d2=1.00

m

Find:

Center of gravity

Find center of gravity of the following system:

m

kg

mkgmkgmkg

mmm

xmxmxm

m

xmx

i

iicg

136.0

0.11

)00.1(00.4)0(00.2)500.0(00.5

321

332211

The wrench is hung freely from two different pivots

The intersection of the lines indicates the center of gravity

A rigid object can be balanced by a single force equal in magnitude to its weight as long as the force is acting upward through the object’s center of gravity

A zero net torque does not mean the absence of rotational motion An object that rotates at uniform angular

velocity can be under the influence of a zero net torque This is analogous to the translational situation

where a zero net force does not mean the object is not in motion

Isolate the object to be analyzed

Draw the free body diagram for that object Include all the

external forces acting on the object

Suppose that you placed a 10 m ladder (which weights 100 N) against the wall at the angle of 30°. What are the forces acting on it and when would it be in equilibrium?

ExampleExample

Given:

weights: w1= 100 Nlength: l=10 mangle: =30° = 0

Find:

f = ?n=?P=?

1. Draw all applicable forces

NP

PN

PLL

mg

6.862

11866.0

2

11000

030sin30cos2

Torques: Forces:

Nn

mgnF

Nf

PfF

y

x

100

0

6.86

0

2. Choose axis of rotation at bottom corner ( of f and n are 0!)

Note: f = s n, so 866.0100

6.86

N

N

n

fs

mg

So far: net torque was zero.What if it is not?

When a rigid object is subject to a net torque (≠0), it undergoes an angular acceleration

The angular acceleration is directly proportional to the net torque The relationship is

analogous to ∑F = ma Newton’s Second Law

so,

:onacceleratitangential

bymultiply,

ra

rmarF

rmaF

t

tt

tt

2mrrFt

torque dependent upon object and axis of rotation. Called moment of inertia I. Units: kg mkg m22

2iirmI

IThe angular acceleration is inversely The angular acceleration is inversely proportional proportional to the analogy of the mass in a rotating to the analogy of the mass in a rotating systemsystem

Image the hoop is divided into a number of small segments, m1 …

These segments are equidistant from the axis 22 MRrmI ii

The angular acceleration is directly proportional to the net torque

The angular acceleration is inversely proportional to the moment of inertia of the object

There is a major difference between moment of inertia and mass: the moment of inertia depends on the quantity of matter and its distribution in the rigid object.

The moment of inertia also depends upon the location of the axis of rotation

I

Example:Example:

Consider a flywheel (cylinder pulley) of mass M=5 kg and radius R=0.2 m and weight of 9.8 N hanging from rope wrapped around flywheel.

What are forces acting on flywheel and weight? Find acceleration of the weight.

mg

Given:

masses: M = 5 kgweight: w = 9.8 Nradius: R=0.2 m

Find:

Forces=?

1. Draw all applicable forces

I

RT

IRT

Forces: Torques:

!Tneed

maTmgF

Tangential acceleration at the edge of flywheel (a=at):

ttt

t

akgam

mkga

I

TRRa

5.22.0

10.0

R

IT

or

aor

2

2

2

2

t

N

T

T

Mg

mg

28.2

5.3

8.9

5.2

5.2

smkg

N

kgm

mga

maakgmg

maTmgF

t

22 10.02

1mkgMRI

A force F is applied to a dumbbell for a time interval t, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed?

1. (a)2. (b)3. no difference4. The answer depends on the

rotational inertia of the dumbbell.

A force F is applied to a dumbbell for a time interval t, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed?

1. (a)2. (b)3. no difference4. The answer depends on the

rotational inertia of the dumbbell.

Return to our example:Return to our example:

Consider a flywheel (cylinder pulley) of mass M=5 kg and radius R=0.2 m with weight of 9.8 N hanging from rope wrapped around flywheel. What are forces acting on flywheel and weight? Find acceleration of the weight.

mg

If flywheel initially at rest and then begins to rotate, a torque must be present:

tt

I

I

00 since,

Define physical quantity:

t

L

intervaltime

momentumang.inchange LI momentumangular

Similarly to the relationship between force and momentum in a linear system, we can show the relationship between torque and angular momentum

Angular momentum is defined as L = I ω

If the net torque is zero, the angular momentum remains constant

Conservation of Linear Momentum states: The angular momentum of a system is conserved when the net external torque acting on the systems is zero. That is, when

t

L

ffiifi IIorLL ,0

t

pF

(compare to )

Return to our example once Return to our example once again:again:

Consider a flywheel (cylinder pulley) of mass M=5 kg and radius R=0.2 m with weight of 9.8 N hanging from rope wrapped around flywheel. What are forces acting on flywheel and weight? Find acceleration of the weight.

mg

Each small part of the flywheel is moving with some velocity. Therefore, each part and the flywheel as a whole have kinetic energy!

2

2

massipulley

ipulleyipulley vKE 2

2

1 IKE pulley

Thus, total KE of the system: 22

2

1

2

1mvIKE pulleytot

An object rotating about some axis with an angular speed, ω, has rotational kinetic energy ½Iω2

Energy concepts can be useful for simplifying the analysis of rotational motion

Conservation of Mechanical Energy

Remember, this is for conservative forces, no dissipative forces such as friction can be present

fgrtigrt PEKEKEPEKEKE )()(

A force F is applied to a dumbbell for a time interval t, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater energy?

1. (a)2. (b)3. no difference4. The answer depends on the

rotational inertia of the dumbbell.

A force F is applied to a dumbbell for a time interval t, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater energy?

1. (a)2. (b)3. no difference4. The answer depends on the

rotational inertia of the dumbbell.