Rotational Motion. 6-1 Angular Position, Velocity, & Acceleration.

Rotational M

6-1 Angular Position, Velocity, & Acceleration

Angular Displacement ()

Measured in radians (rad)

this unit has no dimensions

Counterclockwise is positive

Angular Quantities

oradrev 36021

Angular Velocity ()

Defined in the same way as velocity

Average velocity is displacement divided by time

SI unit

Instantaneous Angular Velocity

1ssrad

Angular Acceleration ()

Defined like acceleration

Angular acceleration is change in angular velocity divide by time

SI unit

Instantaneous Angular

Acceleration

6.2 Rotational Kinematics

All the equations used in kinematics (for conditions of constant acceleration)

Are exactly the same, but with the new quantities

6.2 Rotational Kinematics

attvxx

6.3 Connections Between Linear & Rotational

For an object moving in a circular path

At any given time, the

object is moving with a

linear velocity tangent

to the arc

The speed would be circumference divide by the time for one revolution

Since would be 2/t

Similarly we can determine that

So in a situation where a spinning

object causes linear motion

Circumference of the

Will equal the

displacement

6.4 Rotational K & the Moment of Inertia

In a spinning object, each particle has kinetic energy

Lets assume that we have a rod of uniform mass rotating about its end

Now, just consider a piece of the rod at the very end

To calculate the kinetic energy of the mass

Converting

To rotational

quantities

221 mvK 2

21 )( rmK 2221 )( mrK

So K depends on angular velocity

And it depends on the distribution of mass

This mass distribution is called Moment of Inertia

2221 )( mrK 2

Moment of Inertia varies with shape, mass, and axis of rotation (You will need these to solve

problems)

6.5 Conservation of Energy

We are adding a new type of Kinetic Energy into our existing Energy Equation

Adding Rotational Kinetic Energy

Where Kr is defined as

sgLsgL UUKUUK 000

sgLrsgLr UUKKUUKK 0000

221 IKr

So if a solid ball is rolling down a slope

The equation would become

Using the moment of inertia of a solid sphere

And the relationship between v and

sgLrsgLr UUKKUUKK 0000 Lrg KKU

21 mvImgy 2

21 )( mvmrmgy 2

51 mvmrmgy 2

51 )( mvmrmgy r

v 2212

51 mvmvmgy 2

107 mvmgy

Example: A yo-yo is released from rest and allowed to drop as the top end of the string is kept stationary. The mass of the yo-yo is 0.056kg its radius is 1.5 cm. Assume it acts as a solid disc rotating around its center. What is the angular speed of the yo-yo after it has dropped 0.50m?

sgLrsgLr UUKKUUKK 0000

Example: A yo-yo is released from rest and allowed to drop as the top end of the string is kept stationary. The mass of the yo-yo is 0.056kg its radius is 1.5 cm. Assume it acts as a solid disc rotating around its center. What is the angular speed of the yo-yo after it has dropped 0.50m?

What quantities remain in the equation below?

Expanded equation?

Moment of inertia?

Solve (remember v=r)

Cancel and insert values

sgLrsgLr UUKKUUKK 0000 Lrg KKU

21 mvImgy 2

21 )( mvmrmgy 2

41 mvmrmgy 2

41 )( rmmrmgy 22

41 mrmrmgy 22

43 mrmgy 2243 rgy 2243 )015(.)5)(.8.9( s

6.6 Torque

Rotational Dynamics – the causes of rotational motion

How do we make an object spin

Apply a force away from the pivot

6.6 Torque

The ability to spin increases with force and the distance from the pivot point

If the force is parallel to the distance, no rotation occurs

6.6 Torque

It is the perpendicular component of force that causes rotation

This quantity is called Torque ()

And is measure in Nm

6.6 Torque

As with angular quantities

Counterclockwise torque is positive

Clockwise torques are defined as negative

6.6 Torque

6.7 Torque and Angular Acceleration

If an unbalanced torque is applied to an object

Consider a point where the force is applied

The acceleration of that point is

We can write this equation as

Combining we get

Now, math tricks, multiply by r/r

The of the fraction is torque, and the bottom is inertia

Usually written as

Applying Newton’s second law to rotational motion we get

6.8 Static Equilibrium

Static Equilibrium

How do we calculate the forces needed to support a bridge?

The bridge is at static (not moving) equilibrium

Using a diagram of the bridge

Two conditions for equilibrium

1. The sum of the forces equals zero

021 WPP

2.Sum of the torques equals zero

For torque we need a pivot point – the actual point does not matter because the object is not rotating

Use a pivot point that eliminates a variable

The equation becomes

0FP1 P2

0sinsin 222 wwWrrP

Try the problem with some variable

The bridge is 100 m long with two pylons that are 10 m from each end. The mass of the bridge is 10,000 kg. What is the force on each pylon?

First Condition of Equilibrium

0)8.9)(10000(21 PP NPP 9800021

Second Condition of Equilibrium

0sinsin22 wwWrrP 022 wWrrP 0)40)(98000()80(2 P NP 490002

Combine equations

NPP 9800021 NP 98000490001 NP 490001

A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder)

Diagram?

ladder

Force Diagram?

ladder

Force Equations?

ladder

DLy WWNF 1

Pivot Point for Torque?

ladder

DLy WWNF 1

Distances?

ladder

DLy WWNF 1

20m 10m

Angles?

ladder

DLy WWNF 1

20m 10m

=30o=30o

Torque Equation?

ladder

DLy WWNF 1

20m 10m

=30o=30o

111 sinsinsinsin rNrWrWfr DDDLLLff

Values?

ladder

DLy WWNF 1

20m 10m

=30o=30o

111 sinsinsinsin rNrWrWfr DDDLLLff )30sin()20()30sin()8.9)(75()30sin()10)(8.9)(40()60sin()20( 1Nrf D 0105.36719603.17 1 Nrf D

Value for N1?

ladder

DLy WWNF 1

20m 10m

=30o=30o

111 sinsinsinsin rNrWrWfr DDDLLLff )30sin()20()30sin()8.9)(75()30sin()10)(8.9)(40()60sin()20( 1Nrf D 0105.36719603.17 1 Nrf D

7353920 1 N 11271 N

0)1127(105.36719603.17 Drf 0112705.36719603.17 Drf

Value for f?

ladder

20m 10m

=30o=30o

7353920 1 N 11271 N

0112705.36719603.17 Drf

1Nf )1127)(2(.f 4.225f

0112705.3671960)4.225(3.17 Dr 0112705.36719603899 DrmrD 7.14

6.9 Center of Mass and Balance

An object is balanced, if the net torque is zero

If the rod above is uniform, then when the pivot is in the middle, half the mass is on one side and half on the other

This balance point is called the center of mass

6.9 Center of Mass and Balance

mrmr 21

6.10 Conservation of Angular Momentum

When an object spins, it has angular momentum

Defined as

The conservation of angular momentum is treated just like the conservation of linear momentum

Conservation of L

So the equation for conservation becomes

LL 0 BBAABBAA IIII 0000

Example: A small mass, m, attached to the end of a string revolves in a circle on a frictionless tabletop. The other end of the string passes through a hole in the table. Initially, the mass revolves with a speed v0=2.4m/s in a circle of radius r0=0.80m. The string is then pulled slowly through the hole so that the radius is reduced to r=0.48m. What is the speed, v, of the mass now?

Using conservation of momentum (only one object)

Mass on the end of a string so

Masses cancel and we can convert to v

Enter numbers and solve6.10 Conservation of Angular Momentum

LL 0 II 00 20

20 mrmr 2

020 rr r

vrv rr 22

0 rvvr 00 v)48(.)4.2)(8(. smv 4

Example: A record player with a mass of 1.5 kg is spinning at 33.3 rev/min. The radius of the turntable is 15cm. A bug with a mass of 0.25 kg lands 10cm from the center of the record. What is the new velocity of the record player?

The record player is a solid disk, the bug is a point mass

Convert rev/min to rad/s

Substitute and solve

LL 0 BBRRRR III 00 BBBRRRRRR rmrmrm 22212

srevradrev 49.3))((3.33 60

min12min

BR 2)10)(.25(.

2)15)(.5.1(2

1)49.3(2)15)(.5.1(2

1 BR 0025.0169.0589. v 0025.0169.0589. 0194.0589. srad04.3 min0.2904.3 rev

Rotational Motion. 6-1 Angular Position, Velocity, & Acceleration.

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