physics 201: lecture 18, pg 1 lecture 18 goals: define and analyze torque introduce the cross...

Physics 201: Lecture 18, Pg 1

Lecture 18Goals:Goals:

• Define and analyze torque

• Introduce the cross product

• Relate rotational dynamics to torque

• Discuss work and work energy theorem with respect to rotational motion

• Specify rolling motion (center of mass velocity to angular velocity

• So what causes rotation?


So what forces make things rotate? Which of these scenarios will cause the bar to spin?

Fixed rotation axis.

F

F

F

F

FF

F

F

A

B

C

D

E


Net external torques cause objects to spin

An external force (or forces) properly placed induced changes in the angular velocity. This action is defined to be a “torque”

A force vector or a component of a force vector whose “line of action” passes through the axis of rotation provides no torque

A force vector or the component of a force vector whose “line of action” does NOT pass through the axis of rotation provides torque

The exact position where the force is applied matters. Always make sure the force vector’s line of action contacts the point at which the force is applied.


Net external torques cause objects to spin Torque increases proportionately with increasing force Only components perpendicular to r vector yields torque Torque increases proportionately with increasing

distance from the axis of rotation.

sin|||| Fr This is the magnitude of the torque

is the angle between the radius and the force vector

Use Right Hand Rule for sign

F

r


Force vector line of action must pass through contact point

Force vector cannot be moved anywhere

Just along line of actionF

r

F

r

line of action


Resolving force vector into components is also valid

Key point: Vector line of action must pass through contact point (point to which force is applied)

r

line of action

F t = F sin

F

F r

F = F r + F t

F t

Fr


Exercise Torque

A. Case 1

B. Case 2

C. Same

In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same.

Remember torque requires F, r and sin or the tangential force component times perpendicular distance

L

LF F

axis

case 1 case 2


Torque is constant along the line of action Even though case 2 has a much larger radius vector the

torque remains constant. Case 1 : = L F Case 2 : = r F sin = F r sin Notice that the sin = sin () = - cos sin = sin Case 2 : = r F sin = F r sin = F r sin () = F LHere = 135° and |r| = 2½ L so r sin = 2-1/2 (2½ L ) = L

L

LF

F

axis

case 1 case 2


Torque, like , is a vector quantity

Magnitude is given by (1) |r| |F| sin

(2) |Ftangential | |r|

(3) |F| |rperpendicular to line of action | Direction is parallel to the axis of rotation with respect to the

“right hand rule”

F

r

F

r sin line of action

Torque is the rotational equivalent of force

Torque has units of kg m2/s2 = (kg m/s2) m = N m

FFradial

F

FFtangential


Torque can also be calculated with the vector cross product

The vector cross product is just a definition

Fr

kˆi zjyxr kˆi zyx FjFFF

zyx FFF

zyxFr

kji

k)(

j)(

i)(

xy

zx

yz

yFxF

xFzF

zFyF


Torque and Newton’s 2nd Law Applying Newton’s second law

IrFt

amF

TangentialTangential maF rmarF tt

Imrrra

mrF 22tt


Example: Wheel And Rope A solid 16.0 kg wheel with radius r = 0.50 m rotates

freely about a fixed axle. There is a rope wound around the wheel.

Starting from rest, the rope is pulled such that it has a constant tangential force of F = 8 N.

How many revolutions has the wheel made after 10 seconds?

F

r


Example: Wheel And Rope

m = 16.0 kg radius r = 0.50 m F = 8 N for 10 seconds Constant F Constant constant

Isolid disk = ½ mr2 = 2 kg m2

I = = r F = 4 Nm/ 2 kg m2 = 2 rad/s2

t + ½tt + ½tRev = ( + ½t

Rev = (0.5 x 2 x 100) / 6.28 = 16

F

r


Work

Consider the work done by a force F acting on an object constrained to move around a fixed axis. For an angular displacement dthen ds = r d

dW = FTangential dr = Ft ds

dW = (Ft r) d

dW = dand with a constant torque) We can integrate this to find: W = (f

i

axis of

rotationR

FF

dr =Rdd


Rotation & Kinetic Energy...

The kinetic energy of a rotating system looks similar to that of a point particle:

Point Particle Rotating System

i

iirm 2I

221 IK2

21 vmK

v is “linear” velocity

m is the mass.

is angular velocity

I is the moment of inertia

about the rotation axis.


Work & Kinetic Energy:

Recall the Work Kinetic-Energy Theorem:

K = WNETor WEXT

This is true in general, and hence applies to rotational motion as well as linear motion.

So for an object that rotates about a fixed axis:

NET22

21 I WK if


Example: Wheel And Rope A solid 16 kg wheel with radius r = 0.50 m rotates

freely about a fixed axle. There is a rope wound around the wheel.

Starting from rest, the rope is pulled such that it has a constant tangential force of F = 8 N.

What is the angular velocity after 16 revolutions ?

F

r


Example: Wheel And Rope Mass 16 kg radius r = 0.50 m Isolid disk=½mr2= 2 kg m2

Constant tangential force of F = 8 N. Angular velocity after you pull for 32 rad?

W = F (xf - xi) = r F =0.5 x 8.0 x 32 J = 402 J

K = (Kf - Ki) = Kf = ½ I 2 = 402 J= 20 rad/s

F

r


Exercise Work & Energy

Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same linear distance.

Disk 1, on the left, has a bigger radius, but both have the same mass. Both disks rotate freely around axes though their centers, and start at rest. Which disk has the biggest angular velocity after the

pull?

W = F d = ½ I 2

Smaller I bigger ((A)A) Disk 1

(B)(B) Disk 2

(C)(C) Same

FF

1 2

start

finishd


Home Example: Rotating Rod A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate on a

frictionless pin passing through one end as in the Figure. The rod is released from rest in the horizontal position. What is (A) its angular speed when it reaches the lowest point ?(B) its initial angular acceleration ?(C) initial linear acceleration of its free end ?

L

m


Example: Rotating Rod A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate

on a frictionless hinge passing through one end as shown. The rod is released from rest in the horizontal position. What is

(B) its initial angular acceleration ?

1. For forces you need to locate the Center of Mass

CM is at L/2 ( halfway ) and put in the Force on a FBD

2. The hinge changes everything!L

m

mg

F = 0 occurs only at the hinge

butz = I z = r F sin 90°

at the center of mass and

(ICM + m(L/2)2) z = (L/2) mg

and solve for z




(C) initial linear acceleration of its free end ?

1. For forces you need to locate the Center of Mass

CM is at L/2 ( halfway ) and put in the Force on a FBD

2. The hinge changes everything!L

m

mg

a = L




(A) its angular speed when it reaches the lowest point ?1. For forces you need to locate the Center of Mass CM is at L/2 ( halfway ) and use the Work-Energy Theorem2. The hinge changes everything!

m

mg

W = mgh = ½ I 2

W = mgL/2 = ½ (ICM + m (L/2)2) 2

and solve for

mg

L/2

L


Connection with CM motion

If an object of mass M is moving linearly at velocity VCM without rotating then its kinetic energy is

CM21

RK I

2CM2

12CM2

1 VK MI What if the object is both moving linearly and rotating?

If an object of moment of inertia ICM is rotating in place about its center of mass at angular velocity then its kinetic energy is

2CM2

1T VK M


Rolling Motion

Now consider a cylinder rolling at a constant speed.

VCM CM

The cylinder is rotating about CM and its CM is moving at constant speed (VCM). Thus its total kinetic energy is given by :

2CM2

12CM2

1TOT VK MI


Rolling Motion

Again consider a cylinder rolling at a constant speed.

VCMCM

2VCM


Rolling Motion

Now consider a cylinder rolling at a constant speed.

VCM CM

The cylinder is rotating about CM and its CM is moving at constant speed (VCM). Thus its total kinetic energy is given by :

2CM2

12CM2

1TOT VK MI


Motion Again consider a cylinder rolling at a constant speed.

VCM

CM

Sliding only

CM

Rotation only

vt = R

CM

2VCM

VCM

Both with

|vt| = |vCM |


For Tuesday

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physics 201: lecture 18, pg 1 lecture 18 goals: define and analyze torque introduce the cross...

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