Lecture 17: Torque & Rotational Equilibrium

Questions of YesterdayYou are riding on a Ferris wheel moving at constant speed. 1a) At what point is the net force acting on you the greatest?

a) the top b) the bottom

c) halfway between top and bottomd) the force is the same over the whole motion

1b) Is the net force doing work on you?a) YESb) NO

2) If the mass of the moon were doubled, what would happen to its centripetal acceleration?

a) it would increaseb) it would decreasec) it would stay the same

Force & Angular Acceleration

F = ma

What causes an object to accelerate?

Can an object be accelerating if there is no net force acting on the object?

What causes ANGULAR acceleration?

Force is needed to change the rate of rotation of an object

Is that all?

=t aT=r

F

F

FFF

ANGLE of force with respect to the radial direction affects angular acceleration

r r

POSITION of force with respect to the axis of rotation affects angular acceleration

Force & Angular Acceleration

F

F

FFF

The rate of rotation of an object can’t change unless a FORCE is applied at a certain ANGLE and at a

DISTANCE from the axis of rotation

r r

Force & Angular Acceleration

r

Torque

FT

Fr

= ?F

The rate of rotation of an object can’t change unless a NET FORCE is applied at a certain ANGLE and at a

DISTANCE from the axis of rotation

The rate of rotation of an object can’t change unless the object is acted on by a net TORQUE ()

sinFr

= r*Fsin

Torque

Distance from axis of rotation

Force actingon object

Angle between forceand displacement from axis of rotation

F

F F

F

Direction of Force Vector F

r r

Torque Direction

+

-+

Direction of displacement vector r pointing FROM axis of rotation TO applied force

Direction of Rotation depends on

F F

F

Torque is a VECTOR!!Direction tells you direction of rotation

+ = Counterclockwise Motion- = Clockwise Motion

r r

Torque Direction

+

-+

Right-Hand Rule

Curl your FINGERS of your RIGHT hand in direction of

MOTION

THUMB points in direction of TORQUE Vector

CCW (+) Motion -> Torque = OUT of PAGE

CW (-) Motion -> Torque = INTO PAGE

TORQUE Direction is PERPENDICULAR to plane formed by

FORCE and DISPLACEMENT Vectors

F F

F

r r

+

+ -

Right-Hand Rule

TORQUE Direction is PERPENDICULAR to plane formed by

FORCE and DISPLACEMENT Vectors

F F

F

r r

If you have a FORCE acting on an object and a specific AXIS of rotation…

What is the direction of the TORQUE?

1) POINT the fingers of your right

hand in the direction of r

2) CURL your fingers in direction of F

3) THUMB points in direction of

Torque

Does the seesaw rotate?

What is the TORQUE on the seesaw plank from each person?

What is the NET torque on the seesaw plank?

r r

Fg Fg

What is the net Force on the seesaw plank?

FPmass = m mass = m

Fgs

Torque

r r

Fg Fg

FPmass = m mass = m

Force of gravity of a HOMOGENEOUS, SYMMETRIC body acts at the center of the object

Fgs

What is the TORQUE on the seesaw plank from each person?

What is the NET torque on the seesaw plank?

Equilibrium ConditionsAn object at rest or moving at constant linear velocity

and/or angular velocity (rotation) is in equilibrium

r r

Fg Fg

FPmass = m mass = m

Equilibrium (a = 0, = 0)

conditions

F = 0 = 0

Fgs

Equilibrium ConditionsIs the system in equilibrium?

r/2 r

FgFg

FPmass = m mass = m

Equilibrium (a = 0, = 0)

conditions

F = 0 = 0

Fgs

Equilibrium Conditions

rr

FgFg

FPmass = m mass = m

Is the system in equilibrium now?

Fgs

Equilibrium Conditions

FgFg

mass = m mass = m

Is the system in equilibrium now?

rS

= + Fgr - Fgr + Fgsrs

Fgs

rr

FP

Equilibrium Conditions

Is there a point where the pivot could be placed to make the system in equilibrium?

FgFg

mass = m mass = m

Is the system in equilibrium now?

r3

= + Fgr1 - Fgr2 + Fgsr3 = 0

Fgs

r2r1

FP

Choosing your Axis of Rotation

If we chose a different axis of rotation to calculate our net torque…would the system still be in equilibrium?

FgFg

mass = m mass = m

Do the torques exerted on the object depend on the chosen axis of rotation? What about the net torque?

r3Fgs

r2r1

FP

Choosing your Axis of Rotation

If we chose a different axis of rotation to calculate our net torque…would the system still be in equilibrium?

FgFg

mass = m mass = m

Do the torques exerted on the object depend on the chosen axis of rotation? What about the net torque?

r3 Fgs

r2r1

FP

axis ofrotation

Choosing your Axis of Rotation

FgFg

mass = m mass = m

r3 Fgs

r2r1

FP

axis ofrotation

The net torque of system is independent of the chosen axis of rotation…choose an axis that makes the problem

easiest!!

= +FPr1 - Fg (r1 + r2) - Fgsr3 = 0

Choosing your Axis of Rotation

The net torque of system is independent of the chosen axis of rotation…choose an axis that makes the problem

easiest!!

FgFg

mass = m mass = m

r3 Fgs

r2r1

FP

axis ofrotation

If there is an unknown force F, choose your axis to be at the position of that unknown force (r = 0 -> = 0)

Choosing your Axis of Rotation

A uniform 10.0 N picture frame is supported as shown above.

Find the tension in the cords and the magnitude of F

F

T1T2

10.0 N

60o40.0 cm

30.0 cm

*

Center of Mass

r r

Fg Fg

FPmass = m mass = m

Force of gravity of a HOMOGENEOUS, SYMMETRIC body acts at the CENTER (the axis of symmetry) of the object

Fgs

Center of Mass

r r

Fg Fg

FPmass = m mass = m

CENTER OF MASSThe point on an extended object where a single force Fg

= mg can act to represent the force of gravity acting on the entire extended object

Fgs

C.M.

Center of Mass

r r

Fg Fg

FPmass = m mass = m

CENTER OF MASS

The rotation induced from Fg positioned at the center of

mass is the same as that induced from all the Fg’s

acting on the extended object

Fgs

C.M.

Center of MassCENTER OF MASS of people + seesaw systemPosition of axis of rotation where system is in

equilibrium

FgFg

mass = m mass = m

rSFgs

r2r1

FP

C.M.

In 2-dimensions the center of mass is defined by an x and y coordinate (a single point in x-y space)

Center of Mass

FgFg

mass = m mass = m

rSFgs

r2r1

C.M.

xCM = m1x1 + m2x2 + m3x3….

m1 + m2 + m3….

m1y1 + m2y2 + m3y3….m1 + m2 + m3….

yCM =

=

=

∑mixi

∑mi

∑miyi

∑mi

Practice ProblemA 20.0-m, 500-N uniform ladder rest against a

frictionless wall, making an angle of 60.0o with the horizontal.

Find the horizontal and vertical forces exerted on the base of the ladder by the Earth when an 1000-N

firefighter is 5.0 m from the bottom.

If the ladder is just on the verge of slipping when the firefighter is 10.0 m up, what is the coefficient of static

friction between ladder and ground?

Practice ProblemA hungry 500-N bear walks out on a beam in an attempt to

retrieve some goodies (50 N) hanging at the end of the beam. The beam is attached to a wall by a hinge and supported from the other end by a cable. The beam is uniform, weighs 200 N

and is 10 m long.

-Draw a free-body diagram of the beam

-When the bear is at x = 1.00 m, find the tension in the wire and the

components of the reaction force at the hinge

-If the wire can withstand a maximum tension of 1000 N what is the maximum distance the bear can walk before the

wire breaks?

Questions of the Day1) If an object is rotating at a constant angular speed

which statement is true?a) the system is in equilibriumb) the net force on the object is ZEROc) the net torque on the object is ZEROd) all of the above

2) Student 1 (mass = m) sits on the left end of massless seesaw of length L and Student 2 (mass = 2m) sits at the right end. Where must the pivot be placed so the system is in equilibrium?a) L/2 b) L/3 from the right (from Student 2)c) L/3 from the left (from Student 1)d) the system cant be in equilibrium