Chapter 8

Rotational Equilibrium and Rotational dynamics

Torque and Equilibrium

� First Condition of Equilibrium

� The net external force must be zero

This is a statement of translational

0

0 0x y

or

and

Σ =

Σ = Σ =

F

F F

r

r r

� This is a statement of translational equilibrium

•The Second Condition of Equilibrium states

• The net external torque must be zero

0τΣ =r

A hobbit house

•The magnitude of the force•The position of the application of the force•The angle at which the force is applied

Three Factors affect torque

τ = rF sin θ

τ = rF sin θF sinθ

F cosθ

Torque and Equilibrium

� First Condition of Equilibrium

� The net external force must be zero

This is a statement of translational

0

0 0x y

or

and

Σ =

Σ = Σ =

F

F F

r

r r

� This is a statement of translational equilibrium

•The Second Condition of Equilibrium states

• The net external torque must be zero

0τΣ =r

Torque direction: Right hand rule again

Force turns it in the counterclockwise directioncounterclockwise direction

Force turns it in the clockwise direction

Center of Gravity

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

Center of gravity

i i i icg cg

i i

mx myx and y

m m

Σ Σ= =Σ Σ

A few pointers.� If a body has a

symmetry and it has a uniform density then the cg is on the line of symmetry.

� The center of symmetry coincides with the cg.

� The cg might be outside the object

Example� Find the cg of a 4x8

uniform sheet of plywood with the upper right quadrant removed.

)2,2(),(;2 == yxMm

x

y

4

a

4 8)2,2(),(;2 211 == yxMm 4 8

)1,6(),(; 222 == yxMm

ftM

M

MM

MM

mm

xmxmxcg 3

10

6

10

2

622

21

2211 =⋅⋅=

+⋅+⋅=

++=

ftM

M

MM

MM

mm

ymymycg 3

5

3

5

2

122

21

2211 =⋅⋅=

+⋅+⋅=

++=

Torque and Equilibrium

� First Condition of Equilibrium

� The net external force must be zero

This is a statement of translational

0

0 0x y

or

and

Σ =

Σ = Σ =

F

F F

r

r r

� This is a statement of translational equilibrium

•The Second Condition of Equilibrium states

• The net external torque must be zero

0τΣ =r

Example of a Free Body Diagram--Ladder

� free body diagram shows normal force and force of static friction acting on the ladder at the ground

75N 75N

In-class quiz 18-1Find the force P of the wall on the top of the10 meter ladder that weights 75 N

A. 50 N

B. 25N

C. 30 N

D. 21 N

E. 45 N

40° 40°75N 75N

In-class quiz 18-1Find the force P of the wall on the top of the10 meter ladder that weights 50 N

A. 50 N

B. 25N

C. 30 N

D. 21 N

E. 45 N

40° 40°

A 100-N uniform ladder, 8.0 m long, rests against a smooth vertical wall. The coefficient of static friction between

ladder and floor is 0.40. What minimum angle can the ladder make with the floor before it slips?

A. 42o

B. 22o

C. 18o

D. 51o

E. 39o

A 100-N uniform ladder, 8.0 m long, rests against a smooth vertical wall. The coefficient of static friction between

ladder and floor is 0.62. What minimum angle can the ladder make with the floor before it slips?

A. 42o

B. 22o

C. 18o

D. 51o

E. 39o

Example: a ladder against a wall

� What minimum angle can the ladder make with the floor before it slips?

0

0

=

=

∑

∑

y

x

F

F 0=− Pfs

0=− mgn

P

N

mg

y

x

0

0

=

=

∑

∑τ

yF 0=− mgn

0cos2

sin =− θθ LmgPL

mgnf sss µµ ==

0cos2

sin =− θθµ LmgmgLs µs sinθ − 1

2cosθ = 0

sµθ

2

1tan =

θfs

Torque and Angular Acceleration

analogous to ∑F = ma

I = moment of inertia

Newton’s Second Law for a Rotating Object

Iτ αΣ =

2 2

i iI mr MR= Σ =For Uniform Ring i iI mr MR= Σ =For Uniform Ring

moment of inertia depends on quantity of matter and its distribution and location of axis of rotation

Other Moments of Inertia