Torque Torque and golden rule of mechanics

F 2 r 1 r 2 r 1 d 1 F 1 W 1 =W 2 F 1 d 1 = F 2 d 2 F 1 r 1 = F 2 r 2 r 2 F 1 ||  F 1 F 1  r 1 r 2 F 2 d 2 r 1 /d 1 = r 2 /d 2

Definition of torque

 

F



r



Fr

sin  

Fr

   

r

  

F

r  F

r F Torque and moment of inertia

 

F



r



ma

tan

r



m

 

r



mr

2   

I



I



mr

2

I

 

m i r i

2

Analog of Newton’s Law for rotation Analog of mass for rotation

Example: Two blocks of masses m 1 = 15 kg and m 2 =20 kg are connected through a string that goes through the pulley of mass M = 2.00 kg and radiuss R = 25.0 cm. What is the angular acceleration of the pulley?

T R

1

T m

2

M

2

I

   1 2

MR

2

a R

 

I



T

2

R



T

1

R

 1 2

MR

2

a R T

2 

T

1  1 2

Ma m

1

m

1

g m

2

g

T

1 

m

2

g m

1

g



T

2  

m

1

a m

2

a T

2 

m

2  

T

1

m

1  1 2 

g Ma

 

m

2 

m

1  1 2

M



a

 

a R

 

m

2 

m

2 

m

1 

m

1  1 2 

M



g R

 5 .

45

rad

/

s

2

Example: What is the angular acceleration of a tree as it falls down? Model of a tree as uniform rod of length L. L

  

mg

  2

L

cos 

I



a

   1 3

I



mL

2 

mg

  2

L

1

mL

2 3 cos  

L

 3 2

g

cos   3

g

cos  2

L

mg

The top of the tree will have an acceleration larger than

g

when: 3 2

g

cos  

g

 cos   2 3    48 .

2 



v cm



v cm



v cm



v cm



v cm

+ 

v

2

Rolling



v

3  

v cm



v

1   

v cm



v

4 = 

v

3 45° 

v

2 

v cm

 2 

v cm

45° 

v

4 

v

1  0

Rolling without slipping:

v cm

 

R

Example: The rear wheel on a clown’s bicycle has twice the radius of the front wheel. Is the linear speed at the very top of the rear wheel greater than, less than, or the same as that of the front wheel?

A. Twice greater than B. Twice less than C. Four times greater than D. Four times less than E. The same F. Non of these

Example of Rolling: roller race

Which of these get to the bottom of the ramp first?

KE

 1 2 2

Mv CM

( 1 

k

) 2

KE

 1 2 2

Mv CM

 1 2

I CM

 2  1 2 2

Mv CM

 1 2

kMR

2

v CM R k



I MR

2

Mgh

 1 2 2

Mv CM

( 1 

k

)

Example: Cylinder 1 is released on an incline and rolls down w/o slipping. Cylinder 2 has an initial angular speed at the bottom of the ramp and starts rolling up w/o slipping. What is the direction of the static friction force in each case?

A.

f s

1

Rolling down

B.

Rolling up

V relative C.

f s

2 V relative

Friction must oppose relative motion. In the absence of friction: cylinder 1 would slide down (no rotation), cylinder 2 would keep rotating at the base of the ramp without going up.

In both cases, relative velocity (at the point of contact) is down the incline, but for motion w/o slipping it should be zero, and f S must point up.

The weight and the normal force produce zero torque about the CM. To produce “clockwise” angular acceleration and the appropriate torque direction (into the page), f S must point up in both cases.

f S

Example: A cylinder of mass M and radius R rolls down an incline of angle θ with the horizontal. If the cylinder rolls without slipping, what is its acceleration?

Newton's 2 nd law for rotation: Rolling without slipping:

f s R v CM

 

I CM R

  

a CM



R



I CM

 1 2

MR

2

f s



I CM

 1

R

  2 1

MR

2   

a CM R

  1

R

Newton's 2nd law for translation of the CM:

f s

 1 2

Ma CM Mg

sin  

f s



Ma CM N Mg

sin   3 2

Ma CM a CM

 2 3

g

sin 

compare to g sin θ, the results for an object sliding without friction

mg θ

Example: A disk of radius R and mass M that mounted on a massless shaft of radius r << R and rolling down an incline with a groove. What is its acceleration?

N M

m ~ 0

f S r R I

CM  1 2

MR

2

Mg Mg a

CM sin    

f S r

CM   

Ma

CM

a

CM 

Mg M

 sin 

MR

2

r

2 2  1  1

R

2

r

2 2

g

sin 

a

CM 

Mg M

 sin

I r

 CM 2

Very small if R >> r !