Rotation with Constant Angular Acceleration

When the angular acceleration is constant we can derive simple expressionsthat give us the angular velocity and the angular posit

αω ion as function of time

We could derive these equations in the same way we did in chapter 2 Instead weθ

We could derive these equations in the same way we did in chapter 2. Instead we will simply write the solutions by exploiting the analogy between translationaland rotational motion using the following correspondance between the two motions

Rotational Motion Translational Motion x θ↔

xv

θ↔↔

aωα↔

02

02

(eqs.1)

(eqs.2)

2

2o o o

v v at

atx x v t tt

tω ω α

αθ θ ω

= +

= + +

= +↔

↔ = + +

( ) ( )2 2 22

( q )

2

(2

2 2

o o

o oo o

o

v v xa x ω ω α θ θ− = − − = −↔ eqs.3)

In fig.a we show a body which can rotate about an axis through

i t O d th ti f f li d t i t P di t

Torque

F point O under the action of a force applied at point P a distance

from O. In fig.b we resolve into two componets, ar dial

F

r F and tangential. The radial component cannot cause any rotationrFg p ybecause it acts along a line that passes through O. The tangentialcomponent sin on the other hand causes the rotation of the

r

tF F φ=

object about O. The ability of to rotate the body depends on themagnitude and also on the distance between points P and A.Thus we define as sitorque n

t

rF rF r F

FF r

τ φ ⊥= = =Thus we define as sitorque The distance is known a

ntrF rF r Fr

τ φ ⊥

⊥ s the and it is the

perpendicular distance between point O and the vector

moment arm

FThe algebraic sign of the torque is asigned as follows:

If a force tends to rotate an object in the coubtercF lockwise

di ti th i i iti If f t d t t tFdirection the sign is positive. If a force tends to rotate an object in the clockwise direction the sign is negative.

F

r Fτ ⊥= (10-13)

For translational motion Newton's second law connects Newton's Second Law for Rotation

the force acting on a particle with the resulting accelerationThere is a similar relationship between the torque of a forceapplied on a rigid object and the resulting angular acceleration

This equation is known as Newton's second law for rotation. We will explore q pthis law by studying a simple body which consists of a point mass at the end

of a massless rod of length . A force F is

m

r applied on the particle and rotates

the system about an axis at the origin. As we did earlier, we resolve F into a tangential and a radial component. The tangential component is responsible f th t ti W fi t l N t ' d l f ( 1)F Ffor the

( ) ( )2

rotation. We first apply Newton's second law for . (eqs.1)The torque acting on the particle is: (eqs.2) We eliminate

between equations 1 and 2:

t t t

t t

F F maF r F

ma r m r r mr I

τ τ

τ α α α

==

= = = =( ) ( )between equations 1 and 2:

tma r m r r mr Iτ α α α= = = =

(compare with: )F ma=Iτ α= (10-14)

Analogies between translational and rotational Motion

Rotational Motion Translational Motion

x θ↔

v ω↔

00

av v at t

αω ω α= + = +

↔↔ 0

2

0

2

2

2

o o o o

v v at

atx x v t

t

tt

ω ω α

αθ θ ω

+

= +

+

= +

↔

↔+ +

( ) ( )2 22 2 2 2o oo ov v a x x ω ω α θ θ−↔− − =− =2 2mvK IK ω

== ↔ 2

2

K

mF ma

K

IIτ α=↔=

↔

↔

F maF

P Fv

Iτ ατ=

↔=

↔↔

=

P τω=