Chapter VI Dynamics of Rotational Motion

Chapter VIDynamics of Rotational MotionA. Rotational KinematicsB. Relating Linear and Angular KinematicsC. Moment of Inertia and Rotational Kinetic

EnergyD. Calculating the Moment of InertiaE. TorqueF. Angular MomentumG. Work-Kinetic Energy TheoremH. Torque and Angular Acceleration for a Rigid

BodyI. Rolling Without Slipping J. Conservation Of Angular MomentumK. Static Equilibrium

A. Rotational Kinematics

One rotation, S = 2r2 = angle = S = r = s/r

Rotation with Constant Angular Acceleration

B. Relating Linear and Angular Kinematics

S = rds/dt = (d/dt) rv = rv = x rdv/dt = (d/dt) raT = raT = x ras = v2/r = (v/r)v = vas = x v

aT & v

x

y

Z

as

&

C. Moment of Inertia and Rotational Kinetic Energy

A particleK1 = ½ m1 v1

2 = ½ m1 r12 2

System particleK = K1 + K2 + K3 + ...K = ½ m1 r1

2 2 + ½ m2 r22 2 + ...

K = ½(m1 r12 + m2 r2

2 + ...) 2

K = ½ ( mi ri2) 2

K = ½ I 2

moment of Inertia= I = mi ri2

D. Calculating the Moment of Inertia

moment of Inertia= I = mi ri2 discrete mass distribution

I = r2 dm continuous mass distribution

An engineer is designing a machine part consisting of three heavy disks linked by lightweight struts (Fig.). (a) What is the moment of inertia of this body about an axis through the center of disk A, perpendicular to the plane of the diagram? (b) What is the moment of inertia about an axis through the centers of disks B and C? (c) If the body rotates about an axis through A perpendicular to the plane of the diagram, with angular speed = 4.0 rad/s, what is its kinetic energy?

a)

b)

c)

rr1

r2

dr

M L

r1

r2

I = r2 dmdm = dV = 2rL dr I = r2 2rL dr = 2L r3 drI = 2L ¼ (r2

4 – r14)

I = ½ L (r22 – r1

2) (r22 + r1

2)M = V = ( r2

2L - r12L)

M = L (r22 – r1

2)I = ½ M (r2

2 + r12)

If, r1 = 0, than I = ½ M r2 solid cylinderIf, r1 = r2, than I = M r2 Thin-walled hollow cylinder

The parallelaxis theorem

There is a simple relationship between the moment of inertia Icm of a body of mass M about an axis through its center of mass and the moment of inertia Ip about any other axis parallel to the original one but displaced from it by a distance d. This relationship, called the parallelaxis theorem, states that,

Ip = Icm + M d 2

The moment of inertia Icm of the slice about the axis through the center of mass (at 0) is

The moment of inertia of the slice about the axis through P is

We then expand the squared terms and regroup, and obtain

The second and third sums are roportional to X ern and Yern; these are zero because we have taken our origin to be the center of mass.

E. Torque

x

y

r

r

F

= r x F = r F sin

F. Angular Momentum

x

y

r

r

p

l = r x pI = r p sin

Angular Momentum a Particle

dl/dt = dr/dt x p + r x dp/dtdl/dt = v x mv + r x Fdl/dt =

Angular Momentum System Particle

L = l1 + l2 + l3 + ...dL/dt = dl1/dt + dl2/dt + dl3/dt + ...dL/dt = 1 + 2 + 3 + ...dL/dt = ext

G. Work-Kinetic Energy Theorem

W = F.S = F cos α r = F sin rdW/dt = (d/dt)Power = P =

x

y

r

r

F

Work-Kinetic Energy Theoremw = KdW/dt = dK/dtdW/dt = d(½ I 2 )/dtP = ½I d(2)/dt = ½I (d2/d) (d/dt)P = ½I 2 αP = I αP = = I α

H. Torque and Angular Acceleration for a Rigid Body

For a particle

F1 tan = m1 a1 tan

1 z = F1 tan r1 = m1 a1 tan r1

1 z = m1 r1 αz r1 = m1 r12 αz

For a rigid body

z = i z = mi ri2 αz

z = I αz (rotational analog

of Newton's second law for

a rigid body)

= I α T R = ½ M R2 (a/R) T = ½ M a ................... (1)Fy = m a mg – T = ma .............. (2)(1) (2) mg – ½ M a = mamg = ½ M a + ma = a (½ M + m)a = mg/(½ M + m)

I. Rolling Without Slipping IP = ICM + M R2

K = ½ IP 2

K = ½ (ICM + M R2) 2

K = ½ ICM 2 + ½ M R2 2

The total kinetic energy of a rolling object is the sum of the rotational kinetic energy about the center of mass and the translational kinetic energy of the center of mass.

Conservation of Mechanical Energy

R

M

J. CONSERVATION OF ANGULAR MOMENTUM

Li = Lf Ii i = If f

Decrease in the moment of inertia of the skater causes an increase in the angular speed.

Example.An acrobatic physics professor stands at the center of a turntable, holding his arms extended horizontally with a 5.0-kg dumbbell in each hand (Fig. ). He is set rotating about a vertical axis, making one revolution in 2.0 s. Find the prof's new angular velocity ifhe pulls the dumbbells in to his stomach. His moment of inertia (without the dumbbells) is 3.0 kg . m 2 when his arms are outstretched, dropping to 2.2 kg . m 2 when his hands are at his stomach. The dumbbells are 1.0 m from the axis initially and 0.20 m from it at the end. Treat the dumbbells as particles.

We have 1z = (0.50 rev/s)(2 rad/rev) = 3.14 rad/s and 2z, = (2.5 rev/s)(2 rad/rev) = 15.7 rad/s. The initial kinetic energy is

The extra kinetic energy came from the work that the prof did to pull his arms and the dumbbells inward.

K. Static Equilibrium

Slippery wall

NAA

OfO

NO

WL WP

a

b

½ L

⅓ L

Fx = 0 ; Fy = 0 ; = 0

Fx = 0 NA – fO = 0 ................(1)Fy = 0 NO – WL – WP = 0 .....(2) = 0 We use the convention that the sign of the torque resulting from a force is positive if the turning tendency of the force is counterclockwise and is negative if the turning tendency is clockwise.When we take the torques about an axis through the origin O at the bottom of the ladder, we have

½a WL + ⅓ a WP – b NA = 0 ......(3)