chapter 10 chapter 10 rotational motion rotational motion part 2 part 2

Chapter 10Chapter 10

Rotational motionRotational motion

Part 2Part 2

Parallel axis theoremParallel axis theorem

2MhII COM

Proof:Proof:

Rotational inertia about a given axisRotational inertia about a given axis = = Rotational Inertia about a parallel axis Rotational Inertia about a parallel axis that extends trough body’s Center of that extends trough body’s Center of Mass + MhMass + Mh22

h = perpendicular distance between the given axis and axis h = perpendicular distance between the given axis and axis through COM.through COM.

dmbaydmbxdmadmyx

dmbyaxdmrI

)(22)(

)()(

2222

222

R

222 22 MhIMhbMyaMxdmRI COMCOMCOM

FrFrFrFr t )sin()sin(

Units:Units: NmNm

Tangential component, FTangential component, Ftt:: does cause rotation does cause rotation pulling a pulling a

door perpendicular to its plane. door perpendicular to its plane. FFtt= F sin= F sinφφ

Radial component, Fr : does not cause rotation pulling a door parallel to door’s plane.

TorqueTorque

Torque: Torque:

Twist “Turning action of force F ”.

rr┴┴ : Moment arm of F : Moment arm of F

r : Moment arm of Fr : Moment arm of Ftt

Sign:Sign: Torque >0 if body rotates counterclockwise. Torque >0 if body rotates counterclockwise. Torque <0 if clockwise rotation.Torque <0 if clockwise rotation.

Superposition principle:Superposition principle: When several torques act on When several torques act on a body, the net torque is the sum of the individual a body, the net torque is the sum of the individual torquestorques

Vector quantityVector quantity

Newton’s second law for rotationNewton’s second law for rotation

ImaF

Proof:Proof:

Particle can move only along the circular path only the tangential component of the force Ft (tangent to the circular path) can accelerate the particle along the path.

ImrrrmrmarF

maF

tt

tt

)()( 2

Inet

Kinetic energy of rotationKinetic energy of rotation

Reminder:Reminder: Angular velocityAngular velocity, , ωω is the same for all is the same for all particles within the rotating body. particles within the rotating body.

Linear velocityLinear velocity,, vv of a particle within the rigid body of a particle within the rigid body depends on the particle’s distance to the rotation axis (r).depends on the particle’s distance to the rotation axis (r).

2222

233

222

211

2

1)(

2

1

2

1

...2

1

2

1

2

1

i

ii

i

ii

i

ii rmrmvm

vmvmvmK

Moment of InertiaMoment of Inertia

Rotational Kinetic EnergyRotational Kinetic Energy

We must rewrite our statements of conservation of We must rewrite our statements of conservation of mechanical energy to include KEmechanical energy to include KErr

Must now allow that (in general):Must now allow that (in general):

½ mv2+mgh+ ½ I2 = constant

Could also add in e.g. spring PECould also add in e.g. spring PE

VII. Work and Rotational kinetic energy

Translation Rotation

WmvmvKKK ifif 22

2

1

2

1WIIKKK ifif 22

2

1

2

1

f

i

x

x

FdxW f

i

dW

Work-kinetic energy Theorem

Work, rotation about fixed axis

dFW )( ifW Work, constant torque

vFdt

dWP

dt

dWP Power, rotation about

fixed axis

Proof:

2222222222

2

1

2

1)(

2

1)(

2

1)(

2

1)(

2

1

2

1

2

1ififififif IImrmrrmrmmvmvKKKW

f

i

dWddrFdsFdW tt

dt

d

dt

dWP

Example - Rotational KE

What is the linear speed of a ball with radius 1 cm when it reaches the end of a 2.0 m high 30o incline?

mgh+ ½ mv2+ ½ I2 = constant Is there enough information?

2 m

Example - Rotational KE

So we have that

The velocity of the centre of mass and the

tangential speed of the sphere are the same,

so we can say that:

Rearranging for vf:

2

5

2MRISphere

222

222

5

1

2

15

2

2

1

2

1

ffi

ffi

Rvgh

MRMvMgh

R

v

R

vt

smgh

v

vvvgh

if

fffi

/3.57.0

28.9

7.0

7.05

1

2

1 222

Example: Conservation of KEExample: Conservation of KErr

A boy of mass 30 kg is flung off the edge of a A boy of mass 30 kg is flung off the edge of a roundabout (m = 400 kg, r = 1 m) that is travelling at roundabout (m = 400 kg, r = 1 m) that is travelling at 2 rpm. What is the speed of the roundabout after he 2 rpm. What is the speed of the roundabout after he falls off?falls off?

Roundabout is a disk:Roundabout is a disk: 222 200)1()400(5.02

1kgmmkgMRIR

222 30)1)(30( kgmmkgRMI BB 222 23030200 kgmkgmkgmIII BRtotal

srad

KEKEkgmIKE

JkgmIKE

f

fiffff

iii

/22.02005.0

5

)200(2

1

2

1

560

22)230(5.0

2

1

222

222

Boy has

During a certain period of time, the angular position of a During a certain period of time, the angular position of a swinging door is described by swinging door is described by

θθ= 5.00 + 10.0= 5.00 + 10.0tt + 2.00 + 2.00tt22

where where θθ is in radians and is in radians and tt is in seconds. Determine the is in seconds. Determine the angular position, angular speed, and angular angular position, angular speed, and angular acceleration of the door (a) at acceleration of the door (a) at tt = 0 = 0 and (b) at and (b) at tt = 3.00 s = 3.00 s..

Solution:Solution:

0 5.00 radt

0 00

20

0

10.0 4.00 10.0 rad s

4.00 rad s

t tt

tt

dt

dt

ddt

(a)(a)

During a certain period of time, the angular position of a During a certain period of time, the angular position of a swinging door is described by swinging door is described by

θθ= 5.00 + 10.0= 5.00 + 10.0tt + 2.00 + 2.00tt22

where where θθ is in radians and is in radians and tt is in seconds. Determine the is in seconds. Determine the angular position, angular speed, and angular angular position, angular speed, and angular acceleration of the door (a) at acceleration of the door (a) at tt = 0 = 0 and (b) at and (b) at tt = 3.00 s = 3.00 s..

Solution:Solution:

(b)(b)

3.00 s 5.00 30.0 18.0 53.0 radt

3.00 s 3.00 s3.00 s

23.00 s

3.00 s

10.0 4.00 22.0 rad s

4.00 rad s

t tt

tt

dt

dt

ddt

A rotating wheel requires A rotating wheel requires 3.00 s3.00 s to rotate through to rotate through 37.0 37.0 revolutionsrevolutions. Its angular speed at the end of the . Its angular speed at the end of the 3.00-s3.00-s interval is interval is 98.0 rad/s98.0 rad/s. What is the constant angular . What is the constant angular acceleration of the wheel?acceleration of the wheel?

212fi it t

and fi t

are two equations in two unknowns

i

2 21 12 2fi ff t t t t t

22 rad 137.0 rev 98.0 rad s 3.00 s 3.00 s

1 rev 2

2232 rad 294 rad 4.50 s

22

61.5 rad13.7 rad s

4.50 s

The four particles are connected The four particles are connected by rigid rods of negligible mass. by rigid rods of negligible mass. The origin is at the center of The origin is at the center of the rectangle. If the system the rectangle. If the system rotates in the rotates in the xyxy plane about plane about the the zz axis with an angular axis with an angular speed of speed of 6.00 rad/s6.00 rad/s, calculate , calculate (a) the moment of inertia of the (a) the moment of inertia of the system about the system about the zz axis and (b) axis and (b) the rotational kinetic energy of the rotational kinetic energy of the system.the system.

The four particles are connected by The four particles are connected by rigid rods of negligible mass. The rigid rods of negligible mass. The origin is at the center of the rectangle. origin is at the center of the rectangle. If the system rotates in the If the system rotates in the xyxy plane plane about the about the zz axis with an angular axis with an angular speed of speed of 6.00 rad/s6.00 rad/s, calculate (a) the , calculate (a) the moment of inertia of the system about moment of inertia of the system about the the zz axis and (b) the rotational axis and (b) the rotational kinetic energy of the system.kinetic energy of the system.

(a)(a)2

j jj

I mr

1 2 3 4

2 2

2

2

3.00 m 2.00 m 13.0 m

13.0 m 3.00 2.00 2.00 4.00 kg

143 kg m

r r r r

r

I

The four particles are connected by The four particles are connected by rigid rods of negligible mass. The rigid rods of negligible mass. The origin is at the center of the rectangle. origin is at the center of the rectangle. If the system rotates in the If the system rotates in the xyxy plane plane about the about the zz axis with an angular axis with an angular speed of speed of 6.00 rad/s6.00 rad/s, calculate (a) the , calculate (a) the moment of inertia of the system about moment of inertia of the system about the the zz axis and (b) the rotational axis and (b) the rotational kinetic energy of the system.kinetic energy of the system.

In this case,

(b(b)) 22 21 1

143 kg m 6.00 rad s2 2RK I

32.57 10 J

Find the net torque on the wheel in Figure about the Find the net torque on the wheel in Figure about the axle through axle through OO if if aa = 10.0 cm and = 10.0 cm and bb = 25.0 cm. = 25.0 cm.

0.100 m 12.0 N 0.250 m 9.00 N 0.250 m 10.0 N 3.55 N m

The thirty-degree angle is unnecessary information.

A block of mass m1 = 2.00 kg and a block of mass m2 = 6.00 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10.0 kg.

These blocks are allowed to move on a fixed block-wedge of angle = 30.0 as in the Figure above. The coefficient of kinetic friction is 0.360 for both blocks. Draw free-body diagrams of both blocks and of the pulley. Determine (a) the acceleration of the two blocks, and (b) the tensions in the string on both sides of the pulley.

For m1: y yF ma1 0n m g

1 1 19.6 Nn m g

1 1 7.06 Nk kf n

x xF ma 17.06 N 2.00 kgT a

For pulley, I 2

1 212

aT R T R MR

R

1 21

10.0 kg2

T T a 1 2 5.00 kgT T a

For m2:

2 2 cos 0n m g

22 6.00 kg 9.80 m s cos30.0

50.9 N

n

2 2k kf n 18.3 N

2 2 218.3 N sinT m m a

218.3 N 29.4 N 6.00 kgT a

(a) Add equations for m1, m2, and for the pulley:

17.06 N 2.00 kgT a

1 2 5.00 kgT T a

218.3 N 29.4 N 6.00 kgT a

2

7.06 N 18.3 N 29.4 N 13.0 kg

4.01 N0.309 m s

13.0 kg

a

a

(b)

21 2.00 kg 0.309 m s 7.06 N 7.67 NT

22 7.67 N 5.00 kg 0.309 m s 9.22 NT