a 40-kg mass placed 1.25 m on the opposite side of the support point balances a mass of 25 kg,...

A 40-kg mass placed 1.25 m on the opposite side of the support point balances a mass of 25 kg, placed (x) m from the support point of a uniform beam. What is the value of the unknown distance x?

Example

A large seed initially at rest explodes into two pieces which move off. Which of these could be possible paths the two pieces would take?

(I)

(II)

(III)

A ball is projected straight up. Which graph shows the total energy of the ball as a function of time?

t t t

tt

t

(A) (B) (C) (D)

(E)(F)

Chapter 8Chapter 8

Torque (Torque () ) andand

Angular Momentum (L) Angular Momentum (L)

Rotational Inertia (I)Rotational Inertia (I)

Recall: From Newton’s 2Recall: From Newton’s 2ndnd law of linear motion: law of linear motion:FFnetnet = ma = m(v-v = ma = m(v-voo)/)/t t

mass (m) = measure of inertia of an object.mass (m) = measure of inertia of an object. = measure of how difficult it is to = measure of how difficult it is to change linear velocity (v) of an object.change linear velocity (v) of an object.The larger the mass, the more difficult it is to The larger the mass, the more difficult it is to

change its velocity v.change its velocity v.

For rotation, moment of inertial or rotational For rotation, moment of inertial or rotational inertia (I) = measure of how difficult it is to inertia (I) = measure of how difficult it is to change angular velocity (change angular velocity () of an object.) of an object.


Rotational inertia (I) = measure of how difficult Rotational inertia (I) = measure of how difficult it is to change angular velocity (it is to change angular velocity () of an object.) of an object.

Rotational inertia (I) – Also called Rotational inertia (I) – Also called moment of moment of inertia:inertia:

• Depends on mass, m. (I Depends on mass, m. (I m). m).• Depends on square of radius of rotation (I Depends on square of radius of rotation (I r r22))• Depends on how mass m is distributed.Depends on how mass m is distributed.

Same mass.

Different radius.

Which one easier to get to start rotating?

Rotational InertiaRotational InertiaFor a point object, I = mr2

Point object = one whose size is small c.f. radius r

rm

Units of I = kg-m2

Rotational InertiaRotational InertiaFor a point object, I = mr2

r1m1

For many discrete point objects with different shapes rotated about the same axis,

Total rotational inertia

= sum of I for each object.

Itotal = miri2

Rotational InertiaRotational Inertia

For a point object, I = mr2

For objects of comparable size to radius r, the moment of in inertia depends on distribution of mass.

Rotational Inertia TableRotational Inertia Table


Mass of earth = 5.975 x 10Mass of earth = 5.975 x 102424 kg kgMean radius = 6.37 x 10Mean radius = 6.37 x 1066 m mEarth-sun distance = 1.5 10Earth-sun distance = 1.5 101111 m m

What is the rotational inertia of What is the rotational inertia of the earth’s spin about its axis?the earth’s spin about its axis?

ExampleExample

Two solid spherical balls are joined Two solid spherical balls are joined together by a 4-meter long steel rod. If together by a 4-meter long steel rod. If they are spun about a vertical axis passing they are spun about a vertical axis passing through their center of mass, find the total through their center of mass, find the total rotational inertia. [Hint, treat the spinning rotational inertia. [Hint, treat the spinning balls as point objects with I = mRballs as point objects with I = mR22].].

m1 = 3 kg m2 = 5 kg


Linear Angular (Rotational)

Displacement (x) Angle () in radians

Velocity (v) Angular Velocity ()

Acceleration (a) Angular Acceleration ()

Mass (m) Rotational Inertia (I)

Ktran = ½ mv2 Krot = ½ I 2

Rotational KinematicsRotational Kinematics

Linear Rotationala = constant = constant

v = vo + at = o + t

x = xo + vot + ½at2 = o + ot + ½ t2

v2 = vo2 + 2a(x - xo) 2 = o

2 + 2 ( - o)

Comment on axes and signComment on axes and sign(i.e. what is positive and negative)(i.e. what is positive and negative)

Whenever we talk about rotation, it is implied that there is a rotation “axis”.

This is usually called the “z” axis (we usually omit the z subscript for simplicity).

Counter-clockwise (increasing ) is usuallycalled positive.

Clockwise (decreasing ) is usuallycalled negative.

z

ExampleExampleYou and a friend are playing on a merry-go-round. You stand at the outer edge of the merry-go-round and your friend stands halfway between the outer edge and the center. Assume the rotation rate of the merry-go-round is constant. Who has the greatest angular velocity?

1. You do 2. Your friend does 3. Same

ExampleExampleWho has the greatest tangential velocity (v)?


Rotational Kinetic Energy Rotational Kinetic Energy

mr

v

Mass m in rotational motion.

Its rotational inertia, I = mr2

Since it is moving, it has kinetic energy.

K = ½ mv2

From v = r,

K = ½ mr22 = ½ (mr2) 2 = ½I2

Kinetic Energy of Rotating DiskKinetic Energy of Rotating Disk

Consider a disk with radius R and mass M, spinning with angular frequency Each “piece” of disk has speed vi=ri

Each “piece” has kinetic energy »Ki = ½ mi v2

» = ½ mi 2 ri2

Combine all the pieces » Ki = ½ mi 2 ri

2 » = ½ ( mi ri

2) 2 » = ½ I 2

ri

mi

Rotational Kinetic Energy (KRotational Kinetic Energy (Krotrot) )

A rigid object spinning about a fixed axis (pure rotational motion) has rotational kinetic energy Krot = ½I2 [eg, spinning wheel]

If the rigid object is moving (sliding) with velocity v without any spin (ie pure translational motion), it has only translational kinetic energy K = ½ mv2 [eg, skidding wheel]

Rotational Kinetic Energy (KRotational Kinetic Energy (Krotrot) )

If the rigid object is spinning with angular velocity while its center of mass moves linearly with velocity vcm, it has both translational and rotational kinetic energy.

Its total kinetic energy will be

Ktot = Ktran + Krot

½ mvcm2 + ½ I2

[eg, wheel rolling normally on the ground, ball rolling on the ground]

Ktran = ½ m v2 Linear Motion

Krot = ½ I 2 Rotational Motion

ExampleExample

Who has the greater kinetic energy?

(Assume masses are equal)


ExampleExample

A 10-kg hollow cylindrical shell rolls on the A 10-kg hollow cylindrical shell rolls on the ground at a linear velocity of 5 m/s. Find ground at a linear velocity of 5 m/s. Find its total kinetic energy. its total kinetic energy.

50 cm 50 cm

Torque (Torque (

Force (F) is responsible for change of linear velocity. Net force results in linear acceleration.

Torque ( is responsible for change in angular velocity. Net torque results in angular acceleration.

TorqueTorquer

= rF

= (rsin )F = rFsin

r = rsin = lever arm (moment arm)

Torque = lever arm x force

Unit: m-N

F

r

Larger lever arm r, larger torque if F stays

unchanged

Larger force F, larger torque if lever arm

stays unchanged

3. If r = 0,

If = 0,

If = 90o, rF

F

r

= rF = rsinF

TorqueTorque

Counter clockwise rotation, torque is positive

F

F

Clockwise rotation, torque is negative

Torque is a vector quantity.

A string is tied to a doorknob 0.80 m from the hinge as shown in the figure. At the instant shown, the force applied to the string is 5.0 N. What is the torque on the door?

hinge

EquilibriumEquilibrium

Recall, translational equilibrium – Fnet = 0. Fnet = 0, object at rest – static equilibrium. Fnet = 0, object moving with constant

velocity, - dynamic equilibrium.

Rotational equilibrium, net = 0. An object in equilibrium means it is in both

translational equilibrium and rotational equilibrium.

Conditions for equilibrium:

Fnet = 0 and net = 0

EquilibriumEquilibrium

Conditions for equilibrium:

Fnet = 0 and net = 0

Torque can be calculated about any desired axis.

A judicious choice of axis helps.

Choose axis at a point through which an unknown force acts so that its torque does not appear in the equation.

Is the point through which the force of gravity acts.

Essentially the same as center of mass.

Center of Gravity

1.5 m2.5 m

M1 = 50 kg M2 = 60 kg

pivot

The figure above is a snapshot of two masses on a light beam placed on a pivot.

• What is the net torque about the pivot?

• To be in rotational equilibrium, what should have been the value of M2?

ExampleExampleThe picture below shows two people lifting a

heavy log. Which of the two people is supporting the greatest weight?

1. The person on the left is supporting the greatest weight

2. The person on the right is supporting the greatest weight

3. They are supporting the same weight

ExampleExampleThe picture below shows two people (L &

R) lifting a 20-kg log. The length of the log is 4.0 m and L is 1.0 m from the left end while R is holding from the right end. Find how much force each of them have to exert.

L R

Newton’s 2nd law for linear motion:

Fnet = ma

Eg. What force is required to change velocity of a 6 kg solid spherical object of radius 30 cm object from 15 m/s to 25 m/s in 5 seconds?

Newton’s 2nd law for rotation:

net = IEg. What torque is required to change angular

velocity of a 6 kg solid spherical object of radius 30 cm object from 15 rad/s to 25 rad/s in 5 seconds? [Assume spin about the axis]

Newton’s 2nd Law

Work Done by TorqueWork Done by Torque Recall, work done by constant force

W = F x cos

Also, Wnet = K = ½ mvf2 – ½ mvi

2

For rotation, work done by constant torque:

W = Ftangential s

= Ftangential r = Also, Wnet = Krot = ½ If

2 – ½ Ii2

PowerP = W/t = /t

Work Done by TorqueWork Done by Torque

A 182-kg flywheel has an effective radius of 0.62 m.

(a)How much torque will take it from rest to a rotational speed of 120 rpm within 30.0 s?

(b) How much work will this take?

A rolling object posses both Ktran (½ mv2CM)

and Krot (½ I2)

Rolling Object

E1 = U1 + K1 = mgh + 0

E2 = U2 + K2

0 + ½ mv2 + ½ I2

h

• If it came down rolling: mgh = ½ mvA2 + ½ I2

• If it came down only skidding, mgh = ½ mvB2

Angular Momentum (L)

Recall: Linear Momentum: p = mv

Angular Momentum, L = I

Angular momentum of a rigid object rotating about a fixed axis.

L is a vector quantity.

Units = kg-m2/s

Direction: If rotation is CCW, L is upward.

If rotation is CW, L is downward.

Conservation of Angular MomentumIf net external torque acting on a rotating object

is zero, total angular momentum will be conserved.

If net = 0, Lbefore = Lafter

OR Iii = Iff

EG: Skater spinning with her arms extended away from her body. Lbefore = Iii = (mri

2)i

She then decides to pull her arms close to her body so that her spinning radius becomes smaller

(rf < ri) and Lafter = Iff = (mrf2)f

Conservation of Angular Momentum

Since no external torque acts on her, principle of conservation of angular momentum will hold:

Lbefore = Lafter or Iii = Iff or (mri2)i = (mrf

2)f

Since ri > rf, i < f

ie, the skater will spin faster by just folding her arms.

Example

• A 1000-kg merry-go-round of radius 8.0 m is spinning at a uniform speed of 25 rpm. If a 150-kg mass is gently placed at the edge, what will be the final spin speed of the merry-go-round?

Example• Which of the forces in the figure below

produces the largest torque about the rotation axis indicated? Assume all the four forces have equal magnitudes.

Axis

1

2

3

4

(A) 2(B) 3(C) 4(D) The torques are equal

A 250-kg merry-go-round of radius 2.2 m is spinning at a uniform angular speed of 3.5 rad/s. Calculate the magnitude of its angular momentum. Use I = ½ MR2

A. B. C. D. E.

3% 0%5%

93%

0%

A. 605 kg-m2/s

B. 1.93 x 103 kg-m2/s

C. 7.41 x 103 kg-m2/s

D. 2.12 x 103 kg-m2/s

E. 963 kg-m2/s

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

61 62 63 64 65

A 30.0-cm long wrench is used to generate a torque of 14.6 N-m at a bolt. A force of 70.0 N is applied at the end of the wrench at an angle of to the wrench. The angle is

A. B. C. D.

95%

2%0%2%

A. 44o

B. 0.4o

C. 9.2o

D. 3.6o

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

61 62 63 64 65

A 3.2-kg solid sphere of diameter 28 cm spins at 5.0 rad/s about an axis passing through its center. How much work is needed to bring it to rest within 15 sec? [Use I = 2/5 x mr2]

A. B. C. D. E.

0%

8%

80%

8%5%

A. 0.025 J

B. 1.25 J

C. 1.25x104 J

D. 3.14 x 103 J

E. 0.31 J

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

61 62 63 64 65

Abel and Brian support a 68.0-kg uniform bar that is 4.0 m long. Abel holds the bar 1.0 m from one end, while Brian holds from the other end. How much force does Abel have to exert?

A. B. C. D. E.

50%

10%

30%

5%5%

A. 444 N

B. 45.3 N

C. 1.33x103 N

D. 333 N

E. 222 N

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

61 62 63 64 65

a 40-kg mass placed 1.25 m on the opposite side of the support point balances a mass of 25 kg,...

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