newton’s laws phy1012f rotation gregor leigh gregor.leigh@uct.ac.za

NEWTON’S LAWS

PHY1012

F

ROTATIO

N

Gregor Leighgregor.leigh@uct.ac.za

NEWTON’S LAWS ROTATION OF A RIGID BODYPHY1012F

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ROTATION OF A RIGID BODY

Learning outcomes:At the end of this chapter you should be able to…

Extend the ideas, skills and problem-solving techniques developed in kinematics and dynamics (particularly with regard to circular motion) to the rotation of rigid bodies.

Calculate torques and moments of inertia.

Apply appropriate mathematical representations (equations) in order to solve rotation problems.

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ROTATIONAL KINEMATICS

Rigid body model:A rigid body is an extended object whose shape and size do not change as it moves. Neither does it flex or bend.

Types of motion:Translational motion: Rotational motion: Combination motion:

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ANGULAR ACCELERATION

Linear motion relationship Rotational motion

tds vdt

ddt

tt

dva

dtd

dt

Units: [rad/s2]

A body’s angular acceleration, , is the rate at which its angular velocity changes.

s s = r

vt = r

at = r

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Angular acceleration,

slower

ANGULAR VELOCITY and ANGULAR ACCELERATION

An

gu

lar

velo

city

,

+

+ –

–

> 0

> 0faster

< 0

> 0

faster

> 0

< 0slower

< 0

< 0

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3 6 9 12 t (s)

(rad/s)

0

–3

3

For the first 6 s the

is

VELOCITY GRAPHS ACCELERATION GRAPHS

Angular acceleration is equivalent to the slope of a -vs-t graph.

23 0 0.5 rad/s

6 0 st

Eg: A wheel rotates about its axle…

accelerationslope

(rad/s2)

½

–½

–

03 6 9 12 t (s)

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3 6 9 12 t (s)

(rad/s)

0

–3

3

For the last 6 s the

is

VELOCITY GRAPHS ACCELERATION GRAPHS

Angular acceleration is equivalent to the slope of a -vs-t graph.

23 3 rad/s12 6 st

Eg: A wheel rotates about its axle…

accelerationslope

(rad/s2)

½

–½

–

03 6 9 12 t (s)

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KINEMATIC EQUATIONS

Linear motion Rotational motion

The following equations apply for constant acceleration:

vf = vi + at

xf = xi + vit + ½a (t)2

vf2 = vi

2 + 2ax

f = i + t

f = i + it + ½ (t)2

f2 = i

2 + 2

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CENTRE OF MASS

While the various points on a flipped spanner describe different (complicated) trajectories, one special point, the centre of mass, follows the usual, simple parabolic path.The centre of mass (CM) of a system of particles…

is the weighted mean position of the system’s mass;

is the point which behaves as though all of the system’s mass were concentrated there, and all external forces were applied there;

is the point around which an unconstrained system (i.e. one without an axle or pivot) will naturally rotate.

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xCM

x1 = 0

x2

CENTRE OF MASS

To find the centre of mass of two masses…

m1

m2x

1. place the masses on an x-axis, with one of the masses at the origin;

2. apply the formula: 1 1 2 2CM

1 2

m x m xx

m m

In general, for many particles on any axis:And for a continuous distribution of mass in which mi = M:

CMi i

i

m ss

m

CM1s sdmM

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TORQUE

The rotational analogue of force is called torque, .

Torque may be regarded as…

the “amount of turning” required to rotate a body around a certain point called an axis or a pivot;

the effectiveness of a force at causing turning.

E.g. To push open a heavy door around its hinge (as seen from the top) requires a force applied at some point on the door.

Consider the effectiveness of each of the (equal) forces shown…

hinge4F

1F

3F

2F

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Three factors determine the amount of torque achieved:

magnitude of the applied force;

distance between the point of application and the pivot;

angle at which the force is applied.

TORQUE

y

x

F

pivot

point of application

radial line

Only the tangential component of the applied force produces any turning…

Ft = Fsin

Fr = rFt

Hence: rFsin

Units: [N m] ( joule!)

r

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Torque…is positive if it tries to rotate the object anticlockwise about the pivot; negative if rotation is clockwise;

must be measured relative to a specific pivot point.

TORQUE

Torque can also be defined as the product of the force, F, and the perpendicular distance between the pivot and the line of action of the force, d, known as the torque arm, moment arm, or lever arm:

| | = dF

x

F

pivot

line of action

torque

arm d = r sin

r

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NET TORQUE

Several forces act on an extended object which is free to rotate around an axle.

net = 1 + 2 + 3 + … = i

The axle prevents any translational movement, so…

1F

axle 2F

3F

axleF

4F

And the net torque is given by:

net 1 2 3 axle 0F F F F F

( causes no torque since it is applied at the axle.)axleF

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Mg

pivot

GRAVITATIONAL TORQUE

For an object to be balanced, its centre of mass must lie either directly above or directly below the point of support.

grav = –MgxCM

centre of mass

If this is not so, the body’s own weight, acting through its centre of mass (as if all its mass were concentrated there), causes a net torque due to gravity:

where xCM is the distance

between the centre of mass and the pivot.

0 xCMx

CM

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COUPLES

Rotation without translation is achieved by the application of a pair of equal but opposite forces at two different points on the object.

|net | = lF (sign by inspection)

pivot

1F

2Fd1 d2

l

|net | = d1F + d2F = (d1 + d2)F

The pivot is immaterial – a couple will exert the same net torque lF about any point on the object.

Unless the rotation is constrained to act around a specific pivot, it will occur around the body’s centre of mass.

Notes:

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rod

ROTATIONAL DYNAMICS

Newton II (linear): Force causes linear acceleration.

Linear acceleration is “limited” by inertial mass.

rFt = mr2 = mr2

A rocket is constrained to move in a circle by a lightweight rod…

Ft = mat Ft = mr

netFa

m

y

x

thrustF

pivot

Ft

FrT

Newton II (rotational): Torque causes angular acceleration.

Angular acceleration is “limited” by the particle’s rotational inertia, mr2, about the pivot.

net2mr

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The body’s rotational inertia…is also known as its moment of inertia, I;

is the aggregate of the individual (mr2)’s:

gives an indication of how the mass of the body is distributed about its axis of rotation (pivot);

is the rotational equivalent of mass.

ROTATIONAL INERTIA

A rotating extended object can be modelled as a collection of particles, each a certain distance from the pivot. pivot

m1

m2

m3

r3

r2

r1

I = m1r12 + m2r2

2 + m3r32 + … = miri

2

Thus: net

I

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Linear dynamics Rotational dynamics

ROTATIONAL DYNAMICS

Summary of corresponding quantities and relationships:

force net

m

torque

inertial mass moment of inertia I

acceleration angular acceleration

Newton II

Fnet

Fnet = ma

a

net = INewton II

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MOMENTS OF INERTIA

For an extended system with a continuous distribution of mass, the system is divided into equal-mass elements, m.Then, by allowing these to shrink in size, the moment of inertia summation is converted to an integration:2 2

0i mI r m I r dm

pivot

mr

y

x

y

x

2 2I x y dm

…and, before integration, dm is replaced by an expression involving a coordinate differential such as dx or dy.

For complex distributions of mass, r is usually replaced by x and y components…

NEWTON’S LAWS

MOMENTS OF INERTIA

For simple, uniform distributions of mass, however, the integration can be trivial.

R

In practice, the rotational inertias of certain common shapes (of uniform density) are looked up in tables…

E.g. in a wheel, or hoop, where all the mass lies at a distance R from the axis… 2I r dm becomes

,

2I R dm where the integral is simply the sum of all the mass elements, i.e. the total mass, M, of the wheel.

So for open wheels (hoops) I = MR2.

dm

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PARALLEL AXIS THEOREM

Moments of inertia are always calculated/stated with respect to a specific axis of rotation.

However (assuming the moment of inertia for rotation around the centre of mass is known), the moment of inertia around any off-centre rotation axis lying parallel to the axis through the centre of mass can be found using the parallel-axis theorem:

I = ICM + Md2

d

CM

M (mass)

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ROTATION ABOUT A FIXED AXIS

1. Model the object as a simple shape.

2. Identify the axis around which the object rotates.

3. Draw a picture of the situation, including coordinate axes, symbols and known information.

4. Identify all the significant forces acting on the object and determine the distance of each force from the axis.

5. Determine all torques, including their signs.

6. Apply Newton II: net = I. (I-values from tables and the parallel-axis theorem.)

7. Use rotational kinematics to find angular positions and velocities.

Problem-solving strategy:

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ROTATION ABOUT A FIXED AXISA 70 g metre stick pivoted freely at one end is released from a horizontal position. At what speed does the far end swing through its lowest position?

1-3. Model the stick as a uniform rod rotating around one end, and draw a picture with pivot, x-axis and data.

0 xpivot

M = 0.07 kg

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ROTATION ABOUT A FIXED AXISA 70 g metre stick pivoted freely at one end is released from a horizontal position. At what speed does the far end swing through its lowest position?

4. Identify all significant forces acting on the object and determine the distance of each force from the axis.

Mg

0 xCM = 0.5 m xpivot

M = 0.07 kg

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M = 0.07 kg

ROTATION ABOUT A FIXED AXISA 70 g metre stick pivoted freely at one end is released from a horizontal position. At what speed does the far end swing through its lowest position?

5. Determine all torques, including their signs.

Mg

0 xCM = 0.5 m xpivot

–ve

grav = –MgxCM = –0.07 9.8 0.5 = –0.34 Nm

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ROTATION ABOUT A FIXED AXISA 70 g metre stick pivoted freely at one end is released from a horizontal position. At what speed does the far end swing through its lowest position?

6. Apply Newton II: net = I.

Mg

xpivot

grav = –0.34 Nm

net = I

= –15 rad/s2

210.343

ML M = 0.07 kg

210.34 0.07 13

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ROTATION ABOUT A FIXED AXISA 70 g metre stick pivoted freely at one end is released from a horizontal position. At what speed does the far end swing through its lowest position?

7. Use rotational kinematics to find angular positions and velocities.

pivotf

2 = i2 + 2

f2 = 0 + 2(–15)(–0.5 )

= –15 rad/s2

i = 0 rad i = 0 rad/s

f = –0.5 rad f = ?

f 15 6.8 rad/s

vt = r

vt = –6.8 1 = 6.8 m/s

You canNOT use rotational

kinematics to solve this problem!

Why not?

(Not this time!)

f 15 6.8 rad/s

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CONSTRAINTS DUE TO ROPES and PULLEYS

Provided it does not slip, a rope passing over a pulley moves in the same way as the pulley’s rim, and thus also objects attached to the rope.

As before, the constraints are given as magnitudes. Actual signs are chosen by inspection.

Rnon-slipping rope

rim acceleration = | |R

rim speed = | |R

vobj = | |R

aobj = | |R

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Two blocks are connected by a light string which passes over two identical pulleys, each with a moment of inertia I, as shown. Find the acceleration of each mass and the tensions T1 , T2 and T3.

m1 m2

T1

T2

T3

m1m2

x

y

–ve

+ve

m1g m2g

T1

T3n2T2

T3w

n1T2

T1 w–ve

+ve

F1y = T1 – m1g = m1a (1) F2y = T3 – m2g = –m2a (2)

net = T1R – T2R = –I (3) net = T2R – T3R = –I (4)

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m1 m2

T1

T2

T3

x

y

F1y = T1 – m1g = m1a (1)

F2y = T3 – m2g = –m2a (2)

net = T1R – T2R = –I (3)

net = T2R – T3R = –I (4)

(3) + (4): T1R – T3R = –2I

1 32IT TR

(1) – (2): T1 – T3 + m2g – m1g = m1a + m2a

2 1 1 222I a m m g m m aR

2 1

1 2 22

m m ga

Im mR

…etc

22I aR

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RIGID BODY EQUILIBRIUM

Problem-solving strategy:

1. Model the object as a simple shape.

2. Draw a picture of the situation, including coordinate axes, symbols and known information.

3. Identify all the significant forces acting on the object.

4. Choose a convenient pivot point and determine the moment arm of each force from it.

5. Determine the sign of each torque around the pivot.

6. Write equations for Fx = 0; Fy = 0; net = 0; and solve.

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RIGID BODY EQUILIBRIUMA 3-metre ladder leans against a frictionless wall, making an angle of 60° with the ground. What minimum coefficient of friction is needed to prevent the foot of the ladder from slipping?

1-2. Model the ladder as a rigid rod, and draw a picture with axes, symbols and known information.

x

y

CM

L

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RIGID BODY EQUILIBRIUMA 3-metre ladder leans against a frictionless wall, making an angle of 60° with the ground. What minimum coefficient of friction is needed to prevent the foot of the ladder from slipping?

3. Identify all the significant forces acting on the object.

x

y

CM

sf

w

fn

wn

L

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RIGID BODY EQUILIBRIUMA 3-metre ladder leans against a frictionless wall, making an angle of 60° with the ground. What minimum coefficient of friction is needed to prevent the foot of the ladder from slipping?

4. Choose a convenient pivot point and determine the moment arm of each force from it.

x

y

CM

sf

w

fn

wn

pivot

L

d2

d1

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RIGID BODY EQUILIBRIUMA 3-metre ladder leans against a frictionless wall, making an angle of 60° with the ground. What minimum coefficient of friction is needed to prevent the foot of the ladder from slipping?

5. Determine the sign of each torque around the pivot.

x

y

CM

sf

w

fn

wn

pivotd1

L

–ve

+ve

d2

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RIGID BODY EQUILIBRIUMFx = nw – fs = 0 (1)

6. Write equations for Fx = 0; Fy = 0; net = 0; and

solve.

x

y

CM

sf

w

fn

wn

pivotd1

d2

L

–ve

net = ½(Lcos60°)Mg – (Lsin60°)nw=

0 12

wcos60sin60 2tan60

L Mg Mgn

L

Fy = nf – Mg = 0 (2)

net = d1w – d2nw = 0 (3)

s1 0.29

2tan60

s fn sMgs 2tan60Mg

f

+ve

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ROTATION OF A RIGID BODY

Learning outcomes:At the end of this chapter you should be able to…

Extend the ideas, skills and problem-solving techniques developed in kinematics and dynamics (particularly with regard to circular motion) to the rotation of rigid bodies.

Calculate torques and moments of inertia.

Apply appropriate mathematical representations (equations) in order to solve rotation problems.