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ROTATION OF A RIGID BODYLearning 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 KINEMATICSRigid 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 d
dt
tt
dvadt
ddt
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
Angu
lar v
eloc
ity, +
+ –
–
> 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 GRAPHSAngular acceleration is equivalent to the slope of a -vs-t graph.
23 0 0.5 rad/s6 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 GRAPHSAngular 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 MASSWhile 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 MASSTo find the centre of mass of two masses…
m1
m2 x
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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TORQUEThe 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: rFsinUnits: [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.
TORQUETorque 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 TORQUESeveral 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 TORQUEFor 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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COUPLESRotation 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 INERTIAA 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: netI
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Linear dynamics Rotational dynamics
ROTATIONAL DYNAMICSSummary 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 INERTIAFor 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 INERTIAFor 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 THEOREMMoments 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
dCM
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.
Mgx
pivotgrav = –0.34 Nm
net = I
= –15 rad/s2
210.34 3 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 PULLEYSProvided 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 3
2IT T R
(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 m
R
…etc
22I aR
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RIGID BODY EQUILIBRIUMProblem-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
CML
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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 1
2w
cos60sin60 2tan60
L Mg Mgn L
Fy = nf – Mg = 0 (2)net = d1w – d2nw = 0 (3)
s1 0.292tan60
s fn sMgs 2tan60Mgf
+ve
NEWTON’S LAWS ROTATION OF A RIGID BODYPHY1012F
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ROTATION OF A RIGID BODYLearning 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.