Transcript Rotational Dynamics - curtehrenstrom.com

Rotational Dynamics
Just as the description of rotary motion is
analogous to translational motion, the causes of
angular motion are analogous to the causes of
translational motion:
Newton’s Second Law
Translational Motion
F = ma
This relationship is true, but it disregards the
location of the force on the body!
Accounting for not only the force but also its
location on the object is torque!
• A torque is a twist as a force is
a push or pull
Torque is the vector cross product of the
displacement of the force with respect to an
arbitrary origin and the force that’s acting.
 = r x F
units: N• m
The magnitude of torque can be found:
ø is smallest angle between r and
 = rFsinø
F when tail to tail
ø
r
M
 = rFsinø
The direction of torque
will be found as any
vector cross product
(right hand rule).
F
As viewed from above–
counterclockwise (out of board) is
negative
clockwise (into board) is positive
Find the resultant net torque if:
F1 = 4.20 N, r1 = 1.30 m, ø1 = 75.0˚
F2 = 4.90 N, r2 = 2.15 m, ø2 = 58.0˚
τ = τ1 – τ2
Vector Cross Product- unit vector notation
A = (Axi + Ayj + Azk)
B = (Bxi + Byj + Bzk)
To find A X B use a determinant:
i
j
k
Ax
Ay
Az
Ax
Ay
Az
Ax
Ay
Az
Bx
By Bz
Bx
By Bz
Bx
By
Bz
AXB=
(AyBz – AzBy)i + (AzBx – AxBz)j + (AxBy – AyBx)k
Torque and Moment of Inertia
A net force is directly proportional to inertia and
acceleration: ΣF = ma
An object does not want to change its rotational
motion any more than its translational motion!
A net torque is directly proportional to rotational
inertia and angular acceleration.
• this resistance to angular acceleration is known
as moment of inertia (I ) with units: kg-m2
Net Torque would then be a product of the rotational
inertia and angular acceleration of an object:
Στ = Iα
The pulley shown in the illustration
has a radius of 2.70 m and a moment
of inertia of 39.0 kg∙m2. The hanging
mass is 4.20 kg and it exerts a force
tangent to the edge of the pulley. What
is the angular acceleration of the
pulley?
A string is wound tight around the spindle of a top,
and then pulled to spin the top. While it is pulled,
the string exerts a constant torque of 0.150 N∙m on
the top. In the first 0.220 s of its motion, the top
reaches an angular velocity of 12.0 rad/s. What is
the moment of inertia of the top?
Calculating Moment of Inertia
Newton’s Second Law: F = ma and τ = Iα
For a particle: τ = rF
I= τ
α
= rF
a/r
= rma
a/r
For multiple particles:
I = mr2
I = miri 2
Moment of Inertia of a particle is directly related
to the mass of the particle and its square of the
distance from the arbitrary origin.
Moment of Inertia of a System of Particles
Three particles (m1 = 2.3 kg, m2 = 3.2 kg, m3 = 1.5
kg) are connected by thin rods of negligible mass
and positioned as shown. Find the rotational
inertia through axes perpendicular to the page and
through each of the masses. Also find the moment
of inertia through the center of mass of the system.
m2
5m
m1
m3
3m
4m
A)axis of rotation through m1
I1 = ∑miri2 = (2.3)(02 ) + (3.2)(3.02 ) + (1.5)(4.02 )
= 53 kg•m2
B) axis of rotation through m2
I2 = ∑miri2 = (2.3)(3.02) + (3.2)(02) + (1.5)(5.02)
= 58 kg•m2
C) axis of rotation through m3
I3 = ∑miri2 = (2.3)(4.02) + (3.2)(5.02) + (1.5)(02)
= 117 kg•m2
D) axis of rotation through the cm:
Finding the center of mass (with origin at m1):
xcm = ∑mixi
∑mi
= .86 m
ycm = ∑miyi
∑mi
= 1.37 m
= (2.3)(0) + (3.2)(0) + (1.5)(4)
2.3 + 3.2 + 1.5
= (2.3)(0) + (3.2)(3) + (1.5)(0)
2.3 + 3.2 + 1.5
r12 = xcm2 + ycm2 = (.86)2 + (1.37)2 = 2.62 m2
r22 = xcm2 + (y2- ycm)2 = (.86)2 + (3 - 1.37)2
= 3.4 m2
r32 = (x3 - xcm ) 2 + ycm2 = (4 - .86)2 + (1.37)2 =
= 11.7 m2
Icm = ∑miri2
= (2.3)(2.62) + (3.2)(3.4) + (1.5)(11.7)
= 34.5 kg•m2
The I will always be least if the axis of rotation is
through the center of mass!
Parallel Axis Theorem:
I = Icm + Mh2
M = ∑mi
h = distance from
axis of rotation to cm
For example, from part A:
I1 = ? h2 = 2.62 m2
M = (2.3 + 3.2 + 1.5)kg
Icm = 34.5 kg•m2
= 7.0 kg
I1 = (34.5) + (7.0)(2.62) = 52.8 kg•m2
The masses and coordinates of four particles are
as follows: 50 kg (2 m, 2 m), 25 kg (0 m, 4 m),
25 kg (-3 m, -3 m), and 30 kg (-2 m, 4 m).
Calculate the rotational inertia of this collection
with respect to the A) x axis B) y axis C) z axis.
Rotational Inertia of Solid Bodies
• Any object can be considered to be consisting
of an infinite number of parts.
• If we were to use the parallel axis theorem we
could divide a solid body into parts and find the
rotational inertia in that manner- tedious but valid.
• If we were to do so, we would find that the
rotational inertia of certain objects of regular
geometry would simplify to set equations
depending upon the mass, radius/length, and axis
of rotation!
Table 11.8
A pulley with a mass of 2.5 kg and a radius of 20
cm is mounted on a fixed, frictionless axel. A
block of 1.2 kg hangs from a light cord wrapped
around the pulley (which can be considered a
disk). When the block is released, find the
acceleration of the block, the tension in the cord
as well as the acceleration of the pulley.
mp = 2.5 kg
R = .20 m
mb = 1.2 kg
I = .5mpR2
Equilibrium would mean ∑F = 0 and ∑ = 0 !
a=?
T
∑Fy = ma = mg - T
∑ = TR = I = (1/2)MR2(a /R)
T = (1/2)Ma
ma = mg - (1/2)Ma
T = (1/2)Ma = 6.0 N
mg
a = 2mg
= 4.8 m/s2
M + 2m
 = a / R = 24 rad/s2
In an Atwood’s machine, one block has a mass of
512 g while the other has a mass of 463 g. The
pulley is mounted in horizontal, frictionless
bearings and has a radius of 4.90 cm. When the
masses are released from rest, the heavier block is
observed to fall 76.5 cm in 5.11 s. What must be
the moment of inertia of the pulley?
Rotational Dynamics for a Rigid Body
If we consider a rigid body to be the sum of
particles:
v2
v1

v = r
K1 = .5m1r122
m2
m1
r
2
2
1
K = .5m r 
2
2 2
Ktotal = ∑Ki
r2
K = .5m1r1212 + .5m2r22 22 …………
K = .5miri2 2
Therefore the kinetic energy of a rigid body that
is rotating:
K = .5 I 2
Translational
F = ma
W = FΔs
K = .5mv
Angular
Τ = Iα
W = τΔθ
K = .5 I 2
Combined Translational and Rotational Motion
The total Kinetic Energy of a moving object
consists of two terms-- one associated with the
translational motion and one associated with the
rotational motion.
 The two terms are independent of one
another
 vcm and  can be independent of each
other
K = (1/2)mvcm2 + (1/2)I cm2
A yo-yo with a mass of .023 kg consisting of
two disks of radius 2.6 cm connected by a shaft
of radius .30 cm is spinning at the end of the
string of length .84 m. What angular velocity is
needed for the yo-yo to climb up the string?
Assume the string to be of negligible thickness.

M = .023 kg
R = 2.6 cm
Ro = 0.3 cm
L = .84 m
At the bottom, the E is all Rotational Kinetic,
while at the top it is KR, KT and Ug, so by
conservation of energy:
KR0 = KR + KT + U
.5Io2 = .5I 2 + .5Mv2 + MgL
Min. energy at top is when
v==0
I = (1/2)MR2
=
4gL
R2
= 221 rad/s
A solid cylinder of radius 10.4 cm and mass 11.8
kg starts from rest and rolls without slipping a
distance of 6.12 m down a house roof that is
inclined 27.0˚. A) What is the angular speed of
the cylinder about its center as it leaves the roof?
B) The outside wall of the house is 5.16 m high.
How far from the wall will the cylinder hit the
level ground?