Download Chapter 9. Rotational Dynamics

Chapter 9. Rotational Dynamics
❑ As we did for linear (or translational) motion, we
studied kinematics (motion without regard to the
cause) and then dynamics (motion with regard to
the cause), we now proceed in a similar fashion
❑ We know that forces are responsible for linear
motion
❑ We will now see that rotational motion is caused
by torques
❑ Consider a wrench of length L=r
F
Torque
τ = rF
Units of N m
Not work or energy,
do not use Joules
r
€
r is called the lever arm. Must always be
perpendicular to the force
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If the force is not perpendicular to the level arm,
we need to find the component that is
perpendicular (either the force or the lever arm)
F
φ
F sinφ
τ = rF sin φ
If φ=0, then the torque
is zero
r
❑ Therefore, torques (force times length) are
responsible for rotational motion
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(really from Section 11.4)
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Newton’s 2nd Law for Rotational Motion
❑ The torque for a pointparticle of mass m a
distance r from the
rotation axis is
τ = rFt
Ft = mat, at = rα
2
⇒ τ = mat r = mr α
F=Ft
r
m
Axis of rotation
❑ Define I = mr2 = Moment of Inertia for a point
particle; a scalar, units of kg m2
❑ A rigid body is composed of many, many particles
of mass mi which are ri from the axis of rotation
❑ Each of these masses creates a torque about
the axis of rotation
Rotation axis
2
1 1
2
2 2
τ1 = m r α
τ2 = m r α
....
r1
r2
m1
m2
❑ Sum up all torques due to all particles
α is the same for all particles
2
τ = mr α
∑
i
∑
I = ∑ mi ri
∑τ = Iα
i i
2
I =moment of inertia for the rigid
body. It is different for different
shaped objects and for different
axes of rotation. Fig. 9.18.
❑ For a thin rod of mass M and length L
I =
L/2
I =
1
12
2
ML
1
3
2
ML
L
❑ The last equation is Newton’s 2nd Law for rotation.
Compare to the translational form
!
!
∑τ = Iα
Example Problem
⇔
!
!
∑ F = ma
A rotating door is made of 4 rectangular panes each
with a mass of 85 kg. A person pushes on the outer
edge of one pane with a force of 68 N, directed
perpendicular to the pane. Determine the door’s α.
Given: L = 1.2 m, mpane=85 kg,
F=68 N
L
∑τ = Iα
FL
FL = Iα ⇒ α =
I
From Fig. 9.18, moment of inertia for a thin
F
rectangular rod is (same as a thin sheet)
Since there are 4 panes.
2
2
I pane = ML ⇒ I door = ML
1
3
4
3
FL
FL
α=
= 4
2
I
3 ML
3F
3(68 N)
α=
=
= 0.50 kgNm = 0.50 rad
s2
4 ML 4(85 kg )(1.2 m)
Example Problem
The parallel axis theorem provides a useful way to
calculate I about an arbitrary axis. The theorem
states that I = Icm + MD2, where Icm is the moment
of inertia of an object (of mass M) with an axis that
passes through the center of mass and is parallel to
the axis of interest. D is the perpendicular distance
between the two axes. Now, determine I of a solid
cylinder of radius R for an axis that lies on the
surface of the cylinder and perpendicular to the
circular ends.
Solution: The center of mass of the cylinder is on a
line defining the axis of the cylinder
R
cm
I cm = MR
1
2
From Table 12.2:
2
From the parallel axis theorem with D=R:
2
2
2
I = I cm + MD = MR + MR = MR
1
2
Apply to thin rod:
L/2
I cm =
1
12
2
1
12
I = I cm + MD =
2
ML
3
2
2
L
2
& L#
2
1
ML + M $ ! = 3 ML
%2"
2
Rotational Work
❑ For translational motion, we defined the work as
W = Fs s
❑ For rotational motion
!
Fs
!
s
r
θ
r
s = arc length
= rθ
s
Ft
W = Fs s = Ft rθ = ( Ft r )θ
WR = τ θ Units of N m or J when θ is in radians
Rotational Kinetic Energy
❑ For translational motion, the K was defined
2
K = mv , since v t = rω
2
2 2
1
K = m(rω ) = 2 mr ω
1
2
1
2
❑ For a point particle I = mr2, therefore
1
2
K rot = Iω
2
❑ Or for a rigid body
K rot =
1
2
∑m r ω
2
i i
2
2
1
2
= Iω = K rot
❑ For a rigid body that has both translational and
rotational motion, its total kinetic energy is
2
1
2
1
2
K total = K + K rot = mv + Iω
2
❑ The total mechanical energy is then
1
2
2
1
2
2
1
2
E total = mv + Iω + mgy + kx
2
Rolling Motion (Tires, Billiards)
vt
B
A
vt
vaxis = vcar
C
vt
If tire is suspended, every
point on edge has same vt
R=road
❑ Consider a tire traveling on a road with friction (no
skidding) between the tire and road
❑ First, review concept of relative velocity. What is the
velocity of B as seen by the ground?
❑ Point B has a velocity vt with respect to the A
❑ Now, A (the tire as a whole) is moving to the
right with velocity vcar
v B , A = v t , v A,G = v car
v B ,G = v B , A + v A,G , v t = v car
v B ,G = v t + v car = 2 v car
❑ What is velocity of C as seen by the ground?
v C , A = v t , v A,G = v car
v C ,G = v C , A + v A,G , v t = − v car
v C ,G = v t + v car = 0
Example Problem
A tire has a radius of 0.330 m, and its center moves forward with
a linear speed of 15.0 m/s. (a) What is ω of the wheel? (b)
Relative to the axle, what is vt of a point located 0.175 m from
the axle?
v t v car
15.0
ω=
=
=
= 45.5 rad/s
r
r
0.330
v t = rω = (0.175)(45.5) = 7.96 m/s
Example
A car is moving with a speed of 27.0 m/s. Each
wheel has a radius of 0.300 m and a moment of
inertia of 0.850 kg m2. The car has a total mass
(including the wheels) of 1.20x103 kg. Find (a) the
translational K of the entire car, (b) the total Krot of
the four wheels, and (c) the total K of the car.
Solution:
Given: vcar = 27.0 m/s, mcar = 1.20x103 kg,
0.300 m, Iw = 0.850 kg m2
rw =
a)
b)
c)
1
2
2
car
5
1
2
3
2
K = mcar v = (1.2x10 )(27.0)
K = 4.37x10 J
2
2
#
&
#
&
v
v
2
t
car
1
1
1
K rot = 2 Iω = 2 Iw % ( = 2 Iw %
(
$ rw '
$ rw '
2
1
K rot = 2 (0.850)(27.0 /0.300)
3
= 3.44x10 J
4
K wheels = 4K rot = 1.38x10 J
K total = K + K wheels
5
4
5
= 4.37x10 + 1.38x10 = 4.51x10 J
Example Problem
A tennis ball, starting from rest, rolls down a hill
into a valley. At the top of the valley, the ball
becomes airborne, leaving at an angle of 35° with
respect to the horizontal. Treat the ball as a thinwalled spherical shell and determine the horizontal
distance the ball travels after becoming airborne.
0
3
1.8 m
1
2
θ
4
x
Solution:
Given: v0 = 0, ω0 = 0, y0 = 1.8 m = h, y1 = y2 = y4
= 0, θ2 = 35°, x2 = 0, I=(2/3)MR2
Find: x4 ?
Method: As there is no “friction” or air resistance in
the problem, therefore no non-conservative forces,
we can use conservation2 of mechanical
energy
2
Etotal = 12 mv + 12 Iω + mgy
E0 = mgh
2
2
1
1
E1 = 2 mv1 + 2 Iω1 + mg (0)
2
2
1
1
E2 = 2 mv 2 + 2 Iω 2 + mg (0) = E1
⇒ v1 = v 2 , ω1 = ω2
Since E0 = E1 = E2
2
2
2
2
mgh = mv + Iω , but v t = rω
1
2
vt = v 2
1
2
Velocity of ball equals tangential
velocity at edge of ball
2
2
2
mgh = mv + ( mr )( v 2 / r )
1
2
2
2
1
2
2
2
2
3
gh = v + v = v
1
2
1
3
5
6
2
2
2
⇒ v 2 = 6 gh / 5
= 6(9.80)(1.8)/5 = 4.6 m/s
Now, use 2D kinematic equations for projectile
motion
v 2 x = v 2 cosθ 2 = v 4 x
v
v 2 y = v 2 sin θ 2 = − v 4 y
v
v 4 y = v 2 y − g (t 4 − t 2 ), t 2 = 0
− v 2 y = v 2 y − gt 4
− 2 v 2 y = − gt 4 ⇒ t 4 = 2v 2 y / g
1
x4 = x2 + 2 ( v 2 x + v 4 x )t 4 = v 2 x t 4
x4 = v 2 x 2 v 2 y / g = 2v 2 cosθ 2 v 2 sin θ 2 / g
2
2v 2 cosθ 2 sin θ 2 6 gh 2 cosθ 2 sin θ 2
x4 =
=
g
5
g
12h cosθ 2 sin θ 2 12(1.8) cos 35 sin 35
x4 =
=
5
5
x4 = 2.0 m
2
4
Example Problem
A mass m1 (=15.0 kg) and a mass m2
(=10.0 kg) are suspended by a pulley that
has a radius R (=10.0 cm) and a mass M
(=3.00 kg). The cord has negligible mass
and causes the pulley to rotate without
slipping. The pulley rotates about its axis
without friction. The masses start from rest
at a distance h (=3.00 m) part. Treating
the pulley as a uniform disk, determine the
speeds of the two masses as they pass
each other.