Download General Physics (PHY 2130)

General Physics (PHY 2130)
Lecture 19
•  Rotational dynamics
  rotational kinetic energy
  torque
  moment of inertia
http://www.physics.wayne.edu/~apetrov/PHY2130/
Lightning Review
Last lecture:
1.  Momentum:
  momentum conservation
  collisions in one and two dimensions
Review Problem: A rider in a “barrel of fun”
finds herself stuck with her back to the wall.
Which diagram correctly shows the forces
acting on her?
Rotational kinetic energy and Inertia
For a rotating solid body:
n
1 2 1
1
1 2
2
2
K rot = m1v1 + m2v2 + … + mn vn = ∑ mi vi
2
2
2
i =1 2
vi
ri
vi = riω where ri is the distance from the
rotation axis to the mass mi.
n
1
1 ⎛ n
1 2
2
2 ⎞ 2
K rot = ∑ mi (riω ) = ⎜ ∑ mi ri ⎟ω = Iω
2 ⎝ i =1
2
i =1 2
⎠
K rot
1 2
= Iω
2
n
I = ∑ mi ri
2
i =1
I = rotational inertia or moment of inertia of the body about the axis
SI unit of I = kg.m2
3
mi
4
Example: (a) Find the moment of inertia of the system below. The masses are
m1 and m2 and they are separated by a distance r. Take m1 = 2.00 kg, m2 = 1.00
kg, r1= 0.33 m , and r2 = 0.67 m to compute the moment of inertia numerically.
Assume the rod connecting the masses is massless. (b) What is the moment of
inertia if the axis is moved so that is passes through m1?
ω
m1
ω
r1
Solution:
r2
m2
(a) Note that r1 and r2 are the distances
between mass 1 and the rotation axis
and mass 2 and the rotation axis (the
dashed, vertical line) respectively.
2
I = ∑ mi ri2 = m1r12 + m2 r22
i=1
2
2
= ( 2.00 kg) ( 0.33 m ) + (1.00 kg) ( 0.67 m ) = 0.67 kg m 2
(b) If the axis is moved so that is passes through m1 then r1 and r2 would
change
2
I = ∑ mi ri2 = m1r12 + m2 r22
i=1
2
2
= ( 2.00 kg) ( 0.00 m ) + (1.00 kg) (1.00 m ) = 1.00 kg m 2
Example: Moment of Inertia of a Uniform Ring
•  Image the hoop is divided into a
number of small segments, m1 …
•  These segments are equidistant
from the axis
I = ∑ mi ri = ∑ mi R
2
i
=R
i
2
∑ m = MR
i
i
2
2
6
7
Example: What is the rotational inertia of a solid iron disk of mass 49.0 kg with a
thickness of 5.00 cm and a radius of 20.0 cm, about an axis through its center
and perpendicular to it?
From table 8.1:
1
1
2
2
I = MR = (49.0 kg )(0.2 m ) = 0.98 kg m 2
2
2
Total Energy of Rotating System
•  An object rotating about some axis with an angular
speed, ω, has rotational kinetic energy ½Iω2
•  Energy concepts can be useful for simplifying the
analysis of rotational motion
•  Conservation of Mechanical Energy
( KEt + KEr + PE g )i = ( KEt + KEr + PE g ) f
•  Remember, this is for conservative forces, no dissipative forces
such as friction can be present
ConcepTest
A force F is applied to a dumbbell for a time interval Δt, first
as in (a) and then as in (b). In which case does the dumbbell
acquire the greater energy?
1.
2.
3.
4.
(a)
(b)
no difference
The answer depends on the
rotational inertia of the dumbbell.
ConcepTest
A force F is applied to a dumbbell for a time interval Δt, first
as in (a) and then as in (b). In which case does the dumbbell
acquire the greater energy?
1.
2.
3.
4.
(a)
(b)
no difference
The answer depends on the
rotational inertia of the dumbbell.
Since CM speeds are the same, translational
kinetic energies are the same. But (b) also
rotates, so it also has rotational energy.
Torque
•  Consider force required to open
door. Is it easier to open the door by
pushing/pulling away from hinge or
close to hinge?
Farther from
from hinge,
larger
rotational
effect!
Physics concept: torque
close to hinge
away from
hinge
Torque
τ
•  Torque, , is the tendency of a force to
rotate an object about some axis
Door example:
τ = Fd
• 
τ is the torque
•  d is the lever arm (or moment arm)
•  F is the force
Lever Arm
•  The lever arm, d, is the
shortest (perpendicular)
distance from the axis of
rotation to a line drawn
along the the direction of
the force
•  d = L sin Φ
•  It is not necessarily the
distance between the axis
of rotation and point where
the force is applied
Direction of Torque
•  Torque is a vector quantity
•  The direction is perpendicular
to the plane determined by the
lever arm and the force
•  Direction and sign:
Direction of torque:
out of page
 If the turning tendency of the force
is counterclockwise, the torque will
be positive
 If the turning tendency is clockwise,
the torque will be negative
SI
Units
Newton meter (Nm)
US Customary Foot pound (ft lb)
An Alternative Look at Torque
•  The force could also be
resolved into its x- and ycomponents
•  The x-component, F cos Φ,
produces 0 torque
•  The y-component, F sin Φ,
produces a non-zero torque
τ = FL sin φ
F is the force
L is the distance along the object
Φ is the angle between force and object
L
ConcepTest
You are trying to open a door that is stuck by pulling on the
doorknob in a direction perpendicular to the door. If you
instead tie a rope to the doorknob and then pull with the same
force, is the torque you exert increased? Will it be easier to
open the door?
1. No
2. Yes
ConcepTest
You are trying to open a door that is stuck by pulling on the
doorknob in a direction perpendicular to the door. If you
instead tie a rope to the doorknob and then pull with the same
force, is the torque you exert increased? Will it be easier to
open the door?
1. No
2. Yes
19
Example: The pull cord of a lawnmower engine is wound around a drum of radius
6.00 cm, while the cord is pulled with a force of 75.0 N to start the engine. What
magnitude torque does the cord apply to the drum?
F=75 N
τ = r⊥ F
= rF
R=6.00 cm
= (0.06 m )(75.0 N ) = 4.5 Nm
ConcepTest
You are using a wrench and trying to loosen a rusty nut.
Which of the arrangements shown is most effective in
loosening the nut? List in order of descending efficiency the
following arrangements:
ConcepTest
You are using a wrench and trying to loosen a rusty nut.
Which of the arrangements shown is most effective in
loosening the nut? List in order of descending efficiency the
following arrangements:
2, 1, 4, 3
or
2, 4, 1, 3
What if two or more different forces
act on lever arm?
Net Torque
•  The net torque is the sum of all the torques produced by
all the forces
•  Remember to account for the direction of the tendency for rotation
•  Counterclockwise torques are positive
•  Clockwise torques are negative
Example 1:
N
Determine the net torque:
4m
2m
Given:
weights: w1= 500 N
w2 = 800 N
lever arms: d1=4 m
d2=2 m
Find:
Στ = ?
500 N
800 N
1. Draw all applicable forces
2. Consider CCW rotation to be positive
∑τ = (500 N )(4 m) + (−)(800 N )(2 m)
= +2000 N ⋅ m − 1600 N ⋅ m
= +400 N ⋅ m
Rotation would be CCW
 
Where would the 500 N person have to
be relative to fulcrum for zero torque?
Example 2:
N’
d2 m
y
2m
Given:
weights: w1= 500 N
w2 = 800 N
lever arms: d1=4 m
Στ = 0
Find:
d2 = ?
500 N
800 N
1. Draw all applicable forces and moment arms
∑τ
∑τ
RHS
= − (800 N )(2 m)
LHS
= (500 N )(d 2 m)
−800 ⋅ 2 [ N ⋅ m] + 500 ⋅ d 2 [ N ⋅ m] = 0
⇒ d 2 = 3.2 m
 
According to our understanding of torque there
would be no rotation and no motion!
What does it say about acceleration and force?
∑ F =(−500 N ) + N '+(−800 N ) = 0
i
Thus, according to 2nd Newton’s law ΣF=0 and a=0!
N ' = 1300 N
27
Calculating Work done from the Torque
The work done by a torque τ is
W = τΔθ .
where Δθ is the angle (in radians) the object turns through.
28
Example: A flywheel of mass 182 kg has a radius of 0.62 m (assume the flywheel is
a hoop). (a) What is the torque required to bring the flywheel from rest to a speed
of 120 rpm in an interval of 30 sec? (b) How much work is done in this 30 sec
period?
(a) Let’s recall the definition of force and angular acceleration
ω f = 120
rev ⎛ 2π rad ⎞⎛ 1 min ⎞
⎜
⎟⎜
⎟ = 12.6 rad/sec
min ⎝ 1 rev ⎠⎝ 60 sec ⎠
⎛ Δω ⎞
τ = rF = r (ma ) = rm(rα ) = mr ⎜
⎟
⎝ Δt ⎠
ω f − ωi ⎞
ω f ⎞
2 ⎛
2 ⎛
⎟⎟ = mr ⎜⎜
⎟⎟ = 29.4 Nm
= mr ⎜⎜
⎝ Δt ⎠
⎝ Δt ⎠
2
(a) Use the definition of work for rotational motion:
W = τΔθ = τ (ωav Δt )
⎛ ωi + ω f
= τ ⎜⎜
⎝ 2
⎞
⎛ ω f
⎟⎟Δt = τ ⎜⎜
⎠
⎝ 2
⎞
⎟⎟Δt = 5600 J
⎠