Download Newton`s Second Law for Rotation Newton`s First Law for Rotation

PY105 (C1)
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Newton’s Second Law for Rotation
The equation,
v
v
∑ τ = Iα
v
v
is the rotational equivalent of ∑ F = ma.
Torque plays the role of force.
Rotational inertia plays the role of mass.
Angular acceleration plays the role of the acceleration.
Newton’s First Law for Rotation
An object at rest tends to remain at rest, and an object that is
spinning tends to spin with a constant angular velocity, unless
it is acted on by a nonzero net torque or there is a change in
the way the object's mass is distributed.
Rotational Inertia
How hard it is to get something to spin, or to change an
object's rate of spin, depends on the mass, and on how the
mass is distributed relative to the axis of rotation. Rotational
inertia, or moment of inertia, accounts for all these factors.
The moment of inertia, I, is the rotational equivalent of mass.
The net torque is the vector sum of all the torques acting on
an object.
The tendency of an object to maintain its state of motion is
known as inertia. For straight-line motion mass is the
measure of inertia, but mass by itself is not enough to define
rotational inertia.
For an object like a ball on a string, where all the mass is the
same distance away from the axis of rotation:
I = mr 2
If the mass is distributed at different distances from the
rotation axis, the moment of inertia can be hard to calculate.
It's much easier to look up expressions for I from the table on
page 291 in the book (page 10-15 in Essential Physics).
The parallel axis theorem
A table of
rotational
inertias
If you know the rotational inertia of an object of mass m when
it rotates about an axis that passes through its center of
mass, the object’s rotational inertia when it rotates about a
parallel axis a distance h away is:
I = ICM + mh 2
Take a ring with radius R and mass M as an example:
Axis of rotation
x
I = MR2
x
I = MR2 + MR2 = 2MR2
1
Example 1. Torque on a pulley
Example 1. Torque on a pulley (cont’d)
A constant force of F = 8 N is applied to a string wrapped
around the outside of the pulley. The pulley is a solid disk of
mass M = 2.0 kg and radius R = 0.50 m, and is mounted on a
horizontal frictionless axle. What is the pulley's angular
acceleration?
Solution:
Simulation
v
v
∑ τ = Iα
RF =
1
MR 2
2
1
MRα
2
α=
2 × ( 8.0 N)
2F
=
= 16 rad/s2
MR (2.0 kg) × (0.5 m)
R
F
Example 2. Two identical pulleys (cont’d)
Example 2. Two identical pulleys
Simulation
We take two identical pulleys, both with string wrapped
around them. On the one on the left we apply an 8 N force to
the string. On the one on the right we hang an object with a
weight of 8 N. Which pulley has the larger angular
acceleration?
α1
α2
1. The one on the left
R
R
2. The one on the right
3.
I=
F=
Why are we told that the pulley is a solid disk?
So we know what to use for the rotational inertia. I =
1
MR 2α
2
α
MR2/2
Neither, they're equal
F = 8N
For the pulley on the left, the tension in the string is 8 N.
For the system on the right, let’s draw the free-body diagram
of the weight.
α2
T
τ = RT
R
T
a = (8N-T)/m
mg = 8N
What does the free-body diagram tell us about the tension in
the string?
For the weight to have a net force directed down, the tension
must be less than the force of gravity. So, the tension is less
than 8 N.
mg = 8N
Atwood’s machine
Atwood’s machine involves one pulley, and two objects
connected by a string that passes over the pulley. In
general, the two objects have different masses. We have
encountered this in one of the homework assignments.
a
Re-analyzing the Atwood’s machine
When we analyzed Atwood’s
machine earlier, we found an
expression for the acceleration of the
weights in terms of m, M, and g, but
we neglected the mass of the pulley
(i.e., we assumed that the pulley is
massless). If we include the mass of
the pulley in the analysis, the
acceleration of the masses m and M
should be:
1.
larger than before.
2.
smaller than before.
3.
the same as before.
2
Example 3 Acceleration in an Atwood’s machine
Find an expression for the acceleration of m and M in an
Atwood’s machine given the mass of the pulley is mp.
Solution:
What’s the difference with the earlier case where the
pulley is massless?
The acceleration turns out to be less than what we found
before, because we need to accelerate the pulley – a solid
disc (with I = ½ mpR2, where R is the radius of the pulley).
Example 3 Acceleration in an Atwood’s machine (cont’d)
Step 1: Analyze the lighter object
Sketch a free-body diagram for the lighter object.
Choose a positive direction, and apply Newton’s Second
Law.
Let’s choose positive to be up, in the direction of the
acceleration.
FT1
FT1
FT1
Because of this, the tension force on the right (where the
heavier mass M is) is larger than the tension on the left to
give a net clockwise torque to produce a clockwise
angular acceleration to the pulley.
Example 3 Acceleration in an Atwood’s machine (cont’d)
Step 2: Analyze the heavier object
Sketch a free-body diagram for the heavier object.
Choose a positive direction, and apply Newton’s Second
Law.
Choose positive down this time, to match the object’s
acceleration.
FT2
v
v
∑ F = Ma
a
+Mg − FT 2 = +Ma
v
v
∑ F = ma
FT2
a
FT2
+FT 1 − mg = +ma
mg
Example 3 Acceleration in an Atwood’s machine (cont’d)
Step 3: Analyze the pulley
Sketch a free-body diagram for the pulley.
Choose a positive direction, and apply Newton’s Second
Law for rotation.
Choose positive clockwise, to match the pulley’s angular
acceleration.
v
v
∑ τ = Iα
α
⎛1
⎞
+RFT 2 − RFT 1 = + ⎜ mpR 2 ⎟ α
I = ½mpR2
⎝2
⎠
FT1
FT2
⎛1
⎞a
+FT 2 − FT 1 = + ⎜ mpR ⎟
⎝2
⎠R
1
+FT 2 − FT 1 = + mpa
2
(Note that R cancels out so it is not
needed to be specified.)
Mg
Example 3 Acceleration in an Atwood’s machine (cont’d)
Step 4: combine the equations
Lighter object:
+FT 1 − mg = + ma
Heavier object:
+Mg − FT 2 = +Ma
Pulley:
1
+FT 2 − FT 1 = + mpa
2
Add the equations:
+Mg − mg = +Ma + ma +
Previous result
1
⎛
⎞
+Mg − mg = ⎜ M + m + mp ⎟ a
2
⎝
⎠
a=
Mg − mg
M +m
a=
1
mp a
2
Mg − mg
m
M +m+ p
2
3