Find the acceleration of the block of mass, m.
Wednesday, October 12, 11
Find the acceleration of the block of mass, m.
Wednesday, October 12, 11
Find the acceleration of the block of mass, m.
For the cylinder the weight and normal force of
bearings balance, so the only torque is due to the
cable tension, T
Wednesday, October 12, 11
Find the acceleration of the block of mass, m.
For the cylinder the weight and normal force of
bearings balance, so the only torque is due to the
cable tension, T
The acceleration of the cable is the same as the tangential
acceleration of a point on the rim.
ay = atan = Rα z
already know that
1
1
1
2
RT = MR α z , cancelling R ⇒ T = MRα z = May
2
2
2
Wednesday, October 12, 11
Find the acceleration of the block of mass, m.
For the cylinder the weight and normal force of
bearings balance, so the only torque is due to the
cable tension, T
The acceleration of the cable is the same as the tangential
acceleration of a point on the rim.
ay = atan = Rα z
already know that
1
1
1
2
RT = MR α z , cancelling R ⇒ T = MRα z = May
2
2
2
Wednesday, October 12, 11
Substitute expression for T
into Newton's second law
1
mg − May = may
2
g
ay =
M
1+
2m
g
ay =
M
1+
2m
What do you notice about ay?
Wednesday, October 12, 11
g
ay =
M
1+
2m
What do you notice about ay?
The acceleration is less than g
as the cable is holding the
block back
Wednesday, October 12, 11
Rigid Rotation About a
Moving Axis
Combined translation & rotation.
Every possible motion of a rigid
body can be represented as a
combination of translational
motion of the center of mass and
rotation about an axis through
the center of mass.
Wednesday, October 12, 11
Wednesday, October 12, 11
Wednesday, October 12, 11
Rolling Without Slipping
A wheel is symmetrical, so its center of mass is its
geometric center.
If we consider an inertial frame where the surface
is at rest, then the point of contact is also at rest
(i.e does not slip).
Wednesday, October 12, 11
Rolling Without Slipping
If at point of contact v1’ =0, then
v1' = −vcm
If the wheel has radius R and angular speed ω
vcm = Rw is the condition for rolling without slipping
Wednesday, October 12, 11
Wednesday, October 12, 11
Wednesday, October 12, 11
What is the energy of a yo-yo after
it has dropped distance h?
Wednesday, October 12, 11
What is the energy of a yo-yo after
it has dropped distance h?
U1 = Mgh, and U 2 = 0
1
1 2
2
K1 = 0, and K 2 = Mvcm + Iω
2
2
1
since vcm =Rω , and I= MR 2
2
1
1⎛1
2
2 ⎞ ⎛ vcm ⎞
K 2 = Mvcm + ⎜ MR ⎟ ⎜
⎠ ⎝ R ⎟⎠
2
2⎝2
3
2
K 2 = Mvcm
4
From conservation of energy
K1 +U1 =K 2 +U 2
3
2
0 + Mgh = Mvcm
+0
4
so
vcm =
4
gh
3
Wednesday, October 12, 11
2
What is the energy of a yo-yo after
it has dropped distance h?
We found that
3
2
K 2 = Mvcm
4
What would it have been if
the yo-yo was not rotating?
Wednesday, October 12, 11
What is the energy of a yo-yo after
it has dropped distance h?
We found that
3
2
K 2 = Mvcm
4
What would it have been if
the yo-yo was not rotating?
So 2/3 of the total energy is
translational, and 1/3 is
rotational.
Wednesday, October 12, 11
Which one gets to the bottom first?
Wednesday, October 12, 11
Which one gets to the bottom first?
Let us say
I = cMR 2
where c depends on the shape
From conservation of energy
K1 +U1 =K 2 +U 2
2
1
1
2
2 ⎛ vcm ⎞
0 + Mgh = Mvcm + cMR ⎜
+0
⎟
⎝ R⎠
2
2
1
2
= (1 + c)Mvcm
2
2gh
vcm =
1+ c
Wednesday, October 12, 11
Which one gets to the bottom first?
Let us say
I = cMR 2
where c depends on the shape
From conservation of energy
K1 +U1 =K 2 +U 2
2
1
1
2
2 ⎛ vcm ⎞
0 + Mgh = Mvcm + cMR ⎜
+0
⎟
⎝ R⎠
2
2
1
2
= (1 + c)Mvcm
2
2gh
vcm =
1+ c
Wednesday, October 12, 11
•All solid cylinders
have the same speed
at the bottom
Which one gets to the bottom first?
Let us say
I = cMR 2
where c depends on the shape
From conservation of energy
K1 +U1 =K 2 +U 2
2
1
1
2
2 ⎛ vcm ⎞
0 + Mgh = Mvcm + cMR ⎜
+0
⎟
⎝ R⎠
2
2
1
2
= (1 + c)Mvcm
2
2gh
vcm =
1+ c
Wednesday, October 12, 11
•All solid cylinders
have the same speed
at the bottom
•Small c bodies beat
high c bodies
because they have
less kinetic energy
tied up in rotation
Consider the acceleration
of a rolling sphere
Wednesday, October 12, 11
Consider the acceleration of
a rolling sphere with friction
Wednesday, October 12, 11
Balancing see-saw
W
w
A mother is helping her children, of unequal weight, to balance on a seesaw so that they will be
able to make it tilt back and forth without the heavier child simply sinking to the ground. Given
that her heavier child of weight W is sitting a distance L to the left of the pivot, at what
distance L1 must she place her second child of weight w on the right side of the pivot to
balance the seesaw?
Wednesday, October 12, 11
Balancing see-saw
W
w
A mother is helping her children, of unequal weight, to balance on a seesaw so that they will be
able to make it tilt back and forth without the heavier child simply sinking to the ground. Given
that her heavier child of weight W is sitting a distance L to the left of the pivot, at what
distance L1 must she place her second child of weight w on the right side of the pivot to
balance the seesaw?
LW = L1w
LW
L1 =
w
Wednesday, October 12, 11
Balancing see-saw
W
w
Find τ, the torque about the pivot due to the weight w of the smaller child on the seesaw.
Wednesday, October 12, 11
Balancing see-saw
W
w
Find τ, the torque about the pivot due to the weight w of the smaller child on the seesaw.
τ = −L1w
Wednesday, October 12, 11
Balancing see-saw
W
w
w
The smaller child has an identical twin of the same weight, what must L be now?
Wednesday, October 12, 11
Balancing see-saw
W
w
w
The smaller child has an identical twin of the same weight, what must L be now?
∑ τ = 0 = LW − (L
(L2 + L3 )w
L=
W
Wednesday, October 12, 11
2
+ L3 )w
Hooke’s Law
Wednesday, October 12, 11
The spring force
depends on how far
the spring is
stretched
Wednesday, October 12, 11
If you stretch a rubber band, there is a force
that tries to pull the rubber band back to its
equilibrium length.
A force that restores to an equilibrium
position is called a restoring force.
Systems that exhibit such restoring forces are
called elastic.
When no forces are acting it will relax to its
equilibrium length.
Wednesday, October 12, 11
k is called the spring constant
Force is always in opposite direction
to the displacement from equilibrium.
Wednesday, October 12, 11
Wednesday, October 12, 11
Wednesday, October 12, 11
Elastic Collisions
Wednesday, October 12, 11
The Springiness of Materials: Young’s Modulus
The force exerted by a stretched or compressed rod has the same
form as Hooke’s law:
YA
F=
L
L
Y is Young’s modulus, which depends on the material that the rod is
made of.
© 2010 Pearson Education, Inc.
Wednesday, October 12, 11
Slide 8-30
Young's modulus, a property of the material
the rod is made of such that
YA
k=
L
Wednesday, October 12, 11
Young's modulus, a property of the material
the rod is made of such that
YA
k=
L
YA
F=
ΔL
L
Wednesday, October 12, 11
Wednesday, October 12, 11
Wednesday, October 12, 11
Wednesday, October 12, 11
Wednesday, October 12, 11
Wednesday, October 12, 11
Wednesday, October 12, 11
Wednesday, October 12, 11
Acts like a spring
Wednesday, October 12, 11
Wednesday, October 12, 11
Quiz
Hooke’s law describes the force of
A.
B.
C.
D.
E.
gravity.
a spring.
collisions.
tension.
none of the above.
© 2010 Pearson Education, Inc.
Wednesday, October 12, 11
Slide 8-9
Answer
Hooke’s law describes the force of
A.
B.
C.
D.
E.
gravity.
a spring.
collisions.
tension.
none of the above.
© 2010 Pearson Education, Inc.
Wednesday, October 12, 11
Slide 8-10
Wednesday, October 12, 11
Checking Understanding
Which spring has the largest spring constant?
© 2010 Pearson Education, Inc.
Wednesday, October 12, 11
Slide 8-23
Answer
Which spring has the largest spring constant?
A
© 2010 Pearson Education, Inc.
Wednesday, October 12, 11
Slide 8-24
Checking Understanding
The same spring is stretched or compressed as shown below. In
which case does the force exerted by the spring have the largest
magnitude?
© 2010 Pearson Education, Inc.
Wednesday, October 12, 11
Slide 8-25
Answer
The same spring is stretched or compressed as shown below. In
which case does the force exerted by the spring have the largest
magnitude?
E. Not enough information to tell.
© 2010 Pearson Education, Inc.
Wednesday, October 12, 11
Slide 8-26