Download Linear acceleration of rolling objects Rotational Motion (cont.) R θ

Rotational Motion (cont.)
Linear acceleration of rolling objects
R
Consider a round object (this
could be a cylinder, hoop, sphere
or spherical shell) having mass M,
radius R and rotational inertia I
about its center of mass, rolling without
slipping down an inclined plane.
aCM
θ
What is the linear acceleration of the object’s center of mass, aCM ,
down the incline?
We analyze this as follows:
The force of gravity, Mg, acting straight down
is resolved into components parallel and
perpendicular to the incline.
Since the object rolls without
slipping there is a force of
friction, f, acting on the object,
at it’s point of contact with the
incline, in the direction up
the incline.
f
Mg cos θ
R
θ
a CM
Mg
Mg sin θ
θ
Newton’s 2nd law gives then for acceleration down the incline,
∑ F = Ma

CM
Mg sin θ − f = Ma CM
The force of friction also causes a torque around the center of mass
having lever arm R so we can also write,
τ = Rf = Iα
Solving for the friction,
a CM
f
Mg cos θ
I
f=
α
R
θ
Mg
This is used in the expression
derived from the 2nd law:
Mg sin θ
Mg sin θ − f = Ma CM
I
Mg sin θ − α = Ma CM
R
θ
The objects angular acceleration is related to the linear acceleration
of the edge that contacts the incline by,
a= Rα
Since the object rolls without
slipping this has the same
magnitude as aCM so we have
that,
a CM
α=
R
Using this in,
a CM
f
Mg cos θ
θ
Mg
Mg sin θ
θ
I
Mg sin θ − α = Ma CM
R
I a CM
Mg sin θ −
= Ma CM
R R
I
→ − Ma CM − 2 a CM = − Mg sin θ
R
Multiplying through by –1,
I
Ma CM + 2 a CM =Mg sin θ
R
I 

Ma CM  1 +
= Mg sin θ
2 
 MR 
I 

+
=
a CM  1
g sin θ
2 
 MR 
So that ,finally,
a CM
g sin θ
=
I
1+
MR 2
a CM
f
Mg cos θ
θ
Mg
Mg sin θ
θ
a CM
g sin θ
=
I
1+
MR 2
If the object is a solid
cylinder ,
1
I = MR 2
2
a CM
f
Mg cos θ
θ
Mg
Mg sin θ
g sin θ
g sin θ g sin θ 2
a CM
g sin θ
=
=
= =
2
1 MR
1
3
3
+
1
2
1+
2
2
MR 2
θ
(solid cylinder)
If instead the object is cylindrical shell, with all its mass at the rim
I = MR 2 and,
g sin θ
g sin θ 1
=
a CM
=
=
g sin θ (cylindrical shell)
2
MR
1+1 2
1+
MR 2
Angular Momentum
The rotational analog to linear momentum (p = mv) is
angular momentum,
L = Iω
Recall that linear momentum is important because given a system
of objects, in the absence of external forces, no matter how the
objects of the system interact with each other, their total linear
momentum is conserved. I.e. with,
 n 
P = ∑ pi
For any two times,
i =1
 
Pf = Pi


Thus if there are two objects with linear momenta p1 and p2 so that
their total momentum is at one time,
 

P
=i p1i + p2i
Then no matter how they interact, collide, attract or repel each other,
in the absence of external forces, this total momentum will not change
so we have that,




p1f + p2f = p1i + p2i
Similarly, for rotational motion, in the absence of external torques
(i.e.∑ τ =0 ), the angular momentum is conserved meaning that for
any two times,
Lf = Li
Example
Top view
m
A boy (mass m) runs with speed v and
jumps onto the edge of an initially
stationary merry-go-round (also called a
carousel).
v
R
M
What is the angular velocity of the
carousel (and the boy) after he has
jumped on?
Conservation of angular momentum
requires,
Lf = Li
final
R
ω
M
We must first consider the boy’s
m
angular momentum about the rotation
axis of the carousel when he is running
in a straight line. His angular momentum
is,
Top view
v
r
 B= I BωB
M
Where IB is his rotational inertia about the axis
and ωB is his angular velocity about the axis.
We treat the boy as a point object of mass m, making his rotational
inertia about the carousel axis
I B = mr 2
Where r is his (changing) distance to the carousel axis.
What is the boy’s angular velocity
ωB about the axis?
Top view
m
v
r
When we previously considered motion
on a circular trajectory we had that vt = rω
where the subscript t reminds us that
this was velocity tangent to the circle.
M
Tangent to the circle means the component of the velocity that is
perpendicular to a line drawn to the center of the circle (all of v,
when the trajectory is circular).
For this more general case we must decompose the velocity vector
into components parallel and perpendicular to the line to the rotation
axis.
Then,
v sin θ = rωB
or,
v
sin θ
ω=
B
r
v sin θ
m
v cos θ
θ
θ v
r
Using this and I B = mr 2
in,
2 v
=
 B (mr ) sin θ
r
gives,
=
 B mrv=
sin θ mv(r sin θ)
M
v sin θ
m
v cos θ
θ
θ v
R
r
But notice that along his linear
trajectory r sin θ is also the point of
closest approach to the axis which here is R so that,
=
 B mv(r=
sin θ) mvR
v sin θ
Since all of these m, v and R
are constant what we have just
m
shown is that despite the fact
that r and θ are both changing
as the boy approaches the carousel
his angular momentum about the
axis remains constant and is given
by
v sin θ
 B = mvR
m
Where R is the point of closest
approach to the axis.
v cos θ
θ
θ v
r
M
v cos θ
θ
θ v
R
r
(this must be true for angular momentum to be conserved. I.e.  B
calculated at any two times along this trajectory had better be the
same).
Having shown this once, we
don’t need to re-derive it every
time it comes up.
m
What needs to be remembered is
that an object travelling linearly,
with constant v, on a trajectory to
pass an arbitrary point has a constant
angular momentum about that point
given by,
m
 B = mvR
Where R is the point of closest
approach of the trajectory to the point.
v
r
M
v
R
r
The initial angular momentum of
m
the system is the sum of the initial
angular momenta of the boy  Bi = mvR
and the carousel  Ci = 0 (since ωCi = 0),
Top view
v
R
L i =  Bi +  Ci = mvR
M
After the boy jumps on we have
L f = I f ωf
Where If is the combined rotational inertia
of the boy and the carousel. The carousel
can be treated as a uniform disk with
1
I C = MR 2
2
final
R
ω
M
For rotation about the same axis
rotational inertia are simply additive
so,
Top view
m
v
R
1
M

I f = MR 2 + mR 2 = + m  R 2
2
 2

M
Then,
M
 2
=
ω
=
+
Lf If f 
m  R ωf = L i = mvR
 2

final
Solving for ωf ,
mv
ωf =
M

+
m

R
 2

R
ω
M
Check behavior if M → 0 .
New example
initial
Suppose the boy is on the rim of
carousel at radius R from the center
and the two have angular velocity ωi.
The boy then pulls himself along the
hand rail until he is a distance R/3
from the center.
R
ωi
M
final
What is ωf after this move?
Lf = Li
 Bf +  Cf =  Bi +  Ci
2
1
1
R
m   ωf + MR 2 ωf =
mR 2 ωi + MR 2 ωi
2
2
3
R/3
ωf
M
initial
2
1
1
1
m   R 2 ωf + MR 2 ωf =
mR 2 ωi + MR 2 ωi
2
2
 3
M 2
m M 2

+
ω
=
+
R
m



 R ωi
f
2
9 2

R
ωi
M
M
m M

 +  ωf =  m +  ωi
2
9 2

M

m + 
2

=
ωf
ωi
m M
 + 
9 2
final
R/3
ωf
M
M

m + 
2

=
ωf
ωi
m M
 + 
9 2
initial
R
ωi
Suppose that M = 6m then,
M
6m 

m +

2
ω
=
ωf 
i
m
6m


+


9
2


ω=
f
(1 + 3)
1

3
+


9


ω=i 1.29ωi
final
R/3
ωf
M
So the angular velocity increases when the boy moves in.
If the boy’s mass is m = 25 kg, the carousel’s
mass is 6m = 150 kg the initial angular
velocity was ωi = 2 rad/s and R = 1.5 m.
How much work did the boy do in pulling
himself in to R/3?
initial
R
ωi
Wnc = ∆K + ∆U
M
There is no ∆U here so,
final
Wnc =
∆K =
Kf − Ki
1
1 2
2
Wnc=
If ωf − I i ωi
2
2
Above we found that,
rad
ω=
1.29ω=i 2.58
f
s
R/3
ωf
M
1
I i mR + MR 2
=
2
1

 2
2
Ii =
m
6m
R
4mR
+
=


2


initial
2
R
ωi
2
=
I i 4(25kg)(1.5m)
=
225kg ⋅ m 2
M
2
R 1
2
I
m
MR
=
+
 
f
3 2
1 2 1
=
I f m   R + 6mR 2 = 3.11mR 2
2
9
2
=
I f 3.11(25kg)(1.5m)
=
175kg ⋅ m 2
final
R/3
ωf
M
initial
So with,
=
I i 225kg ⋅ m
2
=
I f 175kg ⋅ m
2
rad
ωi =2.00
s
rad
ωf =2.58
s
R
ωi
M
1
1 2
2
Wnc=
If ωf − I i ωi
2
2
final
Wnc = 582.4 J − 450 J = 132 J
R/3
ωf
M
Summary of translational quantities and their rotational analogs