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Angular Energy and Angular Momentum.
Please refer to figure 10.54 p.399.
The general strategy to solve this problem is to figure out the torque around a point and the direction of the torque will tell you which way it rotates.
Choose the z direction to be perpendicular to the plane shown in the figure and pointing up. If the direction of the torque is up it means that it rotates counterclockwise or it moves to the left. If the direction of the torque is down it means that it rotates clockwise or it moves to the right.
The smartest choice of the point you want to write the torque about is the contact point with the ground because the torque due to friction and normal force is Zero around this point since they are acting at this point. Weight vector is also passing through this point so it doesn’t generate any torque around it. In the following calculation regard the contact point as origin:
−→r contact point =^0
a) F to right; bottom of axle(# 1 ):
−→τ c =^
−→r F c ×
F ∝ − ˆz
(Where c denotes the contact point; −→τc : torque around the contact point c; −→r (^) F c : −→r to force from contact point) So it rotates to the right. Now if you write the torque around center of mass the
only force that has torque around it is
f. Consider z component of
τ (^) cm about center of mass. To roll to right, torque must be in −ˆz, since F gives a +ˆz torque, f (friction) must give −ˆz torque hence points to left.
1
F to right; top of axle(# 2 ):
−→τ c =^
−→r F c ×
F ∝ − ˆz
So it rotates to the right. With the same argument friction can point to the left(if axle radius is small) or to the right (if axle radius is large).
F points up; left of axle(# 3 ):
−→τ c =^
−→r F c ×
F ∝ − ˆz
So it rotates to the right. Friction is the only force acting in hor- izontal direction and is responsible from Newton’s second law in horizontal direction for this acceleration so it points to the right.
b) From the definition:
−→τ c =^
−→r F c ×
F
There are three ways to make −→τ zero: 1 )r=0; 2 )F=0; 3 )φ=
(where φ is the angle between the two vectors −→r (^) F c and
F .) Here you don’t have r=0 or F=0. You can make the angle between them Zero. This configuration is shown in the figure 2.
c) Please see the right triangle shown in the figure two with the
indicated parameters.
sin α =
r R
⇒ α = arcsin(
r R
d) Let’s analyze the Newton’s law
F = m−→a here:
∑ Fy = may = 0 ⇒ N − mg − F cos α = 0 ⇒ N = mg − F cos α
2
v =
10 gh 7
Setting up the familiar kinematic equations for free fall with con- stant gravity:
y(t) = y 0 + v 0 t −
gt^2
x(t) = v 0 t
The time T the ball is in the air is (y(T ) = −y):
T =
2 y g
x ≡ x(T ) = v 0 T =
20 hy 7
b) x does not depend on g, so the result should be the same on the
moon.
c) No system is perfect: energy is lost to noise generation, crushing
dirt on the track air resistance, etc.
d) For dollar coin you have:
Icoin =
mR^2
from equation (2) you have
Kcoin =
mv^2
Repeating the same thing we did for the ball you’ll get:
xcoin =
8 hy 3
(slightly less f ar)
4
3 ) 10.87 Bullet hits pivot rod.
There is no external torque acting on the system of bullet plus rod if you take torques about the pivot point, so that the forces on the axle don’t generate any torque:
d
L P
dt
τ (^) P = 0
We conclude that the total angular momentum about P is conserved
a) Angular momentum
L = −→r × −→p is in z direction. Factor out the ˆz from your equations:
L 1 = mv
L
L 2 = I˜P ω = [IP + m(
L
)^2 ]ω
( I˜P denotes the total IP )
For a rod I around the pivot is
IP =
M L^2
where M is the mass of the rod.
mvL 2
M L^2 ω +
mL^2 ω 4 So we get
ω =
m 2 M 3 +^
m 4
v L
Here m = M 4 :
ω =
v L
b)
Kaf ter =
Iω˜^2 (3)
5
This acceleration is constant and from vx(t) = v 0 + axt we have
vx(t) = μkgt (11)
The equation for ω can be derived from:
∑ τz = Iαz (12)
Where we write τz around the center of mass. Around this point the only force that has torque is fk :
∑ −→τ c =^
−→r f c ×
f (^) k = (−Ryˆ) × (−fk ˆx) = −Rfk ˆz
Combine (12) and (13):
αz = −
Rfk Icm
RμkM g Icm
αz is constant so:
ωz (t) = ω 0 −
Rfk Icm
t (14)
Let’s denote T as the time that we reach the criteria for no sliding:
vx(T ) = Rωz (T ) (15)
Combine (15) with (11) and (14):
μkgT = R(ω 0 −
Rfk Icm
T )
T (μkg +
R^2 fk Icm
) = Rω 0
Replacing fk = μkM g and I = 12 M R^2 you’ll get:
T =
Rω 0 3 μkg
The distance it travels during this time can be derived from kine- matics:
x(T ) = x 0 + v (^0) x T +
axT 2 =
μkgT 2
7
x(T ) =
μkg(
Rω 0 3 μkg
)^2
x(T ) =
R^2 ω 02 18 μkg
From energy conservation equation :
K 0 + U 0 + Wf k
= KT + UT (18)
Where KT denotes the Kinetic energy when there is no slipping:(The equation is similar to 10.78 part d)
KT =
Icm ω^2 +
M v^2 =
M v(T )^2
M (axT )^2 =
M (μkg
Rω 0 3 μkg
)^2 (19)
K 0 =
Icm ω^20 =
M R^2 ω 02 (20)
(Where I dropped ”cm” from vcm throughout) Combine (18) with (19), (20) and U 0 = UT :
Wfk = KT − K 0 =
(μkg
Rω 0 3 μkg
)^2 −
M R^2 ω 02 = −
M R^2 ω 02
Wfk = −
M R^2 ω^20
NOTE: This is more than
−fk x(T ) = −μkM g
R^2 ω^20 18 μkg
M R^2 ω 02
in magnitude because the coin is spinning rapidly (especially at first) so that that relative distance of slipping of the coin edge and the surface is a factor of 3 greater than x(T ).
8
Figure 2: 10.70 part b, c, d
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Figure 3: 10.
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