PHYSICS 8A: PROBLEM SET 8
1.
Angular Motion with Constant Acceleration
(1.1)
θ
(
t
)=
θ
0
+
ω
0
t
+
1
2
αt
2
(A) In (1.1),
θ
is a funciton of time.
(B) In (1.1),
θ
0
is a constant.
(C) In (1.1),
ω
0
is a constant.
(D) In (1.1),
ω
is a funciton of time.
(E)
ω
2
=
ω
2
0
+2
α
(
θ

θ
0
) does not exhibit explicit
t
dependence.
(F)
(1.2)
ω
=
ω
0
+
αt
In (1.2),
t
is the time since the angular velocity was
ω
0
.
(G) Particle
A
’s angular trajectory is given by
(1.3)
θ
A
(
t
)=
θ
0
+
ω
0
t
+
1
2
αt
2
We also know that particle
B
is also undergoing constant acceleration, and that at time
t
1
, it has equal
angular position, half the angular velocity, and twice the angular accerlation, that
A
did at time
t
= 0.
We could ±gure out
θ
B
(
t
) through a lot of calculus and algebra, or we can de±ne
T
=
t

t
1
. Treating
T
like “time,” we quickling conclude that
B
±
s
“initial” angle is
θ
0
, its “initial” angular velocity is
1
2
ω
0
,
and its acceleration is 2
α
. Thus
θ
B
=
θ
0
+
1
2
ω
0
T
+
αT
2
. Substituting back in our de±nition of
T
, we
see that
θ
B
(
t
)=
θ
0
+
1
2
ω
0
(
t

t
1
)+
α
(
t

t
1
)
2
.
(H) Angular velocity being the timederivative of angle, we set
dθ
B
dt
(
t
)=
dθ
A
dt
(
t
)
⇒
1
2
ω
0
+2
α
(
t

t
1
)=
ω
0
+
αt
,
by di²erentiating our answer to (G). Thus we must have
t
=
ω
0
2
α
+2
t
1
, which is
ω
0
2
α
+
t
1
after
t
1
.
2.
Torque and Angular Acceleration
(A) We can ±gure out all the torques, so we can use
Iα
=
τ
to ±nd
α
. We de±ne (+) to be counterclockwise
both because
m
1
>m
2
and because this is the most common convention (though the problem does not
specify this). The magnitude of the torque
τ
1
from
m
1
is
m
1
g
l
2
since the force it exerts is perpendicular
to its lever arm, which is half the length of the seesaw.
τ
1
is positive because it would spin the seesaw
counterclockwise, were it the only torque. Similarly,
τ
2
=

m
2
g
l
2
, since it would spin the seesaw
clockwise. Thus
τ
=
τ
1
+
τ
2
=
gl
2
(
m
1

m
2
). Since the mass of the seesaw itself is stated to be negligible,
I
=
I
1
+
I
2
=
m
1
(
l
2
)
2
+
m
2
(
l
2
)
2
. Thus we arrive at
α
=
τ
I
=
2
g
(
m
1

m
2
)
l
(
m
1
+
m
2
)
.
(B) Since
m
1
>m
2
,
α >
0
and by our convention, this corresponds to a
counterclockwise
rotation. (Had
we used the opposite convention, we would have gotten
α
∝
(
m
2

m
1
) instead, giving
α <
0 and still
ended up with counterclockwise.)
Date
: October 23, 2009.
1
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PHYSICS 8A: PROBLEM SET 8
(C) Now we are told to stop neglecting
m
bar
. A rod rotating about its axis has
I
bar
=
1
12
m
bar
l
2
=
1
3
m
bar
(
l
2
)
2
. Thus the total rotational of the seesaw is now
I
=
l
2
4
(
m
1
+
m
2
+
m
bar
3
)
. The total
torque is the same as in (A) because
m
bar
g
applies at 0 distance from the pivot, so we now Fnd that
α
=
6
g
(
m
1

m
2
)
l
(3
m
1
+3
m
2
+
m
bar
)
.
(D) Changing
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 Fall '07
 SHAPIRO
 Physics, Acceleration, Angular Momentum, Moment Of Inertia, Rotation, Angular velocity, Ibar

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