Answer:To solve this problem, we can use the conservation of angular momentum. The total angular momentum before the stones are dropped must equal the total angular momentum after they are dropped.
The formula for angular momentum (
�
L) is given by:
�
=
�
�
L=Iω
where:
�
L is the angular momentum,
�
I is the moment of inertia, and
�
ω is the angular velocity.
For a rotating object like the merry-go-round, the moment of inertia is given by:
�
=
�
�
2
I=mr
2
where:
�
m is the mass of the object, and
�
r is the radius.
The angular velocity (
�
ω) is related to the linear velocity (
�
v) by:
�
=
�
�
ω=
r
v
Now, let's calculate the angular momentum before and after the stones are dropped.
Initial Angular Momentum (
�
initial
L
initial
):
The merry-go-round is initially rotating freely. The total initial angular momentum (
�
initial
L
initial
) is the sum of the angular momentum of the merry-go-round and the stones.
For the merry-go-round:
�
merry-go-round, initial
=
�
merry-go-round
�
initial, merry-go-round
L
merry-go-round, initial
=I
merry-go-round
ω
initial, merry-go-round
=
(
�
�
2
)
�
initial, merry-go-round
=(MR
2
)ω
initial, merry-go-round
For each stone:
�
stone, initial
=
�
stone
�
initial, stone
L
stone, initial
=I
stone
ω
initial, stone
=
(
�
�
2
)
�
initial, stone
=(mr
2
)ω
initial, stone
Since the stones are initially at rest,
�
initial, stone
=
0
ω
initial, stone
=0, and their initial angular momentum is zero.
So, the total initial angular momentum is just the angular momentum of the merry-go-round:
�
initial
=
�
merry-go-round, initial
L
initial
=L
merry-go-round, initial
Final Angular Momentum (
�
final
L
final
):
After the stones are dropped, the merry-go-round and the stones have a final angular velocity (
�
final
ω
final
). The total final angular momentum (
�
final
L
final
) is the sum of the angular momentum of the merry-go-round and the stones.
For the merry-go-round:
�
merry-go-round, final
=
�
merry-go-round
�
final, merry-go-round
L
merry-go-round, final
=I
merry-go-round
ω
final, merry-go-round
=
(
�
�
2
)
�
final, merry-go-round
=(MR
2
)ω
final, merry-go-round
For each stone:
�
stone, final
=
�
stone
�
final, stone
L
stone, final
=I
stone
ω
final, stone
=
(
�
�
2
)
�
final, stone
=(mr
2
)ω
final, stone
Now, we use the conservation of angular momentum, setting the initial angular momentum equal to the final angular momentum:
�
initial
=
�
final
L
initial
=L
final
�
merry-go-round, initial
=
�
merry-go-round, final
L
merry-go-round, initial
=L
merry-go-round, final
(
�
�
2
)
�
initial, merry-go-round
=
(
�
�
2
)
�
final, merry-go-round
+
2
(
�
�
2
)
�
final, stone
(MR
2
)ω
initial, merry-go-round
=(MR
2
)ω
final, merry-go-round
+2(mr
2
)ω
final, stone
Now, we can substitute the known values and solve for
�
final, merry-go-round
ω
final, merry-go-round
.
Given:
�
=
90.0
kg
M=90.0kg (mass of merry-go-round)
�
=
2.40
m
R=2.40m (radius of merry-go-round)
�
initial, merry-go-round
=
2
�
×
3
rad/s
ω
initial, merry-go-round
=2π×3rad/s (initial angular velocity of merry-go-round)
�
=
45.0
kg
m=45.0kg (mass of each stone)
�
=
1.20
m
r=1.20m (radius where stones are dropped)
Now, substitute these values into the equation and solve for
�
final, merry-go-round
ω
final, merry-go-round
.
(
�
�
2
)
�
initial, merry-go-round
=
(
�
�
2
)
�
final, merry-go-round
+
2
(
�
�
2
)
�
final, stone
(MR
2
)ω
initial, merry-go-round
=(MR
2
)ω
final, merry-go-round
+2(mr
2
)ω
final, stone
Given values:
(
90.0
kg
×
(
2.40
m
)
2
)
×
(
2
�
×
3
rad/s
)
=
(
90.0
kg
×
(
2.40
m
)
2
)
×
�
final, merry-go-round
+
2
×
(
45.0
kg
×
(
1.20
m
)
2
)
×
�
final, stone
(90.0kg×(2.40m)
2
)×(2π×3rad/s)=(90.0kg×(2.40m)
2
)×ω
final, merry-go-round
+2×(45.0kg×(1.20m)
2
)×ω
final, stone
Now, let's calculate:
(
90.0
kg
×
(
2.40
m
)
2
)
×
(
2
�
×
3
rad/s
)
(90.0kg×(2.40m)
2
)×(2π×3rad/s)
≈
3250.94
kg
m
2
/
s
≈3250.94kgm
2
/s
2
×
(
45.0
kg
×
(
1.20
m
)
2
)
×
�
final, stone
2×(45.0kg×(1.20m)
2
)×ω
final, stone
=
2
×
(
45.0
kg
×
1.44
m
2
)
×
�
final, stone
=2×(45.0kg×1.44m
2
)×ω
final, stone
=
129.6
kg
m
2
/
s
×
�
final, stone
=129.6kgm
2
/s×ω
final, stone
Now, subtract the term with
�
final, stone
ω
final, stone
from the total initial angular momentum:
(
3250.94
kg
m
2
/
s
)
−
(
129.6
kg
m
2
/
s
)
×
�
final, stone
=
(
90.0
kg
×
(
2.40
m
)
2
)
×
�
final, merry-go-round
(3250.94kgm
2
/s)−(129.6kgm
2
/s)×ω
final, stone
=(90.0kg×(2.40m)
2
)×ω
final, merry-go-round
Now, solve for
�
final, merry-go-round
ω
final, merry-go-round
:
�
final, merry-go-round
≈
(
3250.94
kg
m
2
/
s
)
−
(
129.6
kg
m
2
/
s
)
×
�
final, stone
90.0
kg
×
(
2.40
m
)
2
ω
final, merry-go-round
≈
90.0kg×(2.40m)
2
(3250.94kgm
2
/s)−(129.6kgm
2
/s)×ω
final, stone
Calculate this expression to find the final angular velocity (
�
final, merry-go-round
ω
final, merry-go-round
).
Note: Make sure to convert revolutions per second to radians per second by multiplying by
2
�
2π, as used in the formula for angular velocity.
Step-by-step explanation: