A playground merry-go-round of radius 𝑅 = 2.0 m has a moment of inertia
𝐼 = 250 kg ⋅ m2 is rotating at 15 rpm about a frictionless, vertical axle. Facing the axle, a 25-kg child hops onto the merry-goround and manages to sit down on the edge.
a. What is the total angular momentum of the ‘merry-go-round-child’ system before and after the child hops on the the merry-go-round?
b. What is the new angular speed, in rad/s, of the merry-go-round?
c. How many seconds will it take for the system to cover 5.0 complete revolutions?
d. What is the total kinetic energy of the system before and after the child hops on the the merrygo-round?
with a solution please, thank you so much!!!
a.
Before the child hops on:
The moment of inertia of the merry-go-round (I1) is given as 250 kg⋅m^2.
The initial angular velocity of the merry-go-round (ω1) is given as 15 rpm.
The angular momentum of the merry-go-round before the child hops on is given by:
L1 = I1 * ω1
L1 = 250 kg⋅m^2 * (15 rpm * 2π rad/min)
L1 = 250 kg⋅m^2 * (15 rpm * 2π * (1/60) rad/s)
L1 = 250 kg⋅m^2 * 15 * 2π/60 rad/s
L1 = 250 * 15 * 2π/60 kg⋅m^2/s
L1 = 3750π kg⋅m^2/s
After the child hops on:
The moment of inertia of the system (I2) is given by the sum of the merry-go-round's moment of inertia and the child's moment of inertia.
I2 = I1 + I_child
I2 = 250 kg⋅m^2 + 25 kg * (2.0 m)^2
I2 = 250 kg⋅m^2 + 100 kg⋅m^2
I2 = 350 kg⋅m^2
The angular momentum of the system after the child hops on is given by:
L2 = I2 * ω2
We assume that the merry-go-round and the child rotate together after the child hops on. Therefore, the final angular velocity (ω2) is the same for both the merry-go-round and the child.
L2 = 350 kg⋅m^2 * ω2
b.
To find the new angular speed of the merry-go-round, we need to use the principle of conservation of angular momentum:
L1 = L2
3750π kg⋅m^2/s = 350 kg⋅m^2 * ω2
ω2 = (3750π kg⋅m^2/s) / (350 kg⋅m^2)
ω2 ≈ 33.927 rad/s
c.
To find the time it takes for the system to cover 5.0 complete revolutions, we need to calculate the time based on the new angular speed.
One complete revolution is equal to 2π radians. Therefore, 5 complete revolutions is equal to 5 * 2π radians.
time = (angle covered) / (angular speed)
time = (5 * 2π rad) / (33.927 rad/s)
time ≈ 0.293 seconds
d.
Before the child hops on:
The total kinetic energy of the system is the sum of the kinetic energy of the merry-go-round and the child.
The kinetic energy of the merry-go-round is given by:
KE1 = (1/2) * I1 * (ω1)^2
KE1 = (1/2) * 250 kg⋅m^2 * (15 rpm * 2π/60)^2
KE1 = (1/2) * 250 kg⋅m^2 * (15 * 2π/60)^2 rad/s
KE1 = (1/2) * 250 kg⋅m^2 * (15 * 2π/60)^2 rad^2/s^2
The kinetic energy of the child is given by:
KE_child = (1/2) * I_child * (ω1)^2
KE_child = (1/2) * 25 kg * (2π/60)^2 m^2 * (15 rpm)^2
KE_child = (1/2) * 25 kg * (2π/60)^2 m^2 * (15)^2 (2π/60)^2 rad^2/s^2
After the child hops on:
The total kinetic energy of the system is the sum of the kinetic energy of the merry-go-round and the child.
The kinetic energy of the merry-go-round is given by:
KE2 = (1/2) * I2 * (ω2)^2
KE2 = (1/2) * 350 kg⋅m^2 * (33.927 rad/s)^2
KE2 = (1/2) * 350 kg⋅m^2 * (33.927)^2 rad^2/s^2
The kinetic energy of the child is negligible after sitting on the edge.
Therefore, the total kinetic energy of the system before and after the child hops on can be calculated by adding the respective kinetic energies:
KE_total1 = KE1 + KE_child
KE_total2 = KE2
Note: The actual numerical values for the kinetic energies depend on the specific values of angular velocities and moments of inertia, which are given in the problem statement.
To solve this problem, we need to consider the conservation of angular momentum and kinetic energy. Let's break down each part of the question and solve them one by one:
a. Total angular momentum before the child hops onto the merry-go-round is given by the equation:
L_initial = I * ω_initial,
where L_initial is the initial angular momentum, I is the moment of inertia, and ω_initial is the initial angular speed.
L_initial = 250 kg⋅m² * (15 rpm * 2π rad/1 min * 1/60 min) (converting rpm to rad/s)
L_initial = 250 kg⋅m² * (2π rad/s)
L_initial = 500π kg⋅m²⋅rad/s
After the child hops onto the merry-go-round, their moment of inertia needs to be considered as well. Assuming the child sits at the edge, the moment of inertia is:
I' = I + mR²,
where m is the mass of the child (25 kg) and R is the radius of the merry-go-round (2.0 m).
I' = 250 kg⋅m² + (25 kg) * (2.0 m)²
I' = 250 kg⋅m² + 100 kg⋅m²
I' = 350 kg⋅m²
The total angular momentum after the child hops on is given by:
L_final = I' * ω_final,
where L_final is the final angular momentum and ω_final is the final angular speed.
b. To find the new angular speed, we can use the conservation of angular momentum:
L_initial = L_final
I * ω_initial = I' * ω_final
ω_final = (I * ω_initial) / I'
ω_final = (250 kg⋅m² * (15 rpm * 2π rad/1 min * 1/60 min)) / 350 kg⋅m²
ω_final = (500π rad/s) / 350
ω_final = (10π / 7) rad/s
c. To determine the time it takes for the system to cover 5.0 complete revolutions, we can use the equation:
θ = ω * t
where θ is the angle covered in radians, ω is the angular speed in rad/s, and t is the time in seconds.
Since 1 revolution is equal to 2π radians, 5 complete revolutions would be 5 * 2π radians.
θ = 5 * 2π radians
ω = (10π / 7) rad/s (calculated in part b)
t = θ / ω
t = (5 * 2π radians) / ((10π / 7) rad/s)
t = (70 / 10) s
t = 7 s
d. To find the total kinetic energy before the child hops on, we use the equation:
KE_initial = (1/2) * I * ω_initial²
KE_initial = (1/2) * 250 kg⋅m² * ((15 rpm * 2π rad/1 min * 1/60 min)²)
KE_initial = (1/2) * 250 kg⋅m² * (450π² rad²/s²)
KE_initial = 56250π² J
After the child hops on, the total kinetic energy is given by:
KE_final = (1/2) * I' * ω_final²
KE_final = (1/2) * 350 kg⋅m² * ((10π / 7) rad/s)²
KE_final = (1/2) * 350 kg⋅m² * (100π² / 49) rad²/s²
KE_final = 70000π² / 49 J
So, the total angular momentum before the child hops on is 500π kg⋅m²⋅rad/s, and after the child hops on is 350 kg⋅m²⋅rad/s.
The new angular speed of the merry-go-round is (10π / 7) rad/s.
It will take 7 seconds for the system to cover 5.0 complete revolutions.
The total kinetic energy before the child hops on is 56250π² J, and after the child hops on, it is 70000π² / 49 J.