A merry-go-round with a a radius of R = 1.95 m and moment of inertia I = 192 kg-m2 is spinning with an initial angular speed of ω = 1.41 rad/s in the counter clockwise direection when viewed from above. A person with mass m = 63 kg and velocity v = 4.7 m/s runs on a path tangent to the merry-go-round. Once at the merry-go-round the person jumps on and holds on to the rim of the merry-go-round. What is the angular speed of the merry-go-round after the person lets go?
This is a conservation of angular momentum question. Momentum will stay the same before and after.
L_initial,merry-go-round: I*ω=(192)(1.41)=193.41
L_initial,person: m*v*r=(63)(4.7)(1.95)=577.395
L_final: (I_merry-go-round+m*r^2)*ω=(192+239.56)(ω)
193.41+577.395=431.56ω
ω=1.79
Hopefully I didn't make any math errors, but this is how to solve it.
Well, it seems like the merry-go-round is going to have quite the surprise when the person jumps on! Let's figure out what happens.
First, let's calculate the initial angular momentum of the merry-go-round. We can use the formula:
L_initial = I * ω_initial
L_initial = 192 kg-m^2 * 1.41 rad/s
L_initial ≈ 270.72 kg-m^2/s
Now, when the person jumps on, there is no external torque acting on the system, so the angular momentum should be conserved. This means that the final angular momentum will be equal to the initial angular momentum.
L_final = L_initial
Now, let's calculate the final moment of inertia of the merry-go-round with the person on it. Since the person is holding onto the rim, they can be considered part of the rotating system, so we add their moment of inertia.
I_final = I_merry-go-round + I_person
We know the moment of inertia of the merry-go-round is I = 192 kg-m^2, and the moment of inertia of a point mass rotating about its axis is given by:
I_person = m * r^2
Since the person jumps onto the merry-go-round, their distance from the axis is equal to the radius of the merry-go-round, R = 1.95 m.
I_person = 63 kg * (1.95 m)^2
I_person ≈ 236.63 kg-m^2
Now, let's calculate the final angular speed of the merry-go-round using the conservation of angular momentum equation:
L_final = I_final * ω_final
L_final = (I_merry-go-round + I_person) * ω_final
ω_final = L_final / (I_merry-go-round + I_person)
Substituting the values we have:
ω_final = 270.72 kg-m^2/s / (192 kg-m^2 + 236.63 kg-m^2)
ω_final ≈ 0.688 rad/s
So, after the person lets go, the merry-go-round will be spinning at approximately 0.688 rad/s. Just remember, the next time you go on a merry-go-round, be prepared for unexpected visitors with their own intertia!
To find the angular speed of the merry-go-round after the person lets go, we can use the principle of conservation of angular momentum.
The angular momentum of the system before the person jumps on the merry-go-round is given by:
L_initial = I_initial * ω_initial
where I_initial is the moment of inertia of the merry-go-round and ω_initial is its initial angular speed.
Let's calculate L_initial:
L_initial = 192 kg-m^2 * 1.41 rad/s
= 271.2 kg-m^2/s
When the person jumps on the merry-go-round, the total angular momentum of the system is conserved. After the person lets go, the angular momentum of the merry-go-round changes, but the total angular momentum remains the same.
The person has angular momentum as well, given by:
L_person = m * v * R
where m is the mass of the person, v is their velocity, and R is the radius of the merry-go-round.
Let's calculate L_person:
L_person = 63 kg * 4.7 m/s * 1.95 m
= 582.255 kg-m^2/s
After the person lets go, the angular momentum of the system consists only of the angular momentum of the merry-go-round. Let's denote the final angular speed of the merry-go-round as ω_final.
The angular momentum of the system after the person lets go is given by:
L_final = I_final * ω_final
Now, since the total angular momentum is conserved:
L_initial + L_person = L_final
Substituting the values we have calculated:
271.2 kg-m^2/s + 582.255 kg-m^2/s = I_final * ω_final
Simplifying:
853.455 kg-m^2/s = I_final * ω_final
Now, we need to find the moment of inertia of the merry-go-round after the person lets go. Since the person is holding on to the rim, their mass does not affect the moment of inertia. Therefore:
I_final = I_initial
Substituting back into the equation:
853.455 kg-m^2/s = I_initial * ω_final
Solving for ω_final:
ω_final = 853.455 kg-m^2/s / I_initial
ω_final = 853.455 kg-m^2/s / 192 kg-m^2
= 4.452 rad/s
The angular speed of the merry-go-round after the person lets go is 4.452 rad/s.
To find the final angular speed of the merry-go-round after the person lets go, we can use the principle of conservation of angular momentum.
Angular momentum is defined as the product of moment of inertia (I) and angular velocity (ω). Mathematically, it is expressed as L = Iω.
Before the person jumps on the merry-go-round, since there is no external torque acting on the system, the angular momentum of the system remains constant. Hence, we can write:
L_before = L_after
Initially, only the merry-go-round is rotating with an angular speed of ω_initial = 1.41 rad/s. So, the initial angular momentum is:
L_before = I_0 * ω_initial
where I_0 is the moment of inertia of the merry-go-round.
Next, when the person jumps onto the merry-go-round and holds on to the rim, their moment of inertia becomes part of the system. Let's denote the moment of inertia of the person as I_person.
At this point, the total moment of inertia (I_total) of the system (merry-go-round + person) is the sum of the moment of inertia of the merry-go-round and the person:
I_total = I_0 + I_person
Finally, once the person lets go, the system consists only of the merry-go-round again. We need to find the final angular velocity (ω_final) of the merry-go-round.
Using the conservation of angular momentum principle, we can set up the equation:
L_before = L_after
I_0 * ω_initial = I_total * ω_final
Substituting the values given in the question, we have:
I_0 * ω_initial = (I_0 + I_person) * ω_final
Now we can solve for ω_final:
ω_final = (I_0 * ω_initial) / (I_0 + I_person)
Plug in the values for I_0, ω_initial, and I_person to calculate the final angular speed.