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?
I tried W=IW+MVR/MR^2+I but I don't think that's right
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.
The initial angular momentum of the system (merry-go-round + person) is given by the product of moment of inertia (I) and initial angular speed (ω):
L_initial = I * ω
When the person jumps onto the merry-go-round, their angular momentum is transferred to the system. Once the person lets go, their angular momentum becomes zero. Thus, the final angular momentum of the system is only due to the merry-go-round:
L_final = I_merry * ω_f
According to the principle of conservation of angular momentum, the initial angular momentum is equal to the final angular momentum:
L_initial = L_final
I * ω = I_merry * ω_f
Rearranging the equation, we can find the final angular speed (ω_f):
ω_f = (I * ω) / I_merry
Now, substituting the given values:
R = 1.95 m (radius of merry-go-round)
I = 192 kg-m^2 (moment of inertia of merry-go-round)
ω = 1.41 rad/s (initial angular speed of merry-go-round)
m = 63 kg (mass of the person)
v = 4.7 m/s (velocity of the person)
Note: We need to calculate the moment of inertia of the merry-go-round, taking into account the moment of inertia of the person. For a person running tangent to the merry-go-round and jumping on, we consider the moment of inertia of the person as Md^2, where M is the mass of the person and d is the distance from the center of the merry-go-round to the person's path (which is the radius of the merry-go-round).
First, find the moment of inertia of the person:
I_person = M * d^2
Where:
M = 63 kg (mass of the person)
d = R = 1.95 m (distance from center of the merry-go-round to the person's path)
Substituting the values:
I_person = (63 kg) * (1.95 m)^2
Next, find the total moment of inertia of the system (merry-go-round + person):
I_total = I_merry + I_person
Now, we can find the final angular speed (ω_f):
ω_f = (I * ω) / I_total
Substituting the values:
ω_f = (192 kg-m^2 * 1.41 rad/s) / (I_total)
I_total = 192 kg-m^2 + (63 kg * (1.95 m)^2)
Evaluate I_total, then substitute it into the equation for ω_f to get the final angular speed of the merry-go-round.