To find the merry-go-round's angular velocity after John jumps on, we need to apply the principle of conservation of angular momentum.
Angular momentum (L) is given by the formula:
L = I * ω
Where:
L is the angular momentum,
I is the moment of inertia,
ω is the angular velocity.
The moment of inertia (I) depends on the mass distribution and shape of the object. For a solid disk like a merry-go-round, the moment of inertia can be calculated using the formula:
I = (1/2) * m * r^2
Where:
m is the mass,
r is the radius.
Given:
Diameter of the merry-go-round (d) = 6.42m, so the radius (r) = d/2 = 6.42m/2 = 3.21m
Mass of the merry-go-round (M) = 648.0kg
Using the formula for the moment of inertia, we find:
I = (1/2) * M * r^2
= (1/2) * 648.0kg * (3.21m)^2
Next, we need to find John's initial angular momentum (L_initial) when he runs tangent to the merry-go-round. Since John is running tangent to the merry-go-round, his initial angular momentum is zero.
L_initial = 0
When John jumps onto the outer edge, he imparts angular momentum to the merry-go-round. The final angular momentum (L_final) is given by:
L_final = (merry-go-round I * merry-go-round ω) + (John's I * John's ω)
However, John's initial angular velocity is not given, so let's calculate it using his tangential velocity:
John's tangential velocity (v) = 6.36m/s
John's radius (r_john) = radius of the merry-go-round = 3.21m
We can use the formula for linear velocity in terms of angular velocity:
v = ω * r
Rearranging the formula, we find:
ω = v / r
John's initial angular velocity (ω_john) = 6.36m/s / 3.21m
Now we can substitute the values into the equation for final angular momentum:
L_final = (I * ω) + (John's I * ω_john)
Lastly, we solve for the final angular velocity (ω) by rearranging the formula for angular momentum:
ω = (L_final - (John's I * ω_john)) / I
Substituting the given values and solving the equation will give us the final angular velocity of the merry-go-round after John jumps on.