A runner of mass m = 48 kg and running at 3.7 m/s runs as shown and jumps on the rim of a playground merry-go-round which has a moment of inertia of 378 kg·m2 and a radius of 2 meters. Assuming the merry-go-round is initially at rest, what is its final angular velocity to three decimal places?
To find the final angular velocity of the merry-go-round, we can apply the law of conservation of angular momentum. The angular momentum before the runner jumps onto the merry-go-round is equal to the angular momentum after the runner jumps onto it.
The formula for angular momentum is given by:
L = Iω
Where:
L = Angular momentum
I = Moment of inertia
ω = Angular velocity
Before the runner jumps onto the merry-go-round, the initial angular momentum is zero since the merry-go-round is initially at rest (ω0 = 0). After the runner jumps onto the merry-go-round, the final angular momentum is given by:
L1 = L2
Here, L1 is the initial angular momentum of the runner, and L2 is the final angular momentum of the combined system (runner + merry-go-round).
The initial angular momentum of the runner can be calculated as follows:
L1 = m * v * r
Where:
m = Mass of the runner
v = Velocity of the runner
r = Radius of the merry-go-round
Substituting the given values:
L1 = 48 kg * 3.7 m/s * 2 m
Now, since the runner jumps onto the rim of the merry-go-round, we can consider the runner and the merry-go-round as a single system with a combined moment of inertia (I2).
The final angular momentum (L2) is given by:
L2 = (m + I2) * ω2
Rearranging the equation, we can isolate the final angular velocity (ω2):
ω2 = L2 / (m + I2)
Since the merry-go-round is initially at rest, we know that the initial angular velocity of the merry-go-round (ω0) is zero. Therefore, the combined moment of inertia (I2) is simply the moment of inertia of the merry-go-round (378 kg·m^2).
Substituting the values into the equation:
ω2 = L2 / (m + I2)
ω2 = (m * v * r) / (m + I2)
ω2 = (48 kg * 3.7 m/s * 2 m) / (48 kg + 378 kg·m^2)
Now, calculate the value of ω2 using the above expression and round it to three decimal places to find the final angular velocity of the merry-go-round.