What is the angular momentum of a figure skater spinning at 2.7rev/s with arms in close to her body, assuming her to be a uniform cylinder with a height of 1.6m , a radius of 14cm , and a mass of 50kg How much torque is required to slow her to a stop in 4.4s , assuming she does not move her arms?
To find the angular momentum of the figure skater, we can use the formula:
Angular momentum (L) = moment of inertia (I) × angular velocity (ω)
The moment of inertia of a uniform cylinder can be calculated using the formula:
Moment of inertia (I) = (1/2) × mass × radius²
In this case, the mass of the figure skater is given as 50 kg, and the radius of the cylinder is 14 cm (or 0.14 m). We can substitute these values to calculate the moment of inertia.
I = (1/2) × 50 kg × (0.14 m)² = 0.49 kg·m²
The angular velocity, ω, is given as 2.7 rev/s. To convert this to radians per second, we need to multiply by 2π since each revolution corresponds to 2π radians.
ω = 2.7 rev/s × 2π = 16.97 rad/s
Now we can calculate the angular momentum:
L = I × ω = 0.49 kg·m² × 16.97 rad/s = 8.35 kg·m²/s
To find the torque required to slow the skater to a stop, we can use the formula:
Torque (τ) = Change in angular momentum / Time
The change in angular momentum can be calculated as the final angular momentum (which is zero since the skater comes to a stop) minus the initial angular momentum. The time is given as 4.4 s.
τ = (0 - 8.35 kg·m²/s) / 4.4 s = -1.89 kg·m²/s²
The negative sign indicates that the torque required to stop the skater is in the opposite direction of her initial angular momentum.