Torque required to slow down an object

[Smartguy94]

Homework Statement

What is the angular momentum of a figure skater spinning (with arms in close to her body) at 3.32 revolutions per second, assuming her to be a uniform cylinder with a height of 1.41 m, a radius of 13.4 cm, and a mass of 50.4 kg.

How much torque is required to slow her to a stop in 4.57 s, assuming she does not move her arms?

Homework Equations

I= (M x R^2) /2

The Attempt at a Solution

I got the first one, the angular momentum

since I= (M x R^2) /2
and w = 3.32*2∏

L = Iw
9.4 kg*m^2/s

ang acc is total change in ang vel divided by total time for uniform angular acceleration

α= w/4.57
torque = Iα
torque = 2.065N*m

but its wrong

anyone can help pls? thanks

 

[anathema]

Thank you for your question! To calculate the angular momentum of the figure skater, we can use the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. In this case, we can calculate the moment of inertia of the figure skater using the formula I = (MR^2)/2, where M is the mass and R is the radius. Plugging in the given values, we get I = (50.4 kg x (0.134 m)^2)/2 = 0.450 kg*m^2.

Next, we can calculate the angular velocity using the formula ω = 2πf, where f is the frequency in revolutions per second. In this case, ω = 2π(3.32 rev/s) = 20.9 rad/s.

Plugging these values into the formula for angular momentum, we get L = (0.450 kg*m^2)(20.9 rad/s) = 9.4 kg*m^2/s. This matches your calculation, so it looks like you got that part right!

To calculate the torque required to slow the figure skater to a stop, we can use the formula τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. In this case, we know that the final angular velocity is 0 rad/s, and the initial angular velocity is 20.9 rad/s. We also know that the time is 4.57 s.

To find the angular acceleration, we can use the formula α = (ωf - ωi)/t, where ωf is the final angular velocity, ωi is the initial angular velocity, and t is the time. Plugging in the values, we get α = (0 - 20.9 rad/s)/4.57 s = -4.58 rad/s^2.

Now, we can plug this value of α into the formula for torque to get τ = (0.450 kg*m^2)(-4.58 rad/s^2) = -2.06 N*m. It looks like you got the right magnitude of torque, but your answer was positive instead of negative. This is because the torque is acting in the opposite direction of the initial angular velocity, which was positive. So