Question

an ice skater is spinning at 6.6 rev/s and has a moment of inertia of 0.24 kg ⋅ m2. > A 33% Part (a) Calculate the angular momentum, in kilogram meters squared per second, of the ice skater spinning at 5.2 rev/s. Grade Summary Deductions 0% Potential 100% L = 1 E sin() cos() tan() cotano asino acos atan acotan() sinh( cosh tanh cotanh0 Degrees Radians ( 7 8 9 HOME 4 5 6 * 1 2 3 - + - 0 . END VO BACKSPACE DEL CLEAR Submissions Attempts remaining: 5 (2% per attempt) detailed view Submit Hint Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 1 Feedback: 0% deduction per feedback. A 33% Part (b) He reduces his rate of rotation by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia (in kilogram meters squared) if his rate of rotation decreases to 0.75 rev/s. A 33% Part (c) Suppose instead he keeps his arms in and allows friction of the ice to slow him to 3.25 rev/s. What is the magnitude of the average torque that was exerted, in N.m, if this takes 19 s?

Answer 1

Part (a):

Given:

Angular velocity, w1 = 6.6 rev/s

Moment of inertia, I = 0.24 kg⋅m^2

We know that the angular momentum (L) of a rotating object is given by:

L = I * w

So, the angular momentum of the skater is:

L1 = I * w1 = 0.24 kg⋅m^2 * 6.6 rev/s = 1.584 kg⋅m^2/s

Now, the skater reduces his rate of rotation to w2 = 5.2 rev/s.

To find his new angular momentum, we use the same equation:

L2 = I * w2 = 0.24 kg⋅m^2 * 5.2 rev/s = 1.248 kg⋅m^2/s

Therefore, the angular momentum of the skater spinning at 5.2 rev/s is 1.248 kg⋅m^2/s.

Part (b):

Let the new moment of inertia be I2.

The conservation of angular momentum tells us that the initial and final angular momenta of the skater must be equal.

So, we can use the equation:

I1 * w1 = I2 * w2

where w1 = 6.6 rev/s, w2 = 0.75 rev/s, and I1 = 0.24 kg⋅m^2.

Solving for I2, we get:

I2 = I1 * w1 / w2 = 0.24 kg⋅m^2 * 6.6 rev/s / 0.75 rev/s = 2.112 kg⋅m^2

Therefore, the value of his moment of inertia is 2.112 kg⋅m^2.

Part (c):

Given:

Initial angular velocity, w1 = 6.6 rev/s

Final angular velocity, w2 = 3.25 rev/s

Time, t = 19 s

We can use the equation:

ΔL = L2 - L1 = I * Δw

where ΔL is the change in angular momentum, L1 and L2 are the initial and final angular momenta, I is the moment of inertia, and Δw is the change in angular velocity.

The skater's initial angular momentum (L1) is given by:

L1 = I * w1 = 0.24 kg⋅m^2 * 6.6 rev/s = 1.584 kg⋅m^2/s

His final angular momentum (L2) is:

L2 = I * w2

We need to find the magnitude of the average torque (τ) that was exerted.

We know that torque (τ) is given by:

τ = ΔL / Δt

where ΔL is the change in angular momentum and Δt is the time over which the change occurred.

So, we can rewrite the equation for angular momentum as:

ΔL = τ * Δt

Substituting this into the equation for torque, we get:

τ = ΔL / Δt = (L2 - L1) / t

Substituting the given values, we get:

τ = (I * Δw) / t = (I * (w2 - w1)) / t

τ = (0.24 kg⋅m^2 * (3.25 rev/s - 6.6 rev/s)) / 19 s

Step-by-step explanation: