. (a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.400 kg ⋅ m2. (b) He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia if his angular velocity decreases to 1.25 rev/s. (c) Suppose instead he keeps his arms in and allows friction of the ice to slow him to 3.00 rev/s. What average torque was exerted if this takes 15.0 s?
(a) L=Iω= I•2πn=0.4•2π•6 = 15.08 J•s
(b) L₁=L₂
I₁•2 πn₁ =I₂•2 πn₂
I₂=I₁•n₁/n₂= 0.4•6/1.25 = 1.92 kg•m²
(c) 2 πn₂= 2 πn₁-εt
ε=2 π(n₁-n₂)/t =
= 2 π(6-3)/15=1.26 rad/s²
τ=Iε =0.4•1.26=0.504 N•m
a) Sure, I'll calculate the angular momentum for you. To find the angular momentum, we use the formula:
Angular Momentum = Moment of Inertia * Angular Velocity
Given:
Angular Velocity (ω) = 6.00 rev/s
Moment of Inertia (I) = 0.400 kg ⋅ m^2
Angular Momentum = 0.400 kg ⋅ m^2 * 6.00 rev/s
But hold on a second, we need to convert the revolutions to radians. There are 2π radians in one revolution, so:
Angular Momentum = 0.400 kg ⋅ m^2 * 6.00 rev/s * (2π radians/1 rev)
Simplifying the expression:
Angular Momentum = 0.400 kg ⋅ m^2 * 6.00 * 2π radians/s
Now we can calculate that!
b) Now let's find the new moment of inertia when the skater reduces his angular velocity to 1.25 rev/s.
We already know the initial angular velocity (ω_i) is 6.00 rev/s, and the initial moment of inertia (I_i) is 0.400 kg ⋅ m^2.
The final angular velocity (ω_f) is 1.25 rev/s. We need to find the new moment of inertia (I_f).
The conservation of angular momentum law states:
Initial Angular Momentum (L_i) = Final Angular Momentum (L_f)
L_i = I_i * ω_i
L_f = I_f * ω_f
Since L_i = L_f, we can set these equations equal to each other:
I_i * ω_i = I_f * ω_f
Solving for I_f:
I_f = (I_i * ω_i) / ω_f
I_f = (0.400 kg ⋅ m^2 * 6.00 rev/s) / 1.25 rev/s
Now you can crunch the numbers!
c) Lastly, let's calculate the average torque exerted if the skater keeps his arms in and allows friction to slow him to 3.00 rev/s in 15.0 s.
Torque is defined as the rate of change of angular momentum. We can use the formula:
Torque = (Final Angular Momentum - Initial Angular Momentum) / Time
But first, we need to calculate the initial angular momentum (L_i) and final angular momentum (L_f).
L_i = I_i * ω_i
L_f = I_f * ω_f
We already know the initial angular velocity (ω_i) is 6.00 rev/s, the initial moment of inertia (I_i) is 0.400 kg ⋅ m^2, and the final angular velocity (ω_f) is 3.00 rev/s.
Now we can calculate the initial and final angular momentum, and then compute the torque.
Torque = (L_f - L_i) / Time
Remember to convert the revolutions to radians using the formula mentioned earlier. Have fun with the calculations!
(a) The formula to calculate the angular momentum is given by:
L = I * ω
Where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Given:
ω = 6.00 rev/s
I = 0.400 kg⋅m^2
Substituting the values into the formula:
L = 0.400 kg⋅m^2 * 6.00 rev/s
To convert rev/s to rad/s, multiply by 2π:
L = 0.400 kg⋅m^2 * (6.00 rev/s * 2π rad/rev)
L = 0.400 kg⋅m^2 * (6.00 * 2π) rad/s
L = 0.400 kg⋅m^2 * 37.70 rad/s
L ≈ 15.08 kg⋅m^2⋅s^-1
Therefore, the angular momentum of the ice skater spinning at 6.00 rev/s is approximately 15.08 kg⋅m^2⋅s^-1.
(b) The formula to calculate the new moment of inertia is given by:
I' = (L / ω')
Where I' is the new moment of inertia, L is the initial angular momentum, and ω' is the final angular velocity.
Given:
ω' = 1.25 rev/s
L = 15.08 kg⋅m^2⋅s^-1
Substituting the values into the formula:
I' = (15.08 kg⋅m^2⋅s^-1) / 1.25 rev/s
To convert rev/s to rad/s, multiply by 2π:
I' = (15.08 kg⋅m^2⋅s^-1) / (1.25 rev/s * 2π rad/rev)
I' = (15.08 kg⋅m^2⋅s^-1) / (1.25 * 2π) rad/s
I' = 15.08 kg⋅m^2⋅s^-1 / 7.85 rad/s
I' ≈ 1.92 kg⋅m^2
Therefore, the value of the moment of inertia when the angular velocity decreases to 1.25 rev/s is approximately 1.92 kg⋅m^2.
(c) The formula to calculate the average torque is given by:
τ = ΔL / Δt
Where τ is the average torque, ΔL is the change in angular momentum, and Δt is the change in time.
Given:
Δt = 15.0 s
ω = 3.00 rev/s (initial angular velocity)
To calculate ΔL, we need to find the initial angular momentum (L) and the final angular momentum (L').
Using the formula from part (a):
L = I * ω
Given:
ω = 3.00 rev/s
I = 0.400 kg⋅m^2
Substituting the values:
L = 0.400 kg⋅m^2 * 3.00 rev/s
To convert rev/s to rad/s, multiply by 2π:
L = 0.400 kg⋅m^2 * (3.00 rev/s * 2π rad/rev)
L = 0.400 kg⋅m^2 * 18.85 rad/s
L ≈ 7.54 kg⋅m^2⋅s^-1
Using the formula for torque:
τ = (L' - L) / Δt
Rearranging the formula:
L' = (τ * Δt) + L
Given Δt = 15.0 s, ω = 3.00 rev/s, and L = 7.54 kg⋅m^2⋅s^-1, we can solve for L' by substituting the values:
L' = (τ * 15.0 s) + 7.54 kg⋅m^2⋅s^-1
From the given information, we can't determine the value of τ.
(a) To calculate the angular momentum of an ice skater spinning at 6.00 rev/s, we can use the formula:
Angular momentum (L) = Moment of inertia (I) * Angular velocity (ω)
Given:
Angular velocity (ω) = 6.00 rev/s
Moment of inertia (I) = 0.400 kg ⋅ m²
Substituting the values into the formula:
L = 0.400 kg ⋅ m² * 6.00 rev/s
To get the answer, we need to find the product of 0.400 kg ⋅ m² and 6.00 rev/s. The units for rev/s can be converted to radians/s by multiplying by 2π (since 1 revolution = 2π radians).
L = 0.400 kg ⋅ m² * 6.00 rev/s * 2π radians/1 rev
Simplifying the equation:
L = 0.400 kg ⋅ m² * 6.00 * 2π radians/s
Now, you can calculate the value of L using a calculator or software, such as:
L = 4.8 kg ⋅ m²/s
Therefore, the angular momentum of the ice skater spinning at 6.00 rev/s is 4.8 kg ⋅ m²/s.
(b) To find the value of the moment of inertia when the ice skater's angular velocity decreases to 1.25 rev/s, we can rearrange the formula:
Angular momentum (L) = Moment of inertia (I) * Angular velocity (ω)
I = L/ω
Given:
Angular velocity (ω) = 1.25 rev/s
Substituting the values:
I = 4.8 kg ⋅ m²/s / 1.25 rev/s * 2π radians/1 rev
Simplifying the equation:
I = 4.8 kg ⋅ m²/s / 1.25 * 2π radians/s
Now, you can calculate the value of I using a calculator or software, such as:
I ≈ 3.05 kg ⋅ m²
Therefore, the moment of inertia when the ice skater's angular velocity decreases to 1.25 rev/s is approximately 3.05 kg ⋅ m².
(c) To find the average torque exerted if the ice skater slows down to 3.00 rev/s with the help of friction over 15.0 s, we can use the formula:
Average torque (τ) = Change in angular momentum (ΔL) / Change in time (Δt)
The change in angular momentum can be calculated using:
ΔL = Final angular momentum (L_final) - Initial angular momentum (L_initial)
Given:
Final angular velocity (ω_final) = 3.00 rev/s
Time taken (Δt) = 15.0 s
First, we need to calculate the initial angular momentum (L_initial) using the formula from part (a):
L_initial = 0.400 kg ⋅ m² * 6.00 rev/s * 2π radians/1 rev
Now, we can calculate the final angular momentum (L_final) using the formula from part (b):
L_final = 3.05 kg ⋅ m² * 3.00 rev/s * 2π radians/1 rev
Substituting the values:
ΔL = (3.05 kg ⋅ m² * 3.00 rev/s * 2π radians/1 rev) - (0.400 kg ⋅ m² * 6.00 rev/s * 2π radians/1 rev)
To simplify the equation, calculate the changes in angular momentum:
ΔL ≈ 18.24 kg ⋅ m²/s - 4.8 kg ⋅ m²/s
ΔL ≈ 13.44 kg ⋅ m²/s
Finally, we can calculate the average torque (τ) using:
τ = ΔL / Δt
τ = 13.44 kg ⋅ m²/s / 15.0 s
τ ≈ 0.896 N⋅m
Therefore, the average torque exerted if the ice skater slows down to 3.00 rev/s over 15.0 s is approximately 0.896 N⋅m.