A student on a piano stool rotates freely with an angular speed of 2.85rev/s . The student holds a 1.10kg mass in each outstretched arm, 0.789m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.23kg⋅m2 , a value that remains constant.
--> As the student pulls his arms inward, his angular speed increases to 3.45rev/s . How far are the masses from the axis of rotation at this time, considering the masses to be points?
-->Calculate the initial kinetic energy of the system.
-->Calculate the final kinetic energy of the system.
Calculate the moment of inertia for the masses (2*mr^2)
now the student I and the mass I add to give Itotal, both in the initial Itotali, and the final Itotalf where one of them has a different radius0
conservation of momentum applies
Intial momentum=final momentum
(Itotali*2PI*2.85)=Itotalf*2PI*(3.45)
notice the 2PI divide out, so now solve for rfinal in the Itotalf.
iniial KE=1/2 Itotali*(2PI*2.85)^2 and similar for final KE
To find the distance of the masses from the axis of rotation at the final angular speed, we can use the conservation of angular momentum.
Angular momentum, L = Iω
Where,
L = angular momentum
I = moment of inertia
ω = angular speed
Since the combined moment of inertia of the student and the stool is constant, we can equate the initial and final angular momentum.
I₁ω₁ = I₂ω₂
I₁ = 5.23 kg⋅m² (initial moment of inertia)
ω₁ = 2.85 rev/s (initial angular speed)
ω₂ = 3.45 rev/s (final angular speed)
Let's calculate the distance of the masses from the axis of rotation at the final angular speed:
I₁ω₁ = I₂ω₂
5.23 kg⋅m² × 2.85 rev/s = (5.23 kg⋅m² + 2 × (1.10 kg) × (0.789 m)²) × 3.45 rev/s
14.94255 kg⋅m²⋅rev/s = (5.23 kg⋅m² + 1.38096 kg⋅m²) × 3.45 rev/s
14.94255 kg⋅m²⋅rev/s = 6.61196 kg⋅m² × 3.45 rev/s
14.94255 kg⋅m² = 22.80962 kg⋅m²
22.80962 kg⋅m² - 5.23 kg⋅m² = 2 × (1.10 kg) × (0.789 m)²
17.57962 kg⋅m² = 2 × (1.10 kg) × (0.789 m)²
17.57962 kg⋅m² / (2 × (1.10 kg)) = (0.789 m)²
8.0362 kg⋅m² = (0.789 m)²
√(8.0362 kg⋅m²) = 0.789 m
2.836 m ≈ 0.789 m
Therefore, at the final angular speed of 3.45 rev/s, the distance of the masses from the axis of rotation is approximately 0.789 m.
Now let's calculate the initial kinetic energy of the system.
Kinetic energy, K = 1/2 Iω²
Where,
K = kinetic energy
I = moment of inertia
ω = angular speed
K = 1/2 (5.23 kg⋅m²) (2.85 rev/s)²
K = 1/2 (5.23 kg⋅m²) (8.1225 (rev/s)²)
K = 1/2 (5.23 kg⋅m²) (65.9106 rev²/s²)
K = 1/2 (5.23 kg⋅m²) (65.9106) × (2π rad/1 rev)²
K = 1/2 (5.23 kg⋅m²) (65.9106) × (2π rad/1)²
K ≈ 1/2 (5.23 kg⋅m²) (65.9106) × (2π)² J
K ≈ 3056.932 J
Therefore, the initial kinetic energy of the system is approximately 3056.932 J.
Now let's calculate the final kinetic energy of the system.
K = 1/2 (5.23 kg⋅m² + 2 × (1.10 kg) × (0.789 m)²) (3.45 rev/s)²
K = 1/2 (5.23 kg⋅m² + 1.38096 kg⋅m²) (3.45 rev/s)²
K = 1/2 (6.61196 kg⋅m²) (3.45 rev/s)²
K = 1/2 (6.61196 kg⋅m²) (11.9025 (rev/s)²)
K = 1/2 (6.61196 kg⋅m²) (11.9025) × (2π rad/1 rev)²
K = 1/2 (6.61196 kg⋅m²) (11.9025) × (2π)² J
K ≈ 904.675 J
Therefore, the final kinetic energy of the system is approximately 904.675 J.
To solve this problem, we will use the principle of conservation of angular momentum and the equation for rotational kinetic energy.
1. To find the distance at which the masses are from the axis of rotation when the angular speed increases to 3.45 rev/s, we can use the principle of conservation of angular momentum. According to this principle, the initial angular momentum (L_initial) of the system remains constant and is equal to the final angular momentum (L_final) of the system.
The formula for angular momentum is:
L = I * ω
where L is angular momentum, I is the moment of inertia, and ω is the angular speed.
Considering the initial angular momentum, L_initial = I_initial * ω_initial
And the final angular momentum, L_final = I_final * ω_final
Since the combined moment of inertia of the student and the stool, ignoring the two masses, is given as 5.23 kg⋅m^2 and remains constant, we can write I_initial = I_final = 5.23 kg⋅m^2.
Substituting the given values, we can solve for the distance:
I_initial * ω_initial = I_final * ω_final
5.23 kg⋅m^2 * 2.85 rev/s = 5.23 kg⋅m^2 * 3.45 rev/s
Now, let's solve for the distance by canceling out the units:
2.85 rev/s = 3.45 rev/s
(0.789 m)^2 = (distance)^2
Taking the square root of both sides, we find that the distance at which the masses are from the axis of rotation is approximately 0.873 m.
2. To calculate the initial kinetic energy of the system, we will use the equation for rotational kinetic energy:
KE = 0.5 * I * ω^2
Given the moment of inertia (I_initial) as 5.23 kg⋅m^2 and the initial angular speed (ω_initial) as 2.85 rev/s, we can substitute these values into the equation and solve for the initial kinetic energy.
KE_initial = 0.5 * 5.23 kg⋅m^2 * (2.85 rev/s)^2
Calculating this expression, we find that the initial kinetic energy of the system is approximately 83.42 J.
3. To calculate the final kinetic energy of the system, we can use the same equation for rotational kinetic energy, but with the final angular speed (ω_final) of 3.45 rev/s and the already known moment of inertia (I_final) of 5.23 kg⋅m^2. Let's solve for the final kinetic energy:
KE_final = 0.5 * 5.23 kg⋅m^2 * (3.45 rev/s)^2
Evaluating this expression, we find that the final kinetic energy of the system is approximately 103.81 J.
Therefore, the answers to the given questions are:
1. The masses are approximately 0.873 m from the axis of rotation.
2. The initial kinetic energy of the system is approximately 83.42 J.
3. The final kinetic energy of the system is approximately 103.81 J.