The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center. When the skater's hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thin-walled hollow cylinder. His hands and arms have a combined mass of 8.50 kg. When outstretched, they span 1.80 m; when wrapped, they form a cylinder of radius 25.0 cm (.25m).
The moment of inertia about the axis of rotation of the remainder of his body is constant and equal to 0.350 kg*m^2 If the skater's original angular speed is 0.450 rev/s, what is his final angular speed?
Use conservation of angular momentum. Assume the outstretched arms have a moment of inertia contribution of
Iarms2 = (1/3) M(arms) L^2, where L = 1.80m m.
[(Ibody + Iarms1)w1] = [(Ibody + Iarms2)w2]
Solve for w2
To solve this problem, we need to apply the conservation of angular momentum. Angular momentum is conserved when there are no external torques acting on a system.
The initial angular momentum (L_initial) of the skater can be calculated using the formula:
L_initial = I_initial * ω_initial
Where:
L_initial is the initial angular momentum
I_initial is the initial moment of inertia of the system
ω_initial is the initial angular speed
The final moment of inertia (I_final) can be obtained by considering the moment of inertia of the skater's arms and the constant moment of inertia of the rest of his body:
I_final = I_arms + I_body
The final angular momentum (L_final) is then given by:
L_final = I_final * ω_final
According to the conservation of angular momentum, L_initial = L_final. So we can equate the two equations and solve for the final angular speed (ω_final).
Let's substitute the given values into the equations:
I_initial = 0.350 kg*m^2 (given)
ω_initial = 0.450 rev/s (given)
I_arms = ½ * m_arms * r^2 = ½ * m_arms * (0.25 m)^2 (moment of inertia of a thin-walled hollow cylinder)
I_body = constant moment of inertia, given as 0.350 kg*m^2
m_arms = 8.50 kg (given)
Substituting these values into the equations, we have:
L_initial = I_initial * ω_initial
L_initial = (0.350 kg*m^2) * (0.450 rev/s)
L_final = I_arms + I_body
L_final = (½ * m_arms * (0.25 m)^2) + 0.350 kg*m^2
Since L_initial = L_final, we can equate the two equations:
(0.350 kg*m^2) * (0.450 rev/s) = (½ * m_arms * (0.25 m)^2) + 0.350 kg*m^2 * ω_final
Now we can solve for ω_final:
ω_final = [(0.350 kg*m^2) * (0.450 rev/s) - (½ * m_arms * (0.25 m)^2)] / (0.350 kg*m^2)
After substituting the given values and calculating this expression, we can find the final angular speed (ω_final).