Question

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 his 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 9.0 kg. When outstretched, they span 1.7 m; when wrapped, they form a cylinder of radius 26 cm. The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.40 kg * m2. If his original angular speed is 0.30 rev/s, what is his final angular speed?

Answer 1

Answer:

$W_f =4.7968 rad/s$

Explanation:

By the law of the conservation of the angular momentum L:

$L_i = L_f$

Where:

$L_i=I_iW_i$

$L_f=I_fW_f$

where $I_i$ is the inicial moment of inertia, $W_i$ is the initial angular velocity,  $I_f$ is the final moment of inertia and $W_f$ is the final angular velocity.

Replacing:

$I_iW_i = I_fW_f$

Now, we have to change the angular velocity from revolutions to radians as:

$W_i=0.3*2\pi rad/s$

$W_i=1.884 rad/s$

Then, we find the initial moment of inertia ($I_i$) as:

$I_i = \frac{1}{12}(M)(R_1)^2+0.4$

$I_i = \frac{1}{12}(9 kg)(1.7m)^2+0.4$

$I_i = 2.5675 kg*m^2$

Where M is the mass of his hands and arms and R1 is the length of his arms and hands whe they are outstretched.

Now we find the final moment of inertia ($I_f$) as:

$I_f = MR_2^2+0.4$

$I_f = (9)(0.26)^2+0.4$

$I_ f = 1.0084 kg*m^2$

Where R2 is the radius of the cylinder formed.

Finally we replace all in the first equation as:

$I_iW_i = I_fW_f$

$(2.5675)(1.884) = (1.0084)W_f$

Solving for $W_f$, we get:

$W_f =4.7968 rad/s$