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 (Figure 1). 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 8.0 kg . When outstretched, they span 1.7 m ; when wrapped, they form a thin-walled hollow cylinder of radius 23 cm . The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.4 kg⋅m2.

If his original angular speed is 0.30 rev/s , what is his final angular speed?

Answer 1

Solution:-

According to law of conservation of momentum

It says that the momentum before an event must be the same as the amount after due to its constant conservation. It is an element of law of inertia.

Iω = (I ω)f

I(initial ) = 1body + 1 arm

0.4kgm² + 1/12 *m*I²

0.4kgm² + 1/12 *(8.0kg)*(1.7m)²

= 2.32 kg-m²

I(initial) = 0.4kgm² + mr²

I(initial) = 0.4kgm² + 8.0*(0.23)²

I (initial) = 0.82kg-m²

= 2.32*0.30= 0.82ὠ (f)

ω(f) = 2.32*0.30 / 0.82

ω (f) = 0.84 rev/s

Final angular speed is ω(f) = 0.84 rev/s