Points will only be given for right answer please show all work! The outstretche
ID: 1447657 • Letter: P
Question
Points will only be given for right answer please show all work!
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. (See the figure below (Figure 1) .) 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 7.50 kg. When outstretched, they span 1.80 m: when wrapped, they form a cylinder of radius 25.0 cm. The moment of inertia about the axis of rotation of the remainder of his body is constant and equal to 0.450 kg middot m^2. If the skater's original angular speed is 0.350 rev/s, what is his final angular speed?Explanation / Answer
This isn't any net force acting so, angular momentum is conserved:
L = I = I' ' = L'
moment of inertia of a cylinder: I = mr²
angular frequency: = 0.35*2 = 2.2 rad/s
I = 0.45 + (mhands)r²
M = (mhands) + mbody
but, .....Ibody = 0.45 kg m²
0.45 = mbody r² = mbody 0.25²
mbody = 0.45/0.25² = 7.2 kg
mhands = M - mbody = 7.5 - 7.2 = 0.3 kg
Now, set up the conservation principle:
I = I' '
0.45*2.2 + 0.3[(1.8/2)²](2.2) = 8.5*0.25² '
solve for the final angular momentum:
' = 1.69 rad/s, ......=> 1.69/2 = 0.26 rev/s