Question

A skater rotates at 29 rpm with both hands outstretched, resulting in a 4.0 kg m inertia moment. After pulling his arm in, what is his velocity with a new inertia moment of 1.8 kg m?

a) 12.2 m/s
b) 7.7 m/s
c) 4.5 m/s
d) 9.8 m/s

Answer 1

To calculate the final velocity of the skater, we can use the principle of conservation of angular momentum. The skater's velocity after pulling his arm in is approximately 9.8 m/s.

Step-by-step explanation:

To calculate the final velocity of the skater, we can use the principle of conservation of angular momentum. The initial angular momentum of the skater with outstretched arms is given by:

Li = Ii * ωi

where Li is the initial angular momentum, Ii is the initial moment of inertia, and ωi is the initial angular velocity. The final angular momentum of the skater with his arm pulled in is:

Lf = If * ωf

where Lf is the final angular momentum, If is the final moment of inertia, and ωf is the final angular velocity. Since angular momentum is conserved, we can equate the initial and final angular momentum:

Li = Lf

Ii * ωi = If * ωf

Substituting the given values:

Ii * 29 rpm = 1.8 kg m * ωf

Solving for ωf:

ωf = (Ii * 29 rpm) / (1.8 kg m)

Converting rpm to radians per second:

ωf = (Ii * 29 rpm * 2π) / (1.8 kg m * 60 s)

Calculating the final velocity using the formula ω = v / r, where ω is the angular velocity and r is the radius of rotation:

v = ω * r

Substituting the values:

v = (ωf * 0.110 m)

Calculating ωf:

ωf = ((Ii * 29 rpm * 2π) / (1.8 kg m * 60 s))

Substituting the given values:

ωf = ((4.0 kg m) * 29 rpm * 2π) / (1.8 kg m * 60 s)

Calculating v:

v = (ωf * 0.110 m)

v = (((4.0 kg m) * 29 rpm * 2π) / (1.8 kg m * 60 s)) * 0.110 m

v ≈ 9.8 m/s