A skater is initially spinning at a rate of 12.4 rad/s with a rotational inertia of 2.8 kg·m2 when her arms ar?
Update:
A skater is initially spinning at a rate of 12.4 rad/s with a rotational inertia of 2.8 kg·m2 when her arms are extended. What is her angular velocity after she pulls her arms in and reduces her rotational inertia to 1.55 kg·m2
More Questions From This User See All
Answers & Comments
Verified answer
momentum before = momentum after
momentum before = 2.8 x 12.4 = 34.72 kg-m^2/sec
momentum after = 34.72 kg-m^2/sec
34.72 = 1.55omega
34.72/1.55 = omega = 22.4 rad/sec
this is a problem in conservation of angular momentum
angular momentum = I w where I is the moment ofinertia and w is the ang velocity
so we have
I(before) w(before) = I(after) w (after)
or w(after) = I(before) w(before)/I(after)
w(after)= 2.8kgm^2*12.4rad/s / 1.55kgm^2 = 22.4rad/s