Respuesta :
Answer:
wf = 3.307 rad/s
Explanation:
Given:-
- The mass of circular platform, mc = 66.1 kg
- The radius of the platform, ro = 3.01 m
- The mass of the student, ms = 56.3 kg
- The combined initial angular velocity of (platform + student), wi = 3.9 rad/s
At the rim
Find:-
Find omega when the student is 2.81m from the center.
Solution:-
- We will consider the circular platform as a disc whose moment of inertia from a central vertical axis is given by:
Idisc = 0.5*mc*ro^2
Idisc = 0.5*66.1*3.01^2
Idisc = 299.436305 kg.m^2
- The Inertia of the student is function of distance r it is from the central vertical axis. Initially the inertia of student would be r = ro = 3.01 at the rim of the platform is given by:
(Istd)_i = ms*ro^2
(Istd)_i =56.3*3.01^2
(Istd)_i = 510.08363 kg.m^2
- The final inertia of the student after he jumps to a point that is 2.81 radially away from the central vertical axis. The new inertia would be defined by r = 2.81 m.
(Istd)_f = ms*r^2
(Istd)_f =56.3*2.81^2
(Istd)_f = 444.55043 kg.m^2
- We apply the principle of conservation of angular momentum ( L ) before and after the jump.
Li = Lf
Li = Initial angular momentum = wi*( Idisc + (Istd)_i )
Lf = Final angular momentum = wf*(Idisc + (Istd)_f )
- The final angular velocity wf can be determined:
wi*( Idisc + (Istd)_i ) = wf*(Idisc + (Istd)_f )
wf = [ ( Idisc + (Istd)_i ) / (Idisc + (Istd)_f ) ] * wi
wf = [ (510.08363 + 299.436305) / ( 510.08363 + 444.55043)]*3.9
wf = ( 809.519935 / 954.63406 ) * 3.9
wf = 0.84798*3.9
wf = 3.307 rad/s
