conservation of angular momentum (the figure skater spins faster when she pulls her arms in
I1 omega1 = I2 omega2
I of platform = (1/2)(143.1)(3.01)^2
original I of student = m r^2 = 75.3(3.01)^2
final I of student = 75.3(2.89)^2
A large horizontal circular platform (M=143.1 kg, r=3.01 m) rotates about a frictionless vertical axle. A student (m=75.3 kg) walks slowly from the rim of the platform toward the center. The angular velocity ω of the system is 2.10 rad/s when the student is at the rim. Find ω (in rad/s) when the student is 2.89 m from the center.
4 answers
something is wrong in the answer, can you explain it again please?..
Ok,
Original I = (1/2)(143.1)(3.01^2 + 75.3*(3.01)^26
= 648 + 682 = 1330 kg m^2
Final I = 648 + 75.3(2.89)^2 = 1276
1330 (2.1) = 1276 omega
so
omega = 2.19 radians/second
Original I = (1/2)(143.1)(3.01^2 + 75.3*(3.01)^26
= 648 + 682 = 1330 kg m^2
Final I = 648 + 75.3(2.89)^2 = 1276
1330 (2.1) = 1276 omega
so
omega = 2.19 radians/second
Thank you very much.
1 answer