A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the centre of the disk. The platform has a mass of `150 kg`, a radius of `2.0 m`, and a rotational inertia of`300 kg m^2` about the axis of rotation. A `60 kg` student walks slowly from the rim of the platform toward the centre. If the angular speed of the system is `1.5 rad//s` when the student starts at the rim, what is the angular speed when she is `0.50 m` from the centre ?

Options

A. `1.2 rad//s`
B. `2.6 rad//s`
C. `1.5 rad//s`
D. `3.6 rad//s`

Answer

B

Explanation

(b) The initial rotational inertia of the system is `I_i = I_(disk) + I_(student)`, where `I_(disk) = 300 kg m^2` (which, incidentally does agree with Table `10-2`(c )) and `I_(student) = mR^2` where `m = 60 kg` and `R = 2.0 m`
The rotational inertia when the student reaches `r = 0.5 m` is `I_f = I_(disk) + mr^2`. Angular momentum conservation leads to
`I_i omega_i = I_f omega_f rArr omega_f = omega_i ((I_(disk) + mR^2)/(I_(disk) + mr^2))`
Which yields, for `omega_i = 1.5 rad//s`, a final angular velocity of `omega_f = 2.6 rad//s`.