a) Using the conservation of angular momentum, we can solve this problem.
Let m be the mass of the beetle and M be the mass of the turntable.
mR^2ω_b + Iω_t=0, where I is the moment of inertia for the turntable, which is 1/2MR^2.
ω_t = - 2m/Mω_b = -0.07 * 2 * 15/ 95=-0.02210rad/s.
In our eyes, beetle is on the turntale. Thereforeω = 0.0700 - 0.02210 = + 0.0579rad/s
b) As we have seen in part a, it is -0.02210rad/s
c) The answer depends on the angular velocity of the beetle RELATIVE TO THE TURNTABLE. Hence
a+b) angular momentum has to be conserved. Since there is no motion at first, the later motions have to sum to 0.
ω=-.022105 rad/s turn table
ω=.04789 rad/s of beetle respect to you
c) we have to use the angular speed of the beetle with respect to the turn table.
Hope that helps