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I have been trying to approach this by finding the gravitational torque of the person on the merry-go-round, however I can't find a way to relate that to angular velocity in the given circumstances.
If this problem requires angular momentum to solve, then I won't need to know the solution because it won't be on my final, however I still want to make sure that I'm not missing something.
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$\begingroup$It looks like an A-Level problem.
Assuming no friction, by conservation of angular momentum
\begin{align*} I_1 \omega_1 &= I_2 \omega_2 \\ (150+65\times 0.5^2) 1.7 &= (150+65\times 1.5^2) \omega_2 \\ \omega_2 &= 0.95 \text{ rad/s} \end{align*}
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