Answer
a) $1.1KJ$
b) $173m$
Work Step by Step
(a) We know that
$K.E=\frac{1}{2}(\frac{1}{12}ML^2)\omega^2$
We plug in the known values to obtain:
$K.E=\frac{1}{2}(0.65)(0.55)^2(\frac{3500rev}{min}\times \frac{2\pi rad}{rev}\times \frac{1min}{60} )^2=1100J=1.1KJ$
(b) As $K.E_r=mgyl_{max}$
This can be rearranged as:
$y_{max}=\frac{K.E_r}{mg}$
We plug in the known values to obtain:
$y_{max}=\frac{1100}{(0.65)(9.81)}=173m$
