Chapter 11 - Rotational Dynamics and Static Equilibrium - Problems and Conceptual Exercises - Page 375: 124

$121^{\circ}$

We know that $sin\theta=(\frac{\omega_f}{rmv})(1+mr^2)$ We plug in the known values to obtain: $sin\theta=\frac{0.272rad/s}{(2.31m)(34.0Kg)(2.80m/s)}[512Kg.m^2+(34.0Kg)(2.31m)^2]$ $\implies sin\theta=0.8576$ $\implies \theta=59.05^{\circ}$ Now $\theta=\pi -59.05^{\circ}$ $\theta=180^{\circ}-59.05^{\circ}=121^{\circ}$