Vertical Circular Motion (Water Bucket) — Practice Problem Solutions

1. $v_{min}=\sqrt{gr}=\sqrt{(9.8)(0.75)}=\sqrt{7.35}\approx2.71\text{ m/s}$.

2. Top of the swing: $N+mg=\dfrac{mv^2}{r}\implies N=\dfrac{mv^2}{r}-mg=\dfrac{(2.0)(4.5)^2}{0.60}-(2.0)(9.8)=67.5-19.6=47.9\text{ N}$. Bottom of the swing: $N-mg=\dfrac{mv^2}{r}\implies N=\dfrac{mv^2}{r}+mg=67.5+19.6=87.1\text{ N}$.