Circular Motion — Practice Problem Solutions

1. $a_c=\dfrac{v^2}{r}=\dfrac{6.0^2}{2.5}=14.4\text{ m/s}^2$.

2. $T=\dfrac{mv^2}{r}=\dfrac{(0.20)(4.0)^2}{0.80}=\dfrac{3.2}{0.80}=4.0\text{ N}$.

3. For an ideally banked curve, $\tan\theta=\dfrac{v^2}{rg}\implies v=\sqrt{rg\tan\theta}=\sqrt{(80)(9.8)\tan25°}\approx\sqrt{365.6}\approx19.1\text{ m/s}$.