Goes in circle because there is acceleration towards the centre. The instantaneous velocity is changing, so there really is acceleration. See proof. $a=\frac{v^2}{r}$ Substitute the circumference divided by period for $v$. Angular Quantities. $\frac{4{\pi }^{2}r}{T^2}={\omega }^{2}r$ To get $F_{net}$, multiply by $m$. $F_t$ doesn’t have direction?

Proof

$a=\frac{v^2}{r}={\frac{2\pi r}{T}}^2\times \frac{1}{r}=\frac{4\pi r}{T^2}$

Relation to cars

The static friction is perpendicular to the direction of travel. See here. When skidding, there is no longer static friction perpendicular to direction of travel, instead kinetic friction that is weaker and opposite direction of travel.

Strings and Gravity

The tension and length of string have to use trigonometry again.

Universal Gravitation Law

$F_g=G\frac{m^1{m}^2}{r^2}$

Radius and Period

Using: $\frac{4{\pi }^{2}r}{T^2}=F_g$ Turn $F_g$ into the universal gravitational law, to get $\frac{4{\pi }^{2}r}{T^2}=G\frac{m^1{m}^2}{r^2}$