Quick questions on Uniform circular motion (horizontal and vertical): VCE Physics Unit 3

$$a_c = \frac{v^2}{r}, \quad F_c = m a_c = \frac{m v^2}{r}$$

The period $T$ is the time for one revolution; the frequency $f = 1/T$.

A mass on a string swung in a horizontal circle so that the string makes angle $\theta$ with the vertical. Two real forces act: tension $T$ along the string, and gravity $mg$ down. The vertical components must balance; the horizontal component of tension supplies the centripetal force.

Consider a ball on a string of radius $r$ in a vertical circle at constant speed $v$. Gravity always acts down. Tension acts along the string toward the centre.

Both tension $T_{top}$ and gravity point down (toward the centre).

Tension $T_{bot}$ points up (toward the centre); gravity points down (away).

Replace tension with the normal force from the rail. The cart feels lightest (sometimes momentarily weightless) at the top, when $N + mg = \frac{m v^2}{r}$.

Normal force points up; $N - mg = \frac{m v^2}{r}$, so apparent weight (felt by the rider) is maximum: $N = mg + \frac{m v^2}{r}$.

Both point toward the centre at the top, so $T_{top} + mg = \frac{mv^2}{r}$. At the bottom they oppose, so $T_{bot} - mg = \frac{mv^2}{r}$.

Centrifugal force is fictitious; only use it in a rotating frame, which VCE does not require.

$a_c = \frac{v^2}{r}$, not $\frac{v^2}{d}$.

$\omega$ is in rad/s, $f$ is in Hz, related by $\omega = 2 \pi f$.