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

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: $N = mg + \frac{m v^2}{r}$.

Derive the minimum speed at the top of a vertical circle for an object to maintain contact with the track. [2 marks]

A $0.5$ kg ball is swung in a vertical circle of radius $1.2$ m at constant speed $4.0$ m s$^{-1}$. Calculate (a) the tension at the top, and (b) the tension at the bottom. [4 marks]

Refer to Luna Park's $8$ m loop. (a) Calculate the minimum speed at the top. (b) Determine the normal force at the bottom for a $70$ kg passenger at $14$ m s$^{-1}$.