AS 91524 · 2023 paper · vertical circular motion

Question 1(b) Merit

Tane has a toy car track: part is a horizontal banked curve of radius 0.750 m, part is a vertical loop of radius 0.250 m. The car has a mass of 0.120 kg. Assume sideways friction on the tyres is negligible.

m	0.120 kg
banked curve r	0.750 m at 1.55 m s⁻¹
loop r	0.250 m

The banked curve has a radius of 0.750 m. Calculate the angle of banking when there is no sideways friction on the wheels as the car goes round at 1.55 m s⁻¹.

The 2023 paper, page 2 ↗ Marking schedule, page 1 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1 How big is the centripetal force the car needs, and how big is its weight?
F_c = mv² / r = 0.120 × 1.55² / 0.750 = 0.3844 N. F_g = mg = 0.120 × 9.81 = 1.1772 N.
2 In your vector diagram from (a), where is the banking angle, and which two sides give it?
The reaction force is tilted from the vertical by the banking angle θ. Its vertical part balances the weight and its horizontal part is F_c, so tan θ = F_c / F_g.
3 Work it out.
tan θ = 0.3844 / 1.1772 = 0.3265, so θ = tan⁻¹(0.3265) = 18.1°. The mass cancels, so it is the same as tan θ = v² / (rg) = 1.55² / (0.750 × 9.81).

Answer θ = 18.1°

What the marker wanted

Achieved One of: the method with one error, such as the wrong trig function; a correct vector diagram; or both F_c and F_g right.
Merit The right answer, 18.1°.

In our words, from NZQA’s 2023 marking schedule.

Common mistake

Using sin or cos, or putting F_g over F_c. NZQA’s 2023 report says many candidates had both forces right and then went wrong rearranging for θ, which the schedule credits at Achieved only.

How did it go?

Got it moves it further out. Review brings it back tomorrow.