AS 91524 · 2023 paper · rotational motion

Question 2(b) Merit to Excellence

Practises Energy methods for rolling and rotating bodies

Tane works weekends unloading barrels. He rolls an empty barrel of mass 5.50 kg and radius 0.280 m down a ramp 1.34 m high. The linear speed of the barrel at the bottom of the ramp is 3.40 m s⁻¹. Parts (c) and (d) move to Tane on a swivelling stool holding a full bottle of water in each hand: he spins more slowly with his arms outstretched than with his arms in.

m	5.50 kg
r	0.280 m
h	1.34 m
v	3.40 m s⁻¹

Calculate the rotational inertia of the barrel. Begin your answer by calculating the gravitational potential energy at the top and the angular velocity of the barrel as it reaches the bottom of the ramp. Assume no energy is lost due to friction.

The 2023 paper, page 4 ↗ Marking schedule, page 4 ↗

Work it out

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

1 How much gravitational potential energy does the barrel have at the top?
E_gp = mgh = 5.50 × 9.81 × 1.34 = 72.30 J.
2 How fast is it spinning at the bottom?
It rolls, so ω = v / r = 3.40 / 0.280 = 12.143 rad s⁻¹.
3 How much of the 72.30 J is linear kinetic energy, and so how much is left for rotation?
E_k(lin) = ½mv² = ½ × 5.50 × 3.40² = 31.79 J, so E_k(rot) = 72.30 − 31.79 = 40.51 J.
4 Which equation gives I from the rotational energy and ω?
E_k(rot) = ½Iω², so I = 2 × 40.51 / 12.143² = 0.549 kg m².

Answer I = 0.549 kg m². The schedule’s 0.553 kg m² comes from rounding ω to 12.1 rad s⁻¹ first; it accepts both. A note on the numbers: they give I ≈ 1.27 mr², more than any real barrel can have (even a thin hoop of this mass and radius is only mr² = 0.431 kg m²). That is the paper’s data, not your working, so just follow the method.

What the marker wanted

Achieved One of: E_gp = 72.3 J; E_k(lin) = 31.8 J; or ω = v / r = 12.1 rad s⁻¹.
Merit All three energies (72.3 J, 31.8 J and 40.5 J); or I found by putting all of the potential energy into rotation (0.987 kg m²).
Excellence All three energies and I = 0.553 kg m² (0.549 kg m² from unrounded values).

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

Common mistake

Putting all 72.3 J into rotation and leaving out ½mv². That gives 0.987 kg m², which the schedule holds at Merit: the barrel moves along as well as spinning, so part of the energy is linear.

How did it go?

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