On the diagram, draw and label the forces acting on the hoop.

Show that the linear acceleration a of the hoop is given by the equation shown.

a = \(\frac{{g \times \sin q}}{2}\)

Calculate the acceleration of the hoop when θ = 20°. Assume that the hoop continues to roll without slipping.

State the relationship between the force of friction and the angle of the incline.

The angle of the incline is slowly increased from zero. Determine the angle, in terms of the coefficient of friction, at which the hoop will begin to slip.

weight, normal reaction and friction in correct direction

correct points of application for at least two correct forces

Labelled on diagram.

Allow different wording and symbols

Ignore relative lengths

ALTERNATIVE 1

ma = mg sin θ – F_f

I\(\alpha \) = F_f x r

OR

mr \(\alpha \) = F_f

\(\alpha \) = \(\frac{a}{r}\)

ma = mg sin θ – mr \(\frac{a}{r}\) → 2a = g sin θ

Can be in any order

No mark for re-writing given answer

Accept answers using the parallel axis theorem (with I = 2mr²) only if clear and explicit mention that the only torque is from the weight

Answer given look for correct working

ALTERNATIVE 2

mgh = \(\frac{1}{2}\)Iω² + \(\frac{1}{2}\) mv²

substituting ω = \(\frac{v}{r}\) «giving v = \(\sqrt {gh} \)»

correct use of a kinematic equation

use of trigonometry to relate displacement and height «s = h sin θ»

For alternative 2, MP3 and MP4 can only be awarded if the previous marking points are present

1.68 «ms^–2»

ALTERNATIVE 1

N = mg cos θ

F_f ≤ μmg cos θ

ALTERNATIVE 2

F_f = ma «from 7(b)»

so F_f = \(\frac{{mg\sin \theta }}{2}\)

F_f = μmg cos θ

\(\frac{{mg\sin \theta }}{2}\) = mg sin θ – μmg cos θ

OR

mg \(\frac{{\sin \theta }}{2}\) = μmg cos θ

algebraic manipulation to reach tan θ = 2μ