Date | November 2017 | Marks available | 2 | Reference code | 17N.3.SL.TZ0.7 |
Level | Standard level | Paper | Paper 3 | Time zone | 0 - no time zone |
Command term | Label and Draw | Question number | 7 | Adapted from | N/A |
Question
A hoop of mass m, radius r and moment of inertia mr2 rests on a rough plane inclined at an angle θ to the horizontal. It is released so that the hoop gains linear and angular acceleration by rolling, without slipping, down the plane.
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 =
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.
Markscheme
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 – Ff
I = Ff x r
OR
mr = Ff
=
ma = mg sin θ – mr → 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 = Iω2 + mv2
substituting ω = «giving v = »
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
Ff ≤ μmg cos θ
ALTERNATIVE 2
Ff = ma «from 7(b)»
so Ff =
Ff = μmg cos θ
= mg sin θ – μmg cos θ
OR
mg = μmg cos θ
algebraic manipulation to reach tan θ = 2μ