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Question 17N.3.SL.TZ0.b

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Date November 2017 Marks available [Maximum mark: 4] Reference code 17N.3.SL.TZ0.b
Level SL Paper 3 Time zone TZ0
Command term Show that Question number b Adapted from N/A
b.
[Maximum mark: 4]
17N.3.SL.TZ0.b

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.

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

a =  g × sin ⁡ q 2

[4]

Markscheme

ALTERNATIVE 1

ma = mg sin θ – Ff

I α = Ff x r

OR

mr  α = Ff

α =  a r

ma = mg sin θ – mr  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 =  1 2 Iω2 +  1 2 mv2

substituting ω =  v r  «giving v = g h »

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

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