| Date | November 2016 | Marks available | 2 | Reference code | 16N.3.SL.TZ0.8 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 0 |
| Command term | Show that | Question number | 8 | Adapted from | N/A |
Question
A flywheel consists of a solid cylinder, with a small radial axle protruding from its centre.
The following data are available for the flywheel.
| Flywheel mass M | = 1.22 kg |
| Small axle radius r | = 60.0 mm |
| Flywheel radius R | = 240 mm |
| Moment of inertia | = 0.5 MR2 |
An object of mass m is connected to the axle by a light string and allowed to fall vertically from rest, exerting a torque on the flywheel.
The velocity of the falling object is 1.89 m s–1 at 3.98 s. Calculate the average angular acceleration of the flywheel.
Show that the torque acting on the flywheel is about 0.3 Nm.
(i) Calculate the tension in the string.
(ii) Determine the mass m of the falling object.
Markscheme
ALTERNATIVE 1
\({\omega _{{\text{final}}}} = \frac{v}{r} = 31.5\) «rad s–1»
«\(\omega = {\omega _o} + \alpha t\) so» \(\alpha = \frac{\omega }{t} = \frac{{31.5}}{{3.98}} = 7.91\) «rad s–2»
ALTERNATIVE 2
\(a = \frac{{1.89}}{{3.98}} = 0.4749\) «m s–2»
\(\alpha = \frac{a}{r} = \frac{{0.4749}}{{0.060}} = 7.91\) «rad s–2»
Award [1 max] for r = 0.24 mm used giving \(\alpha \) = 1.98 «rad s–2».
\(\Gamma = \frac{1}{2}{\rm{M}}{{\rm{R}}^{\rm{2}}}\alpha = \frac{1}{2} \times 1.22 \times {0.240^2} \times 7.91\)
= 0.278 «Nm»
At least two significant figures required for MP2, as question is a “Show”.
i
\({F_T} = \frac{\Gamma }{r}\)
\({F_T} = 4.63\) «N»
Allow 5 «N» if Γ= 0.3 Νm is used.
ii
\({F_T} = mg - ma\) so \(m = \frac{{4.63}}{{9.81 - 0.475}}\)
m = 0.496 «kg»
Allow ECF
