Year: 2008
Paper: 3
Question Number: 11
Course: zNo longer examinable
Section: Moments of inertia
No solution available for this problem.
Most candidates attempted five, six or seven questions, and scored the majority of their total score on their best three or four. Those attempting seven or more tended not to do well, pursuing no single solution far enough to earn substantial marks.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A circular wheel of radius $r$ has moment of inertia $I$ about
its
axle, which is fixed in a horizontal position. A light string is
wrapped around the circumference of the wheel and a particle of
mass $m$ hangs from the free end. The system is released from rest
and the particle descends.
The string does not slip on the wheel.
As the particle descends, the wheel turns through
$n_1$ revolutions, and the string then detaches from the wheel. At this
moment, the angular speed of the wheel is $\omega_0$. The wheel then
turns through a further
$n_2$ revolutions, in time $T$, before coming to rest.
The couple on the wheel due to resistance is constant.
Show that
\[ \frac12 \omega_0 T = 2 \pi n_2\]
and
\[
I =\dfrac {mgrn_1T^2 -4\pi mr^2n_2^2}{4\pi n_2(n_1+n_2)}\;.
\]
Under a twelfth tried this. A number of different correct approaches were successfully applied, and there were very few partially correct solutions.