| Date | May 2016 | Marks available | 1 | Reference code | 16M.3.SL.TZ0.7 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 0 |
| Command term | Outline | Question number | 7 | Adapted from | N/A |
Question
A solid cylinder of mass M and radius R rolls without slipping down a uniform slope. The slope makes an angle θ to the horizontal.
The diagram shows the three forces acting on the cylinder. N is the normal reaction force and F is the frictional force between the cylinder and the slope.
State why F is the only force providing a torque about the axis of the cylinder.
(i) The moment of inertia of a cylinder about its axis is \(I = \frac{1}{2}M{R^2}\). Show that, by applying Newton’s laws of motion, the linear acceleration of the cylinder is \(a = \frac{2}{3}g\sin \theta \).
(ii) Calculate, for θ= 30°, the time it takes for the solid cylinder to travel 1.5 m along the slope. The cylinder starts from rest.
A block of ice is placed on the slope beside the solid cylinder and both are released at the same time. The block of ice is the same mass as the solid cylinder and slides without friction.
At any given point on the slope, the speed of the block of ice is greater than the speed of the solid cylinder. Outline why, using the answer to (b)(i).
The solid cylinder is replaced by a hollow cylinder of the same mass and radius. Suggest how this change will affect, if at all, the acceleration in (b)(i).
Markscheme
because Mg and N act through the axis
OR
only F has a non-zero lever arm «about the axis»
(i) ALTERNATIVE 1
use of Newton’s law for linear motion: Mgsinθ-F=Ma
use of Newton’s law for rotational motion: FR=Iα
combining \(Mg\sin \theta = Ma + \frac{{I\alpha }}{R}\)
substitution of \(I = \frac{1}{2}M{R^2}\) and \(\alpha = \frac{a}{R}\)
to get result
ALTERNATIVE 2
\(Mgh = \frac{1}{2}M{v^2} + \frac{1}{4}M{v^2} \ll {\rm{from}}\frac{1}{2}I{\omega ^2} = \frac{1}{2}\left( {\frac{1}{2}M{R^2}} \right)\frac{{{v^2}}}{{{R^2}}} \gg \)
\({v^2} = \frac{4}{3}gh\)
\({v^2} = 2as = 2a\frac{h}{{\sin \theta }}\)
manipulation to produce given answer
Accept correct use of torques about point of contact.
(ii) rearranging \(s = \frac{1}{2}a{t^2}\) to get \(t = \sqrt {\frac{{2s}}{a}} \)
substitution to get \(t = \ll \sqrt {\frac{{2 \times 1.5}}{{\frac{2}{3} \times 9.81 \times \frac{1}{2}}}} \gg = 0.96{\rm{s}}\)
acceleration of ice is gsinθ whereas for the solid cylinder acceleration is two thirds of this «so speed of ice must always be greater at same point»
Allow answers in terms of energies, eg ice does not use energy to rotate and therefore will have a greater translational speed.
the hollow cylinder has a greater moment of inertia
and hence a smaller acceleration
