A drink can is placed on a rough-surfaced, stationary conveyor belt of a supermarket checkout. The can is modelled as a cylinder consisting of two equal discs, each of mass m1 and radius R, that are the top and bottom, and a curved surface of mass m2, radius R and length h. The thickness of the metal is so small that the can may be modelled as a thin cylinder with a thin disc at each end, of constant surface density. The liquid contents of the can are modelled as a solid cylinder of radius R, length h and mass M.
The conveyor belt is switched on with acceleration f i, where f is a constant, such that the can rolls backwards relative to the direction of motion of the belt, without slipping. The axis of the can remains at right angles to the direction i of the acceleration of the belt. The time that elapses from the instant when the can begins to move is denoted by t, θ is the angle through which the can rotates in this time, and x, taken as positive in the direction of i, is the distance moved by the centre of mass of the can due to the rotation through angle θ.
(a) Use the rolling condition to find an expression for x in terms of R and θ. Hence show that the acceleration of the centre of mass of the can is (f −R¨ θ)i.
(b) Draw a force diagram for the can, expressing each force in vector form.
(c) By applying Newton’s second law, find the equation of motion for the centre of mass of the can.
(d) Write down the torque, relative to the centre of mass of the can, associated with each force acting on the can. Hence use the torque law to determine the equation of relative rotational motion of the can in terms of R, the moment of inertia I of the can about its axis of symmetry, and the magnitude of the frictional force.
(e) Show that the moment of inertia of the can and contents about the axis of symmetry is I = (1/2M +m1 +m2)R2.
(f) Hence show that the acceleration of the centre of mass of the full can is
(M + 2m1 + 2m2)/(3M + 6m1 + 4m2)f i.
Rolling condition = R theta = x
I = Moment of inertia = 1/2MR^2
second law:F = ma (where m is the total mass of the system)
m = (M + 2m1 + m2)
v = dx/dt
w = -d^2x/dt^2
a = f- Rtheta
The Attempt at a Solution
a) When the can rolls anticlockwise, the distance rolled is -x. Using the rolling condition, we have -x= -R theta. The position vector of the centre of mass is Rj-xi. Therefore the velocity is -x' and the acceleration is -x'', since R is a constant. Therefore the acceleration of the centre of mass of the can is (f-R theta'') as required.
b) The force diagram contains N upwards, W downwards and F in the positive i direction.
c) Newton's second law gives F= ma. This gives F= (2m1+m2)(f-R theta), which is the equation of motion of the can (without contents).
d) Since W and N act through the centre of mass, they do not apply any torque. The torque is given by
Torque= -Rj x (2m1+m2)(f-R theta'') i= -R (2m1+m2)(f-R theta) k.
The torque law is torque^rel= Iw'= I theta''.
The moment of inertia of the can without contents is I=(m1+m2)R^2.
So torque^rel= (m1+m2)R^2 theta'' k.
Equating the relative torque and the torque gives f= m1 R theta''/ (2ma+m2)
e) The moment of inertia of the solid cylinder is 1/2 MR^2, and of the can (from part d) is (m1+m2)R^2. Hence the moment of inertia is (1/2 M+ m1+m2) R^2, as required.
f) Now this is where I'm stuck! I don't have a clue where to start and am unsure if I should be using my friction value found in part d)? Any pointers greatly appreciated hehe...
Thanks a lot, Hannah