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CTB 2300
Recent Exams with Solutions
Content Pages
1. Study year 2012/2013
1.1. First opportunity (problems followed by solutions) 2-12
1.2. Resit (problems followed by solutions) 13-20
2. Study year 2013/2014
2.1. First opportunity (problems followed by solutions) 21-28
2.2. Resit (problems followed by solutions) 29-37
3. Study year 2014/2015
3.1. First opportunity (problems followed by solutions) 38-46
3.2. Resit (problems followed by solutions) 47-58
4. Study year 2015/2016
4.1. First opportunity (problems followed by solutions) 59-68
4.2. Resit (problems followed by solutions) 69-79
5. Study year 2017/2018
5.1. First opportunity (problems followed by solutions) 80-90
5.2. Resit (problems followed by solutions) 91-102




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PROBLEMS
Exam: Monday, 24th June 2013

Problem 1 (7.5 exam points).
This problem deals with a playful polar bear jumping on ice as shown in the picture below.




We assume that the bear is jumping on a relatively small ice sheet, which can be schematised
using a mechanical model that is shown in Figure 1.

m, J

L
k c k c


Figure 1. Schematisation of an ice sheet

In Figure 1, the ice sheet is schematized by a rigid bar of mass m , mass moment of inertia (about
the centre of mass) J and length L . The reaction of water on the ice is mimicked with the help of
two spring-dashpot elements that support the ends of the bar. The height of the ice sheet is
assumed to be negligible. The effect of gravity is taken into account in the value of the stiffness of
the springs and should be neglected in the derivation of the equations of motion.

Questions:
1. Assume that the bear jumps right in the middle of the ice sheet and imposes a vertical force
F  t  on the ice as shown in the figure below.
F t 
L2 L 2




2. Derive equation of motion for the vertical displacement of the ice sheet from the
equilibrium (1 exam point).
In the equation of motion derived in the answer to Question 1 assume F  t   F0 cos  t  .
Derive an expression for the steady-state displacement (as a function of time) of the ice
sheet caused by this force (0.5 exam points).




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3. In the equation of motion derived in the answer to Question 1 assume F  t  to be a periodic
function of time with a period T that is shown in the figure below.

F




t
T
Derive the value(s) of T , which will lead to resonance using the following figures for the
system parameters: m  2000 kg, k  28kN m 1 , c  5kN s m 1 (1 exam point).
4. Assume that the bear got tired and at a certain moment just laid right in the middle of the
ice sheet. Assume that at this moment the vertical displacement of the ice sheet from the
equilibrium (downwards) was 10cm and the vertical velocities of the ice sheet and the bear
were zero. The mass of the bear (to be added to the mass of the ice sheet when the bear lies
on it) is 500 kg. The ice sheet parameters are the same as in Question 3. Compute the
velocity of the ice sheet at the time moment t  2s (1 exam point).
5. Assume now that the bear jumps eccentrically (not in the middle of the ice sheet) and
imposes a vertical force F  t  on the ice as shown in the figure below.
F t 
3L 4 L4




Such jumping will cause both the vertical motion of the ice sheet and its rotation about the
centre of its mass. Derive equations of motion (there will be 2 equations) for this situation
assuming that c  0 (2 exam points).
6. Write the stiffness matrix and the mass matrix for the system of equations derived in the
answer to Question 5 (0.5 exam points).
7. In the equation of motion derived in the answer to Question 5 assume F  t   F0 sin  t  .
Derive an expression for the steady-state vertical displacement and the angle of rotation of
the ice sheet caused by this force (1.5 exam points).

Problem 2 (2.5 exam points).

Consider a hydraulic system shown in Figure 2. The system consists of a sea and a harbour that
are connected by a canal.




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Figure 2. A hydraulic system addressed in Problem 2.

The equation of motion for this system, written in terms of the water elevation in the harbour
h  t  and the water elevation at the sea hz  t  can be written as
BMh  BWh  h  hz ,
where B is the horizontal cross-sectional area of the harbour and M  l  gA  is the inertia of
water in the canal and W is the loss factor. At t  0 the system is at rest, i.e. h  0   h  0   0.

Questions:
1. Assume that the water elevation in the sea depends on time as shown in the figure below.
hz
H0




t


Write down an analytical expression for the time dependence hz  t  that corresponds to
the dependence shown in the figure above (0.5 exam points).
2. Derive h  t  that is caused by the sea water elevation hz  t  defined in Question 8 both
for 0  t   and t   (2 exam points).

Problem 3 (bonus question, 1 exam point. This problem has to be addressed by those who
wish to get a grade higher than 9.0. All other students are also invited to answer this
question and earn an additional exam point).

Consider a double pendulum shown in Figure 3. The pendulum
consists of two masses and two massless rods as shown in the figure.
All connections allow the frictionless rotation.

Questions:
1. Derive equations of motion that govern free vibrations of the
pendulum under the assumption that the angles 1 and  2 are
small 1  1,  2  1 .

Figure 3. A double pendulum



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