78AM-1
Sr. No. 5
EXAMINATION OF MARINE ENGINEER OFFICER
APPLIED MECHANICS
CLASS -I
(Time allowed - 3 hours)
INDIA (2001)
N.B. -
Morning Paper
Total Marks 100
(1) Attempt SIX questions only .
(2) All questions carry equal marks.
(3) Neatness in handwriting and clarity in expression carries weightage
1. Replace the force system consisting of three forces shown in the Figure1 by a winch passing through a
point in the yz plane.
Figure 1
2. A semicircle is rotated about its diameter to generate a sphere. Calculate the volume of a sphere of radius R.
4R/3
3. A shell bursts on contact with the ground and pieces fly off in all directions with speeds up to 30 m/s.
A person is standing 40 m away. What is the time duration over which he can be hit by a piece?
4. A force given by F = 3t2 i + 5t j - (8t3 + 400) k N acts from t = 0 to t = 10 s. Determine the impulse of
the force. If this impulse acted at the centre of mass of a body of mass 500 kg and brought it to rest,
estimate the velocity of the body before it acted.
5. A rod of length L and mass M is pivoted at one end to constitute a pendulum. Determine its period of
oscillation and calculate its length if the period is desired to be 1 second. If the rod was instead
suspended from a point at one quarter of its length, what would be the expression for the period and
what would be the length of the rod required for a period of 1 second?
6. Calculate the radius of the largest drop of water that might evaporate at 00C without heat being
supplied to it. The surface energy of water at 00C is 117 x 10–3 Jm–2 and the latent heat capacity of
water is 606 x 103 calorie per kg at 00C.
7. A point particle of mass 0.1 kg. is excuting S.H.M of amplitude 0.1m. When the particle passes
through the mean position, its K.E. is 8 x 10-3 joules. Obtain the equation of motion of the particle, if
the initial phase of oscillation is 450.
8. A cylinder of 0.5 m3 in volume contains air at 60oC and 343 kN/m2 absolute pressure. If the air is
compressed to 0.3 m3, find (1) the pressure inside the cylinder assuming isothermal process (2) the
pressure and temperature assuming adiabatic process.
9. Show that a cylindrical buoy of 1 m diameter and 2.5 m height weighing 10 kN will not float vertically
in sea water. Find the force necessary in a vertical chain attached at the center of base of the buoy that
will keep it vertical ( of sea water 10.35 kN/m3).
10. A heavy mass ‘M’ which can slide along a straight horizontal bar is attached to a fixed point at a
distance c from the bar by a spiral spring of natural length ‘a’ less than ‘c’ such that a mass ‘m’ hung
on to it will stretch it by a length ‘e’. Show that the time of small oscillation of ‘M’ along the bar
when it is slightly disturbed, will be 2[Mec/mg(c–a)]1/2.
78AM-1
Sr. No. 5
EXAMINATION OF MARINE ENGINEER OFFICER
APPLIED MECHANICS
CLASS -I
(Time allowed - 3 hours)
INDIA (2001)
N.B. -
Morning Paper
Total Marks 100
(1) Attempt SIX questions only .
(2) All questions carry equal marks.
(3) Neatness in handwriting and clarity in expression carries weightage
Answers
Answer for Question No.1
Solution The resultant force R = 200 i + 300 j - 150 k N. The moment of the given forces about the origin o
is given by
M= 1.2jx200i+ 1 kx300j + 1.5 ix (-150k) = -300 i + 225 j - 240 k Nm.
Let the wrench-be located at point P(o, y, z) located by the position vector yj + z k. The moment must be
Mi, i.e. normal to yz plane. Then,
(y j + z k) x (200 i + 300 j - 150 k) + Mi = -300 i + 225 j - 240 k.
whence,
300 z. + 150 y = M
200 z = 225
200 y = 240
y = 1.2 m, z = 1.125 m, M = 217.5 Nm
The desired wrench, therefore, consists of a force 200 i + 300 j - 150 k N and a moment 217.5 i Nm at the
point (0, 1.2, 1.125 m).
Answer for Question No. 2
Solution: The theorems of Pappus-Guldinus can be used to great advantage in this case. The centroid of the semicircle is 4R/3
above the diametrical axis. The area of semicircle is R2/2. According to the Pappus-Guldinus, the volume of the body of
revolution generated should be same as that which be obtained if the entire area R2/2 were concentrated at a radius 4R/3. Th
volume of the sphere is therefore,
2(4R/3) x (R2/2) = (4/3) R2
Answer for Question No. 3
Solution With the maximum initial velocity of 30 m/s, the horizontal range of 40 m requires that
302 x sin2
40 = ___________ ;sin2 = 0.435
9.81
whence
2 = 25.8° or 154.2°;  = 12.9° or 77.1°
The first piece that can hit the person has an angle of projection of 12.9° and initial velocity 30 m/s
takes time
t1 =(2 x 30 x sin 12.9)/9.81 = 1.36 sec
The last piece that can hit the person has an angle of projection of 77.1° and initial velocity 30 m/s which
takes time
t2=(2 x 30 x sin 77.1)/9.81 = 5.96 seconds.
How about the maximum time taken by a piece at less than 30 m/s? Well, then the higher value of 
would be less than 77.1° and hence t2 would be less.
The duration over which the person can be hit by a piece is a period of 4.6 seconds, beginning 1.36
seconds and ending 5.96 seconds after the bursting of the shell.
Answer for Question No.4
I =  Fdt
10
10
10
=  3t2dt i +  5t dt j -  (8t3 + 400) dt k
0
0
10 0
10
10
3
2
+
= t 0 i + 2.5t 0 j - 2t +400t 0 k
= 1000 i + 250 j - 24,000 k Ns
From the impulse-momentum principle,
I = m(V2 - V,)
1000 i + 250 j + 24, 000 k = 500(0 – V1)
whence V1, the velocity before the action of impulse should be
(-2 i - 0.5 j + 48 k) m/s
Answer for Question No. 5
Solution For the first case of suspension from the pivot O1
r = L/2
2
I = ML /l2 + M(L/2)2 = ML2/3
and  =2(ML2/3)/(MgL/2) = 2  2L/3g
.
For
=1s L=3g/82 =0. 373 m
For the second case of suspension from the pivot O2
r = L/4
I=ML2/12 +M(L/4)2 =7148 ML2
and
=2 (7/48ML2 )/(MgL/4) =27L/12g
For
=1s, L=12g/28 2=0.426m
An extension of this example leads us to an interesting question: What will  be if the rod is suspended at the cen
If the pivot is located at the centre of mass,
=2(I/mg x 0)
-->
.which shows that the periodic motion is not possible in this case; obviously because the restoring moment disap
body acts at the pivot itself.
Answer for Question No. 6
ANS. Let r be the radius of the drop. Self evaporation will take place when energy repleased due to
decrease in surface area is equal to latent heat of the mass evaparated. Let dr be the decrease in radius.
Volume of water evaporated = 4r2 dr.
Latent heat of water evaporated = 4r2dr rw L.
Decrease in surface area = 4r2 – 4 (r – dr) = 8rdr.
 Energy released = 8rdr T.
 8rdr T = 4r2 dr rw L.
or r 

2T
wL
2  117  10 3
 9 .2 x 10 11 m
1000  4.2  606  10 3
Answer for Question No. 7
ANS.
Equation of S . H. M is  = asin (t + )
When the particle passes through mean position its velocity is maximum and it has only kinetic
energy .
Now Vmax = w. a
So its K. E = ½ m2 . a2 = 8  10-3
½  0.1  2  ( 0.1)2 = 8  10-3 or  =  4
So the equation of motion is  = 0.1 sin (  4t + /4) m .
Answer for Question No.8
Sol: k = 1.4
p1 = 343 kNm2
p2 = ?
2
2
1 = 0.5 m
2 = 0.3 m
T1 = (273 + 60) = 333 K
T2 = ?
During isothermal process, temperature is held constant.
p11 = p22.
0 .5
p2 =
x 343 = 572 kN/m2.
0 .3
During adiabatic process, temperature changes,
p11k = p22k
 0 .5 
p2 = 

 0 .3 
1 .4
p 
T2
=  2 
T1
 p1 
x 343 = 705 kN/m2
0 .4 / 1 .4
0 .4 / 1 . 4
 7.14 
Hence T2 = 333 x 

 3 .5 
= 333 x 1.226 = 409 K.
Answer for Question No. 9
Sol:
Depth of immersion =
1000
= 1.23 m.
 2
 1  10350
4
I
  14  4
BM =
=
= 0.050 m.

64    12  1.23
1.23
BG = OG – OB = 1.25 =
= 0.635 m.
2
Now BG > BM. Therefore, the buoy will not float vertically. Let the tension in the chain attached
to O be T.
Total downwards force = (10000 + T)N
10000  T 3
Volume of sea water displaced =
m
10350
4(10000  T )
Depth of immersion =
10350    12
If the position of new center of buoyancy is B’; then
2(10000  T )
OB’ =
10350  
I
  12  4  10350 
10350
B’M’ =
=
=
2

64(10000  T )
64    1  4(10000  T )
It is necessary to find the new center of gravity G’ due to the self weight acting at G and the
tension T in the chain at O. Taking moment about O,
10000 x 1.25 = (10000 x T)x
12500
x = OG’ =
10000  T
12500
(10000  T )
Now
B’G’ = OG’ – OB’ =
–2
10000  T
10350 
For stable equilibrium B’M’  B’G’
10350 
12500
2(10000  T )


64(10000  T ) 10000  T
10350 
Solving for the condition B’M’ > B’G’
T = 4000 N = 4 kN.
Answer for Question No. 10
Sol: A is the fixed, point O the initial position of M and ‘P’ is position after being disturbed
where OP = x, AO is perpendicular to the bar.



 x
2
x2  c2  a .
Extension of the spring
Tension in the spring is ‘mg’ when the extension is ‘e’.
 Tension in the spring when M is at P will be
[mg/e]
 x
2

 c2  a

Force on M towards O = {[Mg/e](



 c 2  a cos APO

= {mg/e} x 2  c 2 {x/[x2 + c2]}
= [mgx/c] {1 – a/c},
omitting squares and higher powers of x.
Acceleration of M towards O = [mgx/c] {1 – a/c}/M
Hence the motion is simple harmonic period = 2[Mec/mg(c–a)]1/2.