ESC2C5 Summer 2017 Exam Answers

UNIVERSITY OF Warwick Year 2
School of Engineering June 2017
ES2B0 Title of Examination: -
MECHANICS OF SOLIDS AND FLUIDS
Page 1

Questions 3-4 (Section B: Dynamics) Mark
Allocation
Marks
***************************************** Begin 3 ******************************************* *** B3 ***
Q3(a)
i. Conservative forces (e.g. a gravitational force; a spring force) are dependent on position only;
whereas non-conservative forces (e.g. friction; air resistance) also depend on other variables, such as
velocity. (3 marks)

ii. The kinetic energy of the block at ground level is given by:

1
(K.E.)y=0 = mv 2
2
1
= · (2.5) · 202
2
= 500 J
(2 marks)


iii. The work done by gravity on the block Ug in moving it from y=H to y=0, is equal to the change in
kinetic energy:

Ug = (K.E.)
1 1
= m(vy=0 )2 m(vy=H )2
2 2
1 1
= · (2.5) · 202 · (2.5) · 102
2 2
= 500 125 = 375 J (3 marks)

iv. In this case the potential energy of the block at y=H is equal to the work done by gravity in moving
it from y=H to y=0 (the same answer as in iii.)
(1 mark)
(P.E.)y=H = 375 J


























v. The potential energy of the block at y=H is:
(P.E.)y=H = mgH .
Rearranging gives:

(P.E.)y=H
H =
mg
375
= = 15 m
(2.5) · (10) (2 marks)

vi. When considering work done by non-conservative forces (UNC) the following holds:

(P.E.)y=H + (K.E.)y=H + UNC = (P.E.)y=0 + (K.E.)y=0 . (2 marks)

Substitution of the appropriate expressions and values leads to
1 1
(1)(10)H + (1)(5)2 + ( 150) = 0 + (1)(10)2 .
2 2 (1 mark)
Rearrangement tells us: H = 18.75 m.

, Q3(b)

i. The expression for the potential energy of one spring is:
#
(P.E.)spring
= 𝑘𝑠 $ .
$
Noting that the extension s in both cases is equal to x, the potential energy of the two springs (the total)
is:
#
(P.E.)springs
= 2 ∙ 𝑘𝑥 $ = 50𝑥 $ .
$
(2 marks)
ii. The total kinetic energy is composed of the kinetic energy of the translating mass and the kinetic
energy of the rotating pulley.
The kinetic energy of the mass is simply
#
(K.E.)mass
= 𝑚𝑣 $ = 2.5𝑣 $ .
$
Noting that the angular velocity of the pulley (ω) is related to the velocity of the mass by ω= v/r ,
where r is the radius of the pulley, we can express the kinetic energy of the pulley as follows:
# # 12
(K.E.)pulley
= 𝐼𝜔 $ = 𝐼 2 = 6.25𝑣 $ .
$ $ 3
So the total kinetic energy is:
(K.E.)total
= (2.5 + 6.25)𝑣 $ = 8.75𝑣 $ (3 marks)

iii. Principle of work and energy states:

P.E. + K.E. = const.

Therefore,
(2 marks)
8.75v 2 + 50x 2 = const

Differentiate with respect to time:
! dv $ ! dx $
8.75# 2v & + 50 # 2x & = 0
" dt % " dt %

Since v=dx/dt, we can write
dx d 2 x dx
17.5 2
+100x = 0
dt dt dt

Dividing through gives:
d2x
17.5 +100x = 0
dt 2
The characteristic equation is:
d2x
+ 5.71x = 0 (3 marks)
dt 2
i ωt
and assuming vibration of the form x=A e we get:
ω = 5.71 = 2.39 rad/s (1 mark)

****************************************** End 3 ******************************************* *** E3 ***

***************************************** Begin 4 ******************************************* *** B4 ***
Q4(a)

i. Newton’s second law in rotational form:

𝑇 = 𝐼𝛼


where
T torque (or moment), N m
I is the moment of inertia, N m s2 (or kg m2)

(5 marks)
α is the angular acceleration, [rad] s-2