Midterm exam 2014 solution

Midterm examination: Mechanics
Hiroki Okubo
May 21st 2014
z
T
1. A particle moves along the x-axis with an initial ven
locity vx = v0 at the origin when t = 0 where v0 is
α
A
θ
v
r
positive. For the first t = t0 seconds it has no acO
θ
y
celeration, and thereafter it is acted on by a retardr
30
ing force which gives it a constant acceleration −a0
x
where a0 is positive. Calculate the velocity vx and
P
the x-coordinate of the particle and find the maxiFigure 2: A small block
mum positive x-coordinate reached by the particle.
Figure 1: Moment of ten- traveling with constant
(20)
speed.
sion T .
Solution. The velocity of the particle before t = t0
is v0 . The velocity after t = t0 s is computed from
Z vx
Z t
the horizontal range is x = v02 sin 2θ/g. θ = 45◦ gives
dvx = −a0
dt
the maximum range Rmax = v02 /g.
v
t
0
0
vx
= −a0 t + a0 t0 + v0
3. Determine the moment of the tension T about point
P as shown in Fig. 1. (10)
Solution. Unit vectors i, j, k are attached to the
primary inertial system with the ij plane in the vertical and the direction of the j vector up. The tension vector can be written as T = T (cos αi + sin αj).
The moment of T about points P is M P = (r O/P +
r A/O ) × T = r[− cos θi + (1 + sin θ)j] × T (cos αi +
sin αj) = −T r[cos α + sin (α + θ)]k.
The x-coordinate of the particle before t = t0 is v0 t.
The x-coordinate at any time greater than t0 is the
distance traveled during the first t0 plus the distance
traveled after the discontinuity in acceleration occurred. Thus,
Z t
(−a0 t + a0 t0 + v0 )dt
x = v0 t 0 +
t0
= −(1/2)a0 t2 + (a0 t0 + v0 )t − (1/2)a0 t20
4. The small block P travels with constant speed v in
the circular path of radius r on the inclined surface
as shown in Fig. 2. If θ = 0 at time t = 0, determine
the x-, y-, and z-components of velocity as functions
of time. (10)
Solution. Unit vectors i, j, k are attached to
the x-y-z axes. We also attach unit vectors er
and eθ to the inclined surface with the er and
eθ tangential and normal to the circular path, respectively. The unit vectors er and eθ can be expressed as er = cos θ cos 30◦ i+sin θj −cos θ sin 30◦ k,
eθ = − sin θ cos 30◦ i + cos θj + sin θ sin 30◦ k. The
velocity of the small block P can be described as
v = veθ = v(− sin θ cos 30◦ i+cos θj +sin θ sin 30◦ k).
Because θ is a function of θ = (v/r)t, the x, y and
z components
of velocity is v · i = −v sin θ cos 30◦ =
√
(− 3v/2) sin (vt/r), v·j = v cos θ = v cos (vt/r) and
v · k = v sin θ sin 30◦ = (v/2) sin (vt/r).
When the velocity is zero, t = t0 + (v0 /a0 ). The
maximum positive x-coordinate is xmax = v0 t0 +
(v02 /2a0 ).
2. For a given launch speed v0 , the launch angle θ = 45◦
yields the maximum horizontal range R. Determine
the maximum range. Neglect the effect of air resistance. (10)
Solution. We use rectangular coordinates for the
trajectory analysis with the x axis horizontal and
the y axis vertical. The acceleration components are
x
¨ = 0 and y¨ = −g where g is the gravity. Integration
of these accelerations yields the velocity x˙ = v0 cos θ
and y˙ = v0 sin θ − gt where v0 is the initial speed
and θ is the angle of the initial velocity from x axis.
Integrating the velocity components with the initial
conditions of x = 0 and y = 0 gives x = v0 cos θt and
y = v0 sin θt − (1/2)gt2 . When t 6= 0 and y = 0, the
time can be calculated as t = 2v0 sin θ/g from which
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