Tutorial 1
Problem 11.161 The oscillation of rod OA about O is defined by the relation
θ = (2 /π) (sin πt), where θ and t are expressed in radians and seconds,
respectively. Collar B slides along the rod so that its distance from O is
r = 625/(t + 4) where r and t are expressed in mm and seconds, respectively.
When t = 1 s, determine (a) the velocity of the collar, (b) the total acceleration
of the collar, (c) the acceleration of the collar relative to the rod.
Problem 11.165
As rod OA rotates, pin P moves along the parabola BCD. Knowing
that the equation of this parabola is r  2b/(1  cos ) and that   kt ,
determine the velocity and acceleration of P when (a)   0, (b)
  90.
PROBLEM 11.166
The pin at B is free to slide along the circular slot DE and
along the rotating rod OC. Assuming that the rod OC rotates
at a constant rate , (a) show that the acceleration of pin B is
of constant magnitude, (b) determine the direction of the
acceleration of pin B.
PROBLEM 11.169
At the bottom of a loop in the vertical plane, an airplane has a
horizontal velocity of 150 m /s and is speeding up at a rate of
25 m /s2. The radius of curvature of the loop is 2000 m. The
plane is being tracked by radar at O. What are the recorded
values of 𝑟, ̇ 𝑟̈ , 𝜃,̇ and 𝜃̈ for this instant?
PROBLEM 11.170
Pin C is attached to rod BC and slides freely in the slot of
rod OA which rotates at the constant rate . At the instant
when   60, determine (a) r and  , (b) r and . Express
your answers in terms of d and .
APL 100 Engineering Mechanics, Semester II, 2016-17
Department of Applied Mechanics, Indian Institute of Technology Delhi
Tutorial 1
Question 1 (Problem 11.161 of Beer et al.)
The radial location of the collar B along the rod OA is given by r = 625/(t + 4). The rod OA
oscillates with the azimuthal angle given as, θ = π2 sin(πt). Thus,
ṙ =
1250
−625
, r̈ =
(t + 4)2
(t + 4)3
(1)
θ̇ = 2 cos(πt), θ̈ = −2π sin(πt)
The velocity of the collar is given as, v = ṙer + rθ̇eθ .
(a) At t = 1, v = −25er + −250eθ (mm/s).
The acceleration of the collar is given as, a = (r̈ − rθ̇2 )er + (rθ̈ + 2ṙθ̇)eθ .
(b) At t = 1, a = −490er + 100eθ (mm/s2 ).
(c) The acceleration of the collar relative to the rod is r̈er = 10er (mm/s2 ) along the rod.
Question 2 (Problem 11.165 of Beer et al.) The coordinates for the location of the pin are given as,
r = 2b/(1 + cos θ), θ = kt, where k and b are constants. Thus,
ṙ =
2b(θ̈ sin θ + θ̇2 cos θ)
4bθ̇2 sin2 θ
2bθ̇ sin θ
,
r̈
=
+
(1 + cos θ)2
(1 + cos θ)2
(1 + cos θ)3
(2)
θ̇ = k, θ̈ = 0
The velocity of the collar is given as, v = ṙer + rθ̇eθ . The acceleration of the collar is given as,
a = (r̈ − rθ̇2 )er + (rθ̈ + 2ṙθ̇)eθ .
(a) At θ = 0o , r = b, ṙ = 0, r̈ = k 2 b/2
2
v = kbeθ and a = −k2 b er
o
(b) At θ = 90 , r = 2b, ṙ = 2kb, r̈ = 4k 2 b
v = 2kber + 2kbeθ and a = 2k 2 ber + 4k 2 beθ .
Question 3 (Problem 11.166 of Beer et al.) The radial coordinate of the pin B is given in terms of θ
as, r = 2b cos θ. The angular velocity θ̇ = constant. Thus, ṙ = −2bθ̇ sin θ and r̈ = −2bθ̇2 cos θ.
The acceleration of the pin is given as, a = (r̈ − rθ̇2 )er + (rθ̈ + 2ṙθ̇)eθ = −4bθ̇2 (cos θer + sin θeθ ).
Thus, the magnitude of acceleration, 4bθ̇2 , remains constant with time, and the acceleration vector
is directed radially inwards towards point A for any location of pin B (the direction is given by
− cos θer − sin θeθ ).
Question 4 (Problem 11.169 of Beer et al.) The azimuthal angle at the instant shown is tan θ = 3/4
(sin θ = 3/5 and cos θ = 4/5). The radial coordinate r = 1000m. The velocity of the airplane
150m/s is in the horizontal direction (the vertical component is zero), thus, ṙ cos θ − rθ̇ sin θ = 150
and ṙ sin θ + rθ̇ cos θ = 0. So, ṙ = 150 cos θ = 120m/s and θ̇ = −150 sin θ/r = −0.09rad/s.
The acceleration of the airplane is 25m/s2 in the horizontal direction and 11.25m/s2 in the vertically
upward direction. Thus, r̈ − rθ̇2 = 25 cos θ + 11.25 sin θ for the radial direction which yields,
r̈ = 8.1 + 25 ∗ 4/5 + 11.25 ∗ 3/5 = 34.85m/s2 . For the azimuthal direction, rθ̈ + 2ṙθ̇ = −25 sin θ +
11.25 cos θ = −15 + 9 = −6m/s2 . Thus, θ̈ = (−6 + 2 ∗ 120 ∗ 0.09)/1000 = 15.6 ∗ 10−3 rad/s2 .
Question 5 (Problem 11.170 of Beer et al.) The radial coordinate of pin C is given in terms of θ as,
r = 2d cos θ. Since OB = BC = d, the angles ∠BCO = ∠BOC = θ = β/2.
If the arm BC hinged at point B rotates at the given angular velocity ω, then, θ̇ = β̇/2 = ω/2. Thus,
√
ṙ = −2dθ̇ sin θ = −dω/2 and θ̇ = ω/2 at θ = β/2 = 30o . Similarily, r̈ = −2dθ̇2 cos θ = −ω 2 d 3/4
and θ̈ = 0 at θ = β/2 = 30o .
(If the rod OA hinged at point O rotates at the given angular velocity ω, then θ̇ = ω.√Thus,
ṙ = −2dθ̇ sin θ = −dω and θ̇ = ω at θ = β/2 = 30o . Similarily, r̈ = −2dω 2 cos θ = −ω 2 d 3 and
θ̈ = 0 at θ = β/2 = 30o )
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