Engr 240 - Dynamics
Quiz 1
NAME: _SOLUTION_
Instructions: Solve each of the following problems completely.
1. The acceleration of a particle is defined by the relation a k 1 e x , where k is a constant.
Knowing that the velocity of the particle is v 9 m/s
Engr 240 Quiz 2
Oct. 6, 2010
Name:
A small 200-g collar C can slide on a semicircular rod
which is made to rotate about the vertical AB at the
constant rate of o rad/s. The coefficient of static
friction between the collar and the rod if the collar is s
=
Quiz 3
Engr 240 Fall 2010
Name: _
Knowing that at the instant shown bar AB has a constant angular
velocity of 5 rad/s counterclockwise, determine (a) the angular
acceleration of bar BD, (b) the angular acceleration of bar DE.
Geometry.
AD 180 mm
Triangle
Engr 240 - Dynamics Sample Test 1
1. A parachutists acceleration, a , is given by the equation, a g
k
v , where v is its velocity, m is
m
its mass, and k is a constant. If the parachutist starts from rest at t=0, find (a) its velocity as a
function of ti
Dynamics Sample Test 3
1. Two blocks and a pulley are connected by inextensible cords.
Block A has a constant acceleration of 75 mm/s2 and an initial
velocity of 120 mm/s, both directed downward. Determine
a) the number of revolutions executed by the pull
Engr 240 - Dynamics - Sample Test 2 - Solution
NAME: _
1. Two wires AC and BC are tied to a 15-lb sphere which revolves at a
constant speed v in the horizontal circle shown. Knowing that
1 = 50 and 2 = 25 and that d = 4 ft, determine the range of
values o
Engr 240 - Dynamics
TEST 3 - Fall 2006
NAME: _SOLUTION_
1. Block A has an initial downward velocity of 2 m/s and accelerates at a constant rate of 4 m/s2.
Determine:
a) The initial angular velocity of pulley B,
b) The angular acceleration of pulley B, and
Engr 240 - Dynamics: TEST 1A - Fall 2012
NAME: _
dx
dv
dv
; a
v
dt
dt
dx
2
Uniformly Accelerated Motion: v v o at ; x x o v o t 1 at 2 ; v 2 v o 2ax
2
Rectilinear Motion: v
v2
dv
; an
dt
Radial and Transverse: v r er r e ; a r 2 er r 2r e
r
Relative M
Engr 240 - Dynamics
TEST 2A - Fall 2012
NAME: _SOLUTION
You are allowed to use one 4" X 6" index card with formulas only (no solved problems) as a cheat sheet.
Instructions: Solve each problem completely.
1. Block A of mass 2 kg rests on a rough
table and
Engr240-HW15
Amelito Enriquez: Canada College
Chapter 17, Solution 70
Kinematics. Rolling motion. Instantaneous center at C.
v vG r
I mk 2
Moment of inertia.
Kinetics.
Syst. Momenta1
Moments about C:
Syst. Ext. Imp. 1 2
Syst. Momenta 2
0 (mgt )r sin mvr I
Engr240-HW14
Amelito Enriquez: Canada College
Chapter 17, Solution 11
Let v A speed of block A, vB speed of block B, angular speed of pulley.
5
ft,
6
rA
Kinematics.
rB
1
ft
2
5
v A rA
6
1
vB rB
2
5
s A rA
6
1
sB rB
2
(a)
Cylinder A falls to ground.
Engr240-HW13
Amelito Enriquez: Canada College
Chapter 16, Solution 85
0
MC
(b)
L
4
L
mg
4
L
mg
4
( M C )eff : W
I
ma
L
4
1
mL2
12
7
mL2
48
m
L
4
L
4
12g
7L
C
4
mg
7
Reaction at C.
Fy
( Fy )eff : C
mg
ma
C
mg
m
C
mg
L
4
12g
7L
3
mg
7
C
(a)
L
4
a
m
L
4
4
mg
Engr240-HW4
Amelito Enriquez: Canada College
Chapter 12, Solution 13
Load: We assume that sliding of load relative to trailer is impending:
F
Fm
sN
Deceleration of load is same as deceleration of trailer, which is the maximum allowable deceleration a max
Engr240-HW5
Amelito Enriquez: Canada College
Chapter 12, Solution 69
Kinematics
dr
dt
We have
r
k
r
At t
0, r
t
dr
r0 :
kdt
r0
0
or
r
r0
Also,
r
0
kt
ct
c
Now
ar
r
2
r
kt )(ct )2
0 (r0
c 2 (r0
and
a
kt )t 2
r
2r
(r0
kt )(c) 2( k )(ct )
c(r0
3kt )
Kinetics
Engr240-HW7
Amelito Enriquez: Canada College
Chapter 13, Solution 62
(a)
Maximum height is reached when
v2
T1
Position
.
T2
0
V
Thus,
0
Vg
Ve
(Vg )1
0
Total spring deflection from undeflected spring position x1
x1
x1
x1
(Ve )1
mg/k
0.150
(3 kg)(9.81 m/s 2
Engr240-HW8
Amelito Enriquez: Canada College
Chapter 13, Solution 134
The block does not move until P
From t
0 to t
2 s, P
4 lb.
5t.
Thus, the block starts to move when t
(a)
4/5 0.8 s.
0, t , 2 s
For
P
5t
t1
0.8 s t2
mv1
t2
t1
2 s, v1
Pdt W (t2
0
t1 )
mv
Engr240-HW9
Amelito Enriquez: Canada College
Chapter 14, Solution 2
There are no horizontal external forces acting during the impacts. The baggage carrier is to coast between
impacts.
(a)
Conservation of momentum:
2.4(mB 15)
Let vfinal
(1)
40)vfinal
1.2 m
Engr240-HW11
Amelito Enriquez: Canada College
Chapter 15, Solution 78
vD
200 mm/s ,
vB
160 mm/s
vD vB
BD
200 160
90
4 rad/s
vD
200
50 mm
4
CA 60 50 10 mm
CD
(a)
C lies 10 mm to the right of A.
(b)
vA
10
(10)(4)
(c)
vD
vA
240 mm/s
Cord DE is unwrapped at 2
Engr240-HW12
Amelito Enriquez: Canada College
Chapter 16, Solution 7
(a)
Acceleration
Fx
( Fx )eff : 100 N
100 N
ma
(20 kg)a
a
(b)
For tipping to impend :
MB
5.00 m/s 2
A0
( M B )eff : (100 N) h mg (0.3 m)
(100 N)h (20 kg)(9.81 m/s 2 )(0.3 m)
ma (0.9 m)
(
Engr 240 - Dynamics -TEST 1 - Fall 2006 - Solution
NAME: _
Instructions: Solve each of the following problems completely.
1. The acceleration of a particle is defined by the relation a = 9 3t 2 , where a is in m/s2, and
t is in seconds. The particle start