Download Section V

KINETICS OF PARTICLES AND RIGID BODIES USING THE PRINCIPLE OF
IMPULSE & MOMENTUM
1.
A tugboat exerts a constant force of 250 kN on a 200 000- tonne oil tanker.
Neglecting the frictional resistance of the water determine the time required to
increase the speed of the tanker (a) from 1 km/h to 2 km/h, (b) from 2 km/h to 3
km/h.
2.
A light train made of two cars travel t 100 km/h. The mass of car A is 15 Mg.
And the mass of car B is 20 Mg. When the brakes are applied , a constant braking
force of 25 kN is applied to each car. Determine (a) the time required for the train
to stop after the brakes are applied, (b) the force in the coupling between the cars
while the train is slowing down.
3.
A steel-jacket bullet of mass 20 g is fired with a velocity of 600 m/s toward a steel
plate; the bullet ricochets along the path CD with a velocity of 500 m/s. Knowing
that the bullet caused a 50-mm scratch on the surface of the plate, determine the
average impulsive force exerted on the bullet during its contact with the plate.
(Hint Assume an average speed of 550 m/s during contact)
4.
1.
A 60-Mg railroad car is to be coupled to a second car which has a mass of 40 Mg.
If initially the speed of the 60-Mg car is 1 km/h and the 40-Mg car is at rest,
determine (a) the final speed of the coupled cars. (b) the average impulsive force
acting on each car if the coupling is completed in 0.5s.
Principle of linear impulse and linear momentum:
F t  mv 2  mv1
 2 1 
FX t  mv 2 x  mv 1x  250x10 3 t  200x10 6 

36
t  3.7 min
2.
a) Consider entire train
y
A
B
x
+
A
B
=
A
F t
mv1
B
mv2 = 0
V1  100 km / hrs  100 / 3.6 m / s i  27.78 m / s i
F  2x 25x10 3 (i ) N
m = mA + mB =(15 + 20) x103 KG = 35x103 Kg
V2 =0
Principle of linear impulse and linear momentum:
F t = m(V2-V1)
-50x103 i t = 35x103(-27.78 i)  t = 19.44 sec
b)
Consider the first car (A)
y
A
x
+
A
P t
=
A
mAv2 = 0
F t
mAv1
P: Coupling force
Principle of Linear impulse and linear momentum:
F t + P t = mA(V2 – V1)
-25x103 x 19.44i – P x 19.44i = 15x103(-27.78i)
 P = -3.57 kN or P = 3.57i kN
3.
Consider bullet
mv2
y
+
x
mv1 = – (0.02)(600)i kgm/s
= –12i kgm/s
=
i Fx t
i Fy t
30o
Principle of linear impulse and linear momentum:
mV1 + F t = mV2
i (mV1) + i Fx t + d Fyt = i mV2Cos30° + d mV2sin30°
i -components : -12 + Fxt = -(0.02)(500)Cos30°
Fxt = 12-8.66 = 3.34 Ns
j - components: Fy t = (0.02)(500) sin30° = 5Ns
Ft = (3.34 i + 5 d ) Ns
 Ft
=
tan  =
(FX t ) 2  (Fyt ) 2 = 6.013 Ns
Fyt
5

 1.497    56.26
FX t 3.34
F t

But x = VAverage . t  t =
F=
0.05
 90.91x10 6 s
550
6.013
 66.1 KN
90.91x10 6
The 6-kg block, shown in Fig. 19-4a, is attached to a cord
which is wrapped around the periphery of a 20-kg disk. If
the block is initially moving downward with a speed of 2
m/s, determine its speed in 3s. Neglect the mass of the cord
in the calculation.
The 100-kg spool shown in Fig. 19-5a has a radius
of gyration of kG =0.35 m. A cable is wrapped
around the central hub of the spool and a variable
horizontal force having a magnitude of P = (t _ 10)
N is applied, where t is measured in seconds. If the
spool rolls without slipping.
The 5kg slender rod shown is pinned at 0 and is initially at
rest. If a 4g bullet is fired into the rod with a velocity of
400 m/s, as shown, determine the angle max through
which the rod swings. After impact the bullet becomes
embedded in the rod.
171
The 75-kg flywheel has a radius of gyration about its shaft
axis of k  0.50 m and is subject to the torque M = 10 (1 e-1) N.m, where t is in seconds. If the flywheel is at rest at
time t - 0, determine its angular velocity w at t - 3 s. Ans.
w = 1.093 rad/s
172
The constant tensions of 200 N and 160N are applied to
the hoisting cable as shown. If the velocity t of the load is
2 m/s down and the angular velocity w of the pulley is 8
rad/s counterclockwise at time 1 = 0, determine v and w
after the cable tensions have been applied for 5 s. Note the
independence of the results.
174
The 30-g bullet has a horizontal velocity of 500 mi,
as it strikes the 25-kg compound pendulum, which
has a radius of gyration ko = 0.95 m. If h = 1.1 m
calculate the angular velocity w of the pendulum with
its embedded bullet immediately after the impact.
Ans. w = 0.730 rad/
175If the bullet of Prob. 6/174 takes 10-4 s to imbed itself in
the pendulum calculate the time average of the
horizontal force O? exerted by the pin on the bar at O
during the impact. Use the result cited Prob. 6/174.
183
If the system shown is released from rest,
how much time passes before the velocity
of the 10-kg block reaches 2 m/s? The
radius of gyration of the 12-kg pulley at O
is Ko = 150 mm and the mass of the lower
pulley is negligible. Neglect all friction.
Ans. t = 1.586 s