2020HW4soln
pdf
keyboard_arrow_up
School
North Carolina State University *
*We aren’t endorsed by this school
Course
208
Subject
Aerospace Engineering
Date
Dec 6, 2023
Type
Pages
7
Uploaded by Amonanny
MAE 208 Engineering Dynamics
Dr. Saul
March 2 2020
Total Possible Points: 30
Check your work using the informationprovided. Boxyour answer to the underlined question.
- 2 -
K. Saul MAE 208
1.
The baseball has a horizontal speed of 35 m/s when it is struck by the bat. It then travels away at an
angle of 60
o
from the horizontal and reaches a maximumheight of 50 m as measured from the height of
the bat (at B). What is the velocity of the ball wheen it leaves the bat? D
e
t
e
r
m
i
n
e
t
h
e
m
a
g
n
i
t
u
d
e
o
f
t
h
e
n
e
t
i
m
p
u
l
s
e
o
f
t
h
e
b
a
t
o
n
t
h
e
b
a
l
l
.
The ball has a mass of 400 g. Neglect the weight of the ball during
the time the bat strikes the ball. Note that the net impulse has an x and a y component that contribute
to its magnitude.
CHECK: v = 36.17 m/s
SOLUTION:
Impulse = 24.6 Ns
Impulse is a vector (just like force and velocity). The impulse momentum equationmust be evaluated
for the x and y directions separately. The impulse will have an x and a y component which can be
used to calculate the total net impulse. Units ofimpulse are Ns.
Energy is a SCALAR.
Some of you used conservation of energy to solve for the velocity of
the ball after it was hit with the bat.
This is valid and works, but you can not just consider the
y_component of the velocity when using the equation. The velocity in the kinetic energy is the velocity
of the ball. Period. If you were to use conservation of energy, the initial velocity of the ball would be
v2, what you are looking for, and the final velocity of the ball would be v2cos60, since you know that
when the ball reaches max height it has no y-velocity.
- 3 -
K. Saul MAE 208
2.
The 1-lb ball A is thrown so that when it strikes the 10-lb block B it is traveling horizontally at 20 ft/s.
The coefficient of restitution between A and B is e = 0.6. What is the final velocity of the ball A?
D
e
t
e
r
m
i
n
e
t
h
e
a
v
e
r
a
g
e
n
o
r
m
a
l
f
o
r
c
e
e
x
e
r
t
e
d
b
e
t
w
e
e
n
A
a
n
d
B
i
f
t
h
e
i
m
p
a
c
t
o
c
c
u
r
s
i
n
0
.
0
2
s
.
CHECK: -9.09 ft/s
SOLUTION: F = 45.3 lb
In this problem, be careful about your signs. In the conservation oflinear momentum equation, do not
make assumptions about the direction of any unknown velocities. For known velocities, you should
use your established coordinates system to establish the appropriate sign before putting it in the
equation. But the math willindicate the sign of the unknown velocities. Note that when using the
momentum/impulse equation, this solution alreadymade the presumption that the force of the block on
ball A was in the negative direction, thus their answer came out to be positive. If you had included this
unknown force as positive (a fine thing to do) you would have gotten a negative solution, indicating that
this force acts in the negative coordinate direction.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
- 4 -
K. Saul MAE 208
3.
Two smooth billiard balls A and B each have a mass of 200 g. Ball B is originally at rest and the
coefficient of restitution is e = 0.85. If A strikes B with a velocity (v
A
)
1
=1.5 m/s as shown, determine
the final velocity of A just after collision. W
h
a
t
i
s
t
h
e
f
i
n
a
l
v
e
l
o
c
i
t
y
o
f
B
?
CHECK: v
A
= 0.968 m/s, Note that only the magnitude of the velocity is given. In your answer,
also indicate the directions.
SOLUTION: v
B
= 1.06 m/s
- 5 -
K. Saul MAE 208
4.
The 10-lb block is moving on a surface for which
μ
k
=0.5. It is acted on by a radialforce of 2 lb and a
horizontalforce of 7 lb, always directed at 30
o
from the tangent to the path as shown. Neglect the size
of the block for the calculation. If the block is initiallymoving in a circular path with a speed v
1
=2 ft/s
at the instant the forces are applied, what is the velocity when the tension in cord AB is 20 lb?
D
e
t
e
r
m
i
n
e
t
h
e
t
i
m
e
r
e
q
u
i
r
e
d
b
e
f
o
r
e
t
h
e
t
e
n
s
i
o
n
i
n
c
o
r
d
A
B
b
e
c
o
m
e
s
2
0
l
b
.
CHECK: 13.67 ft/s
SOLUTION: 3.41 s
- 6 -
K. Saul MAE 208
5.
A child having a mass of 50 kg holds her legs up as shown as she swings downward from rest at
θ
1
=
30
o
. Her center ofmass is located at point G1 (a radius of 2.8 mfrom the origin). When she is at the
bottom position of
θ
=0
o
, she suddenly lets her legs come down, (approximately instantaneously)
shifting her center ofmass to position G
2
(a radius of 3 mfrom the origin). Determine her speed in the
upswing immediately following the change in the position of her center ofmass. (Think about what
quantity has changed now that her center ofmass is further from origin.) F
i
n
d
t
h
e
a
n
g
l
e
θ
2
t
o
w
h
i
c
h
s
h
e
s
w
i
n
g
s
b
e
f
o
r
e
m
o
m
e
n
t
a
r
i
l
y
c
o
m
i
n
g
t
o
r
e
s
t
.
Treat the child’s body as a particle.
CHECK: v
2
= 2.53 m/s
SOLUTION:
θ
2
=27
o
.
1)
Initial height with a velocity of v1 = 0.;
2)
Lowest position with a velocity v2.
3)
Lowest
position, but changes her COM, NEW velocity v3.
4)
Final height with velocity v4 = 0. (Same
approach as from 1 to 2)
Stage 1 to 2:
We know initial velocity, acted on by forces, and then velocity, v1. We don't know
time, so we can't use momentum-impulse.We do know how far she traveled, and normalforce does
no work (perpendicular to motion) and weight is conservative.So we can use conservation of energy
(T1+V1=T2+V2) instead of work and energy.
Stage 2 to 3:
The girl has a velocity that is an initial distance from the point of rotation, then she
changes the distance instaneously. This affects the angular momentum, H = r xmv. Looking at the
FBD during this transition, we see that both the weight force and the normalforce pass through the
pivot point, so NEITHER of them contributes an angular impulse. Angular momentumis conserved.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
- 7 -
K. Saul MAE 208
6.
The figure shows the internal gearing of a “spinner” used for drilling oil wells. With constant angular
acceleration, the motor M rotates the shaft S to 100 rev/min in t = 3 s starting from rest. Determine
the angular acceleration of the drill-pipe connection D and t
h
e
n
u
m
b
e
r
o
f
r
e
v
o
l
u
t
i
o
n
s
i
t
m
a
k
e
s
d
u
r
i
n
g
t
h
e
3
-
s
s
t
a
r
t
u
p
.
CHECK: 1.39 rad/s
2
SOLUTION: 1.0 rev