Dynamics Problem Solutions
Problem D1
Use the following equation:
1
y   gt 2  v0 y t  y0 .
2
Substitute g=32.2, y0=200, v0y=400sin(30) and y=0. Solve the resulting equation for t.
The result is t=13.3 seconds. Then use the following equation:
x : v0 x t  400 cos(30) * (13.3)
and find the distance covered. The answer is: x=4626 feet.
Problem D2
Find an as a function of time using:
v2
v  at t and an  ,
r
Where at=7 and r=300. Substitute into the expression for the magnitude of the
acceleration:
a  an2  at2
The magnitude of acceleration is given as g/4. Solve for t. The answer is t=4.93 sec.
And the velocity is v=34.5 ft/sec.
Problem D3
v2
given r=9 and =10. Using at  r given , find the
r
tangential acceleration. Finally find the magnitude of the acceleration vector:
Using v  r find an 
a  an2  at2 . The answer is v=7.5 ft/sec, and a=80.8 ft/s2.
Problem D4
Use the polar coordinates formulation: ar  r  r 2 and a  r  2r and find the
magnitude of acceleration vector by combining the two using:
a  an2  at2
The important point is that the two components of the tangential acceleration point in
opposite directions and must be subtracted (refer to lecture notes for helpful graphics).
Substitute for:
18
r
r  15 r  180   10   40
12
The answer is a=242 ft/sec2
Problem D5
v2
which evaluates to 84.5 ft/s2 using v=13 ft/sec
r
and r=2 ft. The equation of motion in the radial (normal) direction is:
W
 Fn  man  T  W cos(30)  ( g )(84.5)
where g=32.2 and W=25 lbs. We can solve for T. The answer is T=87.25 lbs. The
equation of motion in tangential direction is:
W
 Ft  mat  W sin( 30)  ( g )at .
We can solve for the tangential acceleration and find the angular acceleration using:
at  r . The answer is a=16.1 ft/s2 and =8.05 rad/s2.
The ball’s normal acceleration is an 
Problem D6
1
1
Using conservation of energy: T  U  0  mv 2  mgh  kx2  0 . In this
2
2
formula, x is the difference between the stretched length and the free length of the spring
and is calculated using the Pythagorean formula: x  12  1.52 1. The change is
kinetic energy is calculated to be 0.087v2. The change is potential energy is calculated to
be -5.244. We can solve for v. The answer is v=8.2 ft/s.
Problem D7
The two positions for writing the energy equation is the point of release and the position
where the block comes to a rest. The change in kinetic energy is
1
m(v22  v12 )  29.14 in  lb .
2
The change in gravitational P.E. is
U g  mg (h2  h1 )  25(10  ) sin(  )  25(10  )(0.6).
where  is the maximum amount the spring gets compressed. The change in elastic
(spring) PE is:
1
1
U s  kx2  (100)2 .
2
2
There is also work done on the particle by friction:
U12   fd  N (10  ).
Where f is the friction force,  is the coefficient of friction, and N is the normal force
equal to Wcos(). This work is always negative. Now, using the energy relationship:
T  U  U12   29.14  502  15(10  )  6(10  ).
We can find . The answer is=1.637 in and Fmax=163.7 lbs.
Problem D8 :
The change in KE as well the change in PE are the same as in the notes.
T  0.393 2
U  7 ft  lbs
The work done by the external force on the roller is U=Fd=(-1)(3)=-3 ft-lbs. Therefore:
U12  T  U   3  0.393 2  7    3.2 rad / s
The answer is w=3.2 rad/s
Problem D9
The change in K.E. is:
1
1
1
T  mb v 2  m p v 2  I p 2 .
2
2
2
The pulley moment of inertia can be found from its radius of gyration and its mass using:
I p  mp k 2 .
The velocity of the pulley center is related to the pulley’s angular velocity: v  r .
Therefore the change in K.E. is a function of .
The change in gravitational P.E. is
U g  (mb  m p ) gh.
The change in spring potential energy is:
1
U s  k ( x22  x12 ) .
2
Where, x1 and x2 are the initial and final length differences with respect to the free length.
When the block falls 9 inches, the combined length of the rope and the spring need to
increase 18 inches which is all due to spring stretching as the rope remains the same
length. Therefore: x1=0, and x2=18.
Setting up the energy equation
T  U g  U s  0
we can find =12.12 rad/s and v=4.04 ft/s.
Problem #D10
By running the physical event in reverse, we can find the bullet’s velocity. From the
energy relation and the swing of the pendulum, we can find the bar’s angular velocity just
after impact. Then, from the angular momentum relationship before and after impact we
can find the bullet’s speed.
T  U  0
1
1
1
I o 2f  I oi2   I oi2
2
2
2
1
L
1
1 1
where I o  mL2  m( ) 2  mL2  (
)( 2) 2  0.041 slugs  ft 2
12
2
3
3 32.2
2
Therefore : T  0.0205i
T 
L L
U  mg (h)  mg (  sin(  ))  1(1  sin( 30))  1.5
2 2
ft  lb
Substituting into the energy equation and solving for , we get i=8.55 rad/sec. This is
the angular velocity of the pendulum after impact. From angular momentum relationship
before and after impact:
.05
G A  GB where Gb  mb rv 
(1)v
32.2
Ga  I o (a )  0.41(8.55)
Equating angular momentums and solving for v we get: v=226 ft/sec.