Ch 12 problems¶
In [1]:
from numpy import *
dtr = pi/180
def mag_dir(vec):
# returns magnitude and direction (deg) of 2D vector
direc = arctan2(vec[1],vec[0]) / dtr
mag = sqrt( vec[0]**2 + vec[1]**2 )
return mag,direc
In [2]:
a0,a1 = 50,75
Da = a0-a1
s1 = 1000
#
v = sqrt(2*a1*s1 + Da*s1)
t = -12.649*( arcsin( sqrt(1-0.000166667*s1)) - arcsin(1) )
t
Out[2]:
5.3193444638093785
In [3]:
vx = 10 # m/s
x0 = 1.3 # m
y0 = sqrt(1-x0**2/4) # m
A,B = 1/4,1
y0 = sqrt(1-A*x0**2)
vy = -A*x0/y0*vx
v = sqrt(vx**2+vy**2)
print ('v = {:.2f} m/s'.format(v))
ay = (-A*vx**2-vy**2)/y0
print ('a = {:.2f} m/s'.format(ay))
v
v = 10.88 m/s a = -56.97 m/s
Out[3]:
10.87612262205807
In [4]:
A = 1.5e-3 # ft^-1
B = 15 # ft
v0 = 75 # ft/s
x0 = 50 # ft
#vy = v0*( 1 + 1/(4*A**2*x**2))**(-0.5)
vy = v0/sqrt( 1 + 1/(4*A**2*x0**2) )
vx = -sqrt(v0**2-vy**2)
print ('v = {:.2f} i + {:.2f} j ft/s'.format(vx,vy))
ay = -2*A*vx**2/( 1 + 4*A**2*x0**2)
ax = ay*2*A*x0
print ('a = {:.2f} i + {:.2f} j ft/s2'.format(ax,ay))
v = -74.17 i + 11.13 j ft/s a = -2.42 i + -16.14 j ft/s2
In [5]:
dtr = pi/180
g = 32.2 # ft/s2
v0,Q = 14,30*dtr # ft/s, rad
A = 0.04
vox,voy = v0*cos(Q), v0*sin(Q)
x = voy/vox / ( g/(2*vox**2) - A )
print('vox=',vox)
print('voy=',voy)
y = -A*x**2
print ('hits ground at x=',x,'ft, and y=',y,'ft')
vox= 12.124355652982143 voy= 6.999999999999999 hits ground at x= 8.304353186974069 ft, and y= -2.758491274160255 ft
In [6]:
%matplotlib inline
import matplotlib.pyplot as plt
x = linspace(0,10)
yy = voy/vox*x - (0.5*g*x**2)/vox**2
yslope = -A*x**2
plt.plot(x,yy,label='flight')
plt.plot(x,yslope,label='slope')
plt.legend()
plt.grid()
F12-30¶
When $x$ = 10 ft the crate has a speed of 20 ft/s which is increasing at 6 ft/s$^2$. Determine the direction of the crate's velocity and the magnitude of the crate's acceleration at this instant.
hand written solution (png)
In [7]:
A = 1/24 # 1/ft
x = 10 # ft
v = 20 # ft/s
at = 6 # ft/s2
rho = (1+(2*A*x)**2)**(3/2) / (2*A)
an = v**2/rho
a = sqrt( at**2 + an**2 )
print('acceleration = {:.2f} ft/s2'.format(a))
Q = arctan(2*A*x)/dtr
print('angle of velocity (below horizontal) = {:.1f} deg'.format(Q))
acceleration = 16.26 ft/s2 angle of velocity (below horizontal) = 39.8 deg
In [8]:
R = 3 # ft
w = 0.5 # rad/s
A = 3 # ft
B = 4 #
t1 = 3 # s
In [9]:
vq = R*w
vz = A*B*w*cos(B*w*t1)
v = sqrt(vq**2+vz**2)
print ('velocity magnitude =',v,'ft/s')
ar = -R*w**2
az = -A*B**2*w**2*sin(B*w*t1)
a = sqrt(ar**2+az**2)
print ('acceleration magnitude =',a,'ft/s2')
velocity magnitude = 5.953097618650719 ft/s acceleration magnitude = 3.4358426872108927 ft/s2
In [10]:
A = 10 # m/s
B = 5 # m/s2
C = 18.5 # m/s
D = 2 # m/s2
Q = 45*dtr # rad
rho = 100 # m
vAB = array(( A*cos(Q)-C , -A*sin(Q) ))
aAB = array(( (B-A**2/rho)*cos(Q)-D, (-B-A**2/rho)*sin(Q) ))
print( 'relative vel. of A relative to B = {:.3f} m/s at {:.3f} deg'.format(*mag_dir(vAB)))
print( 'relative acc. of A relative to B = {:.3f} m/s2 at {:.3f} deg'.format(*mag_dir(aAB)))
relative vel. of A relative to B = 13.440 m/s at -148.255 deg relative acc. of A relative to B = 4.323 m/s2 at -78.951 deg
In [11]:
h = 2.0 # m
vB = 6. # m/s
aB = 2 # m/s2
sA = 1.5 #m
vA = -vB / sA * sqrt( sA**2+h**2)
print ('velocity of A = {:.2f} m/s'.format(vA))
x = sqrt( sA**2 + h**2 )
aA = -aB/sA*x + vB*vA/sA**2*x - vB*vA/x
print ('acceleration of A = {:.2f} m/s'.format(aA))
velocity of A = -10.00 m/s acceleration of A = -46.00 m/s
12-220¶
hand written solution (jpg)
The only trick here is solving for $\theta$. Need to use quadratic formula.
In [12]:
v0 = 19 # km/h
A = 4 # km/h
p = 70*dtr # rad
alpha = v0*sin(p)/cos(p)
cc1 = (2*A*v0 + sqrt( (2*A*v0)**2 - 4*(v0**2+alpha**2)*(A**2-alpha**2) ))/( 2*(v0**2+alpha**2))
Q1 = arccos(cc1)/dtr
print ('point at angle = {:.2f} deg'.format(Q1))
point at angle = 15.87 deg
In [13]:
vA,vB = 10,15 # m/s
QA,QB = 120*dtr, 45*dtr # rad
vrx = vA*cos(QA) - vB*cos(QB)
vry = vA*sin(QA) - vB*sin(QB)
t = 600 / sqrt(vrx**2 + vry**2)
print ('time is = {:.2f} s'.format(t))
time is = 38.15 s
In [14]:
tan(45*dtr)
arccos( 2*cos(45*dtr) -1 )/dtr
Out[14]:
65.5301994792978