# CHAPTER10 : ROTATIONAL MECHANICS¶

## Example E1 : Pg 167¶

In [1]:
#example 10.1
#calculation of the number of revolutions made
#given data
import math
wzero=100.*2.*math.pi/60.#initial angular velocity(in rad/s) of the motor
w=0#final angular velocity(in rad/s) of the motor
t=15.#time interval(in s)

#calculation
alpha=(w-wzero)/t#equation of angular motion
theta=(wzero*t)+(alpha*t*t/2.)#equation of angular motion

print '%s %.2f' %("the number of revolutions the motor makes before coming to rest is",theta/(2*math.pi))

the number of revolutions the motor makes before coming to rest is 12.50


## Example E2 : Pg 168¶

In [2]:
import math
#example 10.2
#calculation of the time taken by the fan to attain half of the maximum speed

#given data
wzero=0#initial angular velocity(in rad/s) of the fan
w=400.*(2.*math.pi/60.)#final angular velocity(in rad/s) of the fan
t=5#tiem(in s) taken

#calculation
alpha=(w-wzero)/t#equation of angular motion
wdash=w/2.#half of maximum speed
t1=(wdash-wzero)/alpha#equation of angular motion

print '%s %.2f %s' %("the time taken by the fan to attain half of the maximum speed is",t1,"s")

the time taken by the fan to attain half of the maximum speed is 2.50 s


## Example E3 : Pg 168¶

In [3]:
import math
#example 10.3
#calculation of the angular velocity and angular acceleration of the pulley

#given data
v=20.#linear speed(in cm/s) of the bucket
a=4.*10.**2.#linear acceleration(in cm/s**2) of the pulley

#calculation
w=v/r#formula of angular velocity
alpha=a/r#formula of angular acceleration

print '%s %.2f %s %s %.2f %s' %("the angular velocity of the pulley is",w,"rad/s","and angular acceleration of the pulley is",alpha,"rad/s**2")

the angular velocity of the pulley is 2.00 rad/s and angular acceleration of the pulley is 40.00 rad/s**2


## Example E5 : Pg 171¶

In [4]:
import math
#example 10.5
#calculation of the moment of inertia of the wheel

#given data
F=5.#force(in N) of pulling
aplha=2.#angular acceleration(in rad/s**2) of the wheel

#calculation
tau=F*r#net torque
I=tau/aplha#moment of inertia

print '%s %.2f %s' %("the moment of inertia of the wheel is",I,"kg-m**2")

the moment of inertia of the wheel is 0.25 kg-m**2


## Example E : Pg 182¶

In [5]:
import math
#example 10.12
#calculation of the kinetic energy of the sphere

#given data
M=200.*10.**-3.#mass(in kg) of the sphere
vcm=2.*10.**-2.#speed(in m/s) of the sphere

#calculation
#kinetic energy is  K = (Icm*w*w/2) + (M*vcm*vcm/2)
#taking Icm = (2*M*r*r*w*w/5) and w=vcm/r
K=(M*vcm*vcm/5.)+(M*vcm*vcm/2.)#kinetic energy

print '%s %.5f %s' %("the kinetic energy of the sphere is",K,"J")

the kinetic energy of the sphere is 0.00006 J


# WORKED EXAMPLES¶

## Example E2w : Pg 183¶

In [6]:
import math
#example 10.2w
#calculation of the angle rotated during the next second

#given data
theta=2.5#angular displacement(in rad) of the wheel
t=1.#time(in s) required

#calculation
alpha=(theta*2.)/(t*t)#equation of angular motion
theta1=(alpha*(t+1.)*(t+1.)/2.)#angle rotated during first two seconds
thetar=theta1-theta#angle rotated during next second

print '%s %.2f %s' %("the angle rotated during the next second is",thetar,"rad\n")

the angle rotated during the next second is 7.50 rad



## Example E3w : Pg 183¶

In [7]:
import math
#example 10.3w
#calculation of the torque required to stop the wheel in one minute

#given data
wzero=50.*(2.*math.pi/60.)#initial angular velocity(in rad/s) of the wheel
w=0#final angular velocity(in rad/s) of the wheel
t=60.#time(in s) taken to stop the wheel
I=2.#moment of inertia(in kg-m**2) of the wheel

#calculation
alpha=(w-wzero)/t#equation of angular motion
tau=I*abs(alpha)#torque

print '%s %.2f %s' %("the torque required to stop the wheel in one minute is",tau,"N-m")

the torque required to stop the wheel in one minute is 0.17 N-m


## Example E4w : Pg 184¶

In [8]:
import math
#example 10.4w
#calculation of the angular velocity of the wheel

#given data
F=20.#force(in N) of pull applied
I=.2#moment of inertia(in kg-m**2)
t=5.#time(in s) interval
wzero=0#initial angular velocity(in rad/s) of the wheel

#calculation
tau=F*r#torque applied to the wheel
alpha=tau/I#angular acceleration
w=wzero+(alpha*t)#equation of angular motion

print '%s %.2f %s' %("the angular velocity of the wheel after 5 s is",w,"rad/s")

the angular velocity of the wheel after 5 s is 100.00 rad/s


## Example E7w : Pg 184¶

In [9]:
import math
#example 10.7w
#calculation of the position of second kid on a balanced seesaw

#given data
ma=10.#mass(in kg) of kid A
mb=15.#mass(in kg) of kid B
l=5.#length(in m) of the seesaw
la=(l/2.)#distance of A kid from fulcrum as he is sitting at an end

#calculation
#taking torque about fulcrum...........(mb*g*x) = (ma*g*)
x=(ma*la)/mb

print '%s %.2f %s' %("the second kid should sit at a distance of",x,"m from the centre")

the second kid should sit at a distance of 1.67 m from the centre


## Example E8w : Pg 185¶

In [12]:
import math
#example 10.8w
#calculation of the normal force and the frictional force that the floor exerts on the ladder

#given data
g=9.8#gravitational acceleration(in m/s**2) of the earth

#calculation
#taking horizontal and vertical components
#N1 = f........................(1)
#N2 = W........................(2)
W=m*g
N2=W#from equation (2)
f=(W*math.sin(theta)*57.3/2.)/(math.cos(theta)*57.3)#from equation (1)

print '%s %.2f %s' %("the normal force that the floor exerts on the ladder is",N2,"N\n")
print '%s %.2f %s' %("the frictional force that the floor exerts on the ladder is",f,"N\n")

the normal force that the floor exerts on the ladder is 98.00 N

the frictional force that the floor exerts on the ladder is -21.13 N



## Example E9w : Pg 185¶

In [11]:
import math
#example 10.9w
#calculation of the contact force exerted by the floor on each leg of ladder

#given data
theta=60.#angle(in degree) between the two legs
m=80.#mass(in kg) of the person
g=9.8#gravitational acceleration(in m/s**2) of the earth

#calculation
N=m*g/2.
T=(N*2.*math.tan(90-theta)*57.3)/1.

print '%s %.2f %s' %("the contact force exerted by the floor on each leg of ladder",N,"N\n")
print '%s %.2f %s' %("the tension in the crossbar is",T,"N\n")

the contact force exerted by the floor on each leg of ladder 392.00 N

the tension in the crossbar is -287747.97 N



## Example E13w : Pg 186¶

In [13]:
import math
#example 10.13w
#calculation of the kinetic energy and angular momentum of the disc

#given data
M=200.*10.**-3.#mass(in kg) of the disc

#calculation
I=(M*r*r)/4.#moment of inertia
K=(I*w*w/2.)#kinetic energy
L=I*w#angular momentum

print '%s %.2f %s' %("the kinetic energy of the disc is",K,"J")
print '%s %.2f %s' %("\nthe angular momentum of the disc is",L,"J-s\n")

the kinetic energy of the disc is 0.00 J

the angular momentum of the disc is 0.00 J-s



## Example E14w : Pg 186¶

In [14]:
import math
#example 10.14w
#calculation of the work done by the torque in first two seconds
#given data
wzero=20.#initial angular velocity(in rad/s) of the motor
w=0#final angular velocity(in rad/s) of the motor
t=4.#time(in s) taken to attain rest position
I=.20#moment of inertia(in kg-m**2) of the disc about axis of rotation
t1=2.#time(in s)

#calculation
alpha=(wzero-w)/t#equation of angular motion in case of deceleration
tau=I*alpha#torque
theta=(wzero*t1)-(alpha*t1*t1/2)#equation of angular motion
W=tau*theta#work done by the torque

print '%s %.2f %s' %("the work done by the torque in first two seconds is",W,"J")

the work done by the torque in first two seconds is 30.00 J


## Example E19w : Pg 188¶

In [15]:
import math
#example 10.19w
#calculation of the moment of inertia of the system about the axis perpendicular to the rod passing through its middle point

#given data
m=1.2#mass(in kg) of the sphere
sep=50.*10.**-2.#separation(in m) between the two spheres

#calculation
d=sep/2.#distance of each sphere from centre
I=Icm+(m*d*d)#by parallel axis theorem,moment of inertia about given axis
#since second sphere has same moment of inertia
Isys=2.*I#moment of inertia of the system

print '%s %.2f %s' %("the moment of inertia of the system about the axis perpendicular to the rod passing through its middle point is",Isys,"kg-m**2")

the moment of inertia of the system about the axis perpendicular to the rod passing through its middle point is 0.16 kg-m**2


## Example E22w : Pg 189¶

In [16]:
import math
#example 10.22w
#calculation of the number of revolutions made by the wheel per second

#given data
p=220.*10.**-2.#perimeter(in cm) of the wheel
v=9.*10.**3./(60.*60.)#linear speed(in m/s) of wheel on the road

#calculation

the number of revolutions made by the wheel per second is 1.14 rev/s