Mechanical Engineering Assignment: Shafts, Flywheels, and Engines

This document presents solutions to a mechanical engineering assignment. The assignment includes four tasks: determining bearing forces on a rotating shaft with multiple cranks, calculating the initial kinetic energy and time to rest for a flywheel with an applied torque, finding the mass of a flywheel to maintain engine speed within a specified range, and calculating the common speed and loss of kinetic energy when two flywheels are connected. The solutions involve applying principles of rotational dynamics, including moment of inertia, angular velocity, torque, and conservation of angular momentum, to solve for the required parameters. The document provides detailed steps and calculations for each task, demonstrating the application of relevant formulas and concepts.

Task 1 a rotating shaft of length 4.5mm supported by a bearing at each end it carries three cranks A,B
and C mounted on its 20°, 80° and 150° respectively from each other.
A has a mass of 1 kg acting at a radial distance of 400mm from the crank centre and is 1 m from the
left hand bearing.
B has a mass of 2.5 kg acting at a radial distance of 500mm from the crank centre and is 2.5 m from
the left hand bearing.
C has a mass of 4 kg acting at a radial distance of 250mm from the crank centre and is 3.5 m from the
left hand bearing.
Determine the forces that are exerted on the bearings when the shaft is rotating at4 rev/s.
Solution:
Given data mA=1 kg, mB=2.5 kg, mC=4 kg, n = 4 rev/s
L=4.5m, l1=0.4m, l2=0.5m, l3=0.25m
ØA=20°, ØB=80°, ØC=150°
Crank Distance Mass Angle
A 0.4 1 20
B 0.5 2.5 80
C 0.25 4 150
Task 2 a flywheels has a moment of inertia of 20000 kgm2 and is rotating at 600rpm. If constant
resistant torque of 900 Nm is applied to the wheel determine
a) The initial kinetic energy of the flywheel.
b) The time taken to bring it to rest.
Solution: given data I= 20000 kgm2
n = 600 rpm, T= 900 Nm
Initial Kinetic Energy (K) = ½ Iω2 (1)
ω= 2 π n

= 2*3.14*600/60 rad/sec
=62.8 rad/sec
Keep this value in the equation 1
Kinetic energy = (1/2)*20000*(62.8)2
(a) =10000*3943.84 =39.44*106 j Answer-a
Torque (T) = I*α
α = T/I
=900/20000 =45*10-3 rad/sec2
So for final angular velocity ω2=0
ω2= ω1-αt
0=62.8-45*10-3*t
t= 62.8/ (45*10-3)
=62.8*1000/45 =1.395*1000
(b) t= 1395 sec Answer-b
Task 3 find the mass of the flywheel required to keep the speed of engine between 397 and 403 rpm if
Its radius of gyration is 600mm and the greatest fluctuation energy is 9KJ.
Solution: given data Nmin= 395rpm, Nmax= 403rpm
R=600mm, ØE=9KJ
I =mr2 (rim type flywheel)
I =1/2*mr2 (disc type flywheel)
Let’s assume rim type flywheel
ØE= EMAX-EMIN
9000= ½*I*ω max2 -½*I*ω min2
9000=1/2 * I*(ω max2 -ω min2)

9000= ½ * m *r2 * (ω max2 -ω min2)
9000=1/2 *m*(600*10-3)*{4* π 2 / (602)( Nmax2- Nmin2)}
9000=1/2 *m*(0.6)*{4*3.142 / (3600) ((403)2-(395)2)}
m= 7.149 * 10-3+5
Mass = 714.9 kg Answer
Task 4 a flywheel and shaft have a mass of 60kg and a effective radius of gyration k of 500mm and
rotation of 300 rpm. By means of clutch another identical flywheel and shaft on the same axis is
suddenly connected. If this flywheel is initially at rest what will be:
a) The common speed of the rotation after two are connected
b) The loss of kinetic energy.
Solutions =
Given data mass m=60kg, radius of gyration k=500mm=0.5m, n1=300 rpm
When the common speed of the rotation after two are connected
Apply conservation of angular momentum
L1=L2
I1ω1= I1ω1+ I2ω2
I1= I2, after connected, ω1= ω2= ω
Iω1= (I1+ I2) ω
Iω1=2I ω
ω1= 2*ω
ω = ω1/2 =2*3.14*300/2
2*3.14*n= 2*3.14*300/2
N=150 rpm Answer
Loss of kinetic energy
Before connected kinetic energy
K= ½* I1ω12 =1/2* Iω12 (1)

After connected
Total kinetic energy
K2= ½*I1ω12 + ½* I2ω22
As we know after connected, I1= I2, ω1= ω2= ω
K2=2*1/2* Iω2 = Iω2 (2)
Use equation 1 and 2
Loss of kinetic energy
ØK= K2-K1
= Iω2- ½* Iω1 2
=I (ω2- ½*ω1 2)
=m*k2*(ω2- ½*(2ω) 2)
=60*(0.5)2*(ω2)
=60*0.25*(2*3.14*150/60)2
=15*(15.7)2
=3697.35 J =3.7KJ Answer