Task 1. Part 1. Provided Data:. Radius of the wheel rim

Task 1
Part 1
Provided Data:
Radius of the wheel rim R = 0.6 m
Torque (τ ) required to spin the wheel = 4 Nm
Time taken (t) to spin the wheel from rest to final velocity = 24 seconds
Final angular velocity (ω2 ¿ = 27 rad/s
Moment of inertia = m R2
Distance from chain to centre of wheel = 0.12 m
Required: Mass of the wheel, angular momentum and axial torque in the system.
Solution
Torque = Force * Radius
τ =FR
F= τ
R = 4
0.6 =6.6667 N
Angular acceleration α = ω2−ω1
∆ t
α = 27−0
24 =1.125 rad /s2
Force F=mRα
Making m the subject;
m= F
Rα = 6.6667
0.6∗1.125 =9.877 kg
Therefore, the mass of the wheel is 9.877 kg.
Angular momentum L=Iω

Moment of inertia I =m R2
I =9.877∗0.62=3.5556 kg m2
ω=27 rad / s
Substituting in the formula, we get;
L=3.5556∗27=96 kg m2 / s
Thus, the angular momentum is 96 kg m2 / s
Torque on the axle producing the gyroscopic perception is given by;
⃗
τ = ⃗ r m ⃗ g
Where;
⃗
r =0.12 m
m=9.877 kg
⃗
g=9.8 m/s2
Substituting back;
⃗
τ =0.12∗9.877∗9.81=11.627 Nm
⃗
τ =11.627 Nm
Energy transfer explanation

Figure 1: Gyroscope precession sample
From figure 1 shown above, we observe how gyroscopic precession looks like. When the top is
positioned on a level surface near to the surface of the earth at angle to the vertical and is not
rotating, force of gravity will cause it to fall over. The force of gravity generates a torque acting
on its centre of mass. When spinning on its axis, it precesses about the vertical and does not fall
over. This is made possible by the torque created on its center of mass which results to change in
angular momentum. The momentum is conserved in all the cases. This is the principle applied in
gyroscopic precession (Ungar, 2012).
Task 2
Provided data
l1=70 mm