Question

3. An experiment is set up as shown in Figure Q3 to measure the moment of inertia of a circular flywheel. The mass of 30 kg is attached to a light inextensible string which is wrapped around the outer diameter of the flywheel which has a radius of 0.4 m. The mass is released when the system is at rest and drops 5.4 m in a time of 3 seconds. Friction in the bearings can be neglected Flywheel Bearing 30 kg Figure Q3. A system for determining the moment of inertia of a flywheel a) Considering the kinematics of the 30 kg mass, calculate the downwards acceleration of the 20 kg mass from the information givern 4 marks] b) Draw free body diagrams for the 30 kg mass and the flywheel labelling all external forces applied [6 marks] c) Determine the equation of motion of the 30 kg mass and the angular equation of motion for the flywheel and generate a constraint equation to relate the angular motion of the flywheel to the translational motion of the 30kg mass. Hence, or otherwise, determine the moment of inertia of the flywheel [8 marks] d) Determine the vertical reaction at the bearing of the flywheel under these conditions given that the moment of inertia, I, of a circular flywheel of mass m and radius r about its axis is given by 1-ļmr2 [7 marks]

Answer 1

a)

y = ut + (1 /2)at'

with given values:

-5.4 = 0 * 3 + (1//2)a * 32

1.2m/s a-- Cl

b)

c)

Torque on the flywheel:

$\tau = T*R = 0.4T$

rotational equation of motion:

$\tau = 0.4T =I \alpha$

Equation of motion for the 30 kg mass

-mq = ma

we already found acceleration

$\Rightarrow T = mg-1.2m$

T-30 * 9.8-1.2 * 30

$\Rightarrow T = 258N$

From the figure:

$y = R\theta$

differentiating it twice

$\ddot y = R\ddot \theta$

$a = R\alpha$

Now consider:

0.4TIa

$0.4*258 =I (\frac{a}{R})$

$0.4*258 =I (\frac{1.2}{0.4})$

$\Rightarrow I = 34.4kg-m^2$

d)

$\Rightarrow I = 34.4kg-m^2 = \frac{1}{2}mr^2$

$\Rightarrow I = 34.4 = \frac{1}{2}m*0.4^2$

vertical reaction is given by: