Given:
• Rotational inertia = 40 kg.m²
,• Initial angula speed = 10 rev/s
,• Mass, m = 4 kg
,• Diameter, d = 1.2 m
Let's find the angular speed of the wheel.
To find the angular speed, apply the formula:
[tex]L_i=(I+md^2)*w_f[/tex]Where wf is the final angular speed
I is the rotational inertia
m is the mass
d = 1.2
Li is the angular momentum.
To find the angular momentum, we have:
[tex]\begin{gathered} L_i=40*10*2\pi \\ L_i=2513.27\text{ kg.m}^2\text{ rad/s} \end{gathered}[/tex]Now, to find the final angular speed, wf, plug in values in the first equation and solve for wf:
[tex]\begin{gathered} Li=(I+md^2)w_f \\ \\ 2513.27=(40+4*1.2^2)w_f \\ \\ 2513.27=45.76w_f \\ \\ w_f=\frac{2513.27}{45.76} \\ \\ w_f=54.9\approx55\text{ rad/s} \end{gathered}[/tex]Therefore, the final angular speed is 55 rad/s.
ANSWER:
1.) 55 rad/s
