Answer:
[tex]17.3 rad/s[/tex]
Explanation:
using conservation of energy:
let's say the brick weighs= m
the pulley weighs= M
and the radius is =R
the rotational inertia of the pulley is =J
Imagine the pulley is rotating at w rad/s. Then the total kinetic energy would be
[tex]K.E =1/2 J w^2 + 1/2 m (wR)^2[/tex]
(the brick travels at wR m/s)
[tex]K.E= 1/2 (J+mR^2)w^2[/tex]
[tex]K.E= 1/2(1/2 MR^2 +mR^2)w^2[/tex]
[tex](1/2MR^2+mR^2)[/tex] term is like a referred inertia of the system.
the loss in potential energy when the brick drops:
[tex]5 rad *1.5m =7.5m[/tex]
now
[tex]45N*7.5m =1/2Iw^2[/tex]
[tex]I=1/2 mr^2[/tex]
[tex]w=\sqrt{4*45*7.5/(2*1.5^2)}[/tex]
angular speed is:
[tex]w=17.3 rad/s[/tex]
Angular speed was measured as the rate of increase of angular displacement, and the following are the calculation of the angular speed of the pulley:
utilizing energy efficiency:
brick weight= m
weight of the pulley= M
radius=R
The pulley's rotational inertia is =J
Assume the pulley is rotating at [tex]w \frac{rad}{s}[/tex], Therefore the total kinetic energy:
[tex]K.E= \frac{1}{2} Jw^2+\frac{1}{2} m (wR)^2\\\\\text{(The brick moves at wR} \frac{m}{s})\\\\K.E= \frac{1}{2} w^2(J+ m R^2)\\\\K.E= \frac{1}{2} w^2(\frac{1}{2} MR^2+ m R^2)[/tex]
[tex]=(\frac{1}{2} MR^2+ m R^2)[/tex]
The above equation refers to the system's inertia, and when the brick falls, there is a loss of potential energy:
[tex]\to 5 rad \times 1.5 \ m = 7.5 \ m\\\\[/tex]
now
[tex]\to 45 \ N \times 7.5 \ m= \frac{1}{2}I w^2\\\\\to I=\frac{1}{2} mr^2\\\\\to w=\sqrt{ \frac{4 \times 45\times 7.5}{2\times 1.5^2} }[/tex]
[tex]=\sqrt{ \frac{4 \times 45\times 7.5}{2\times 1.5 \times 1.5 } }\\\\=\sqrt{ \frac{4 \times 45\times 75 \times 100 }{2\times 15 \times 15 \times 10} }\\\\=\sqrt{ 2 \times 3\times 5 \times 10 }\\\\=\sqrt{ 300 }\\\\=17.32[/tex]
Therefore, the final angular speed" [tex]w=17.3 \ \frac{rad}{s}[/tex]".
Learn more:
brainly.com/question/10756532
