Respuesta :
Answer:
[tex]\theta=12.5\ rad[/tex]
Explanation:
Given:
- mass of disk, [tex]m=4\ kg[/tex]
- radius of the disk, [tex]r=2\ m[/tex]
- rate of increase in kinetic energy, [tex]\dot {KE}=20\ J.s^{-1}[/tex]
- time of observation, [tex]t=5\ s[/tex]
Now the moment of inertia of the disk:
[tex]I=\frac{1}{2} m.r^2[/tex]
[tex]I=\frac{1}{2}\times 4\times 2^2[/tex]
[tex]I=8\ kg.m^2[/tex]
We know that Kinetic energy:
[tex]KE=\frac{1}{2}\times I.\omega^2[/tex]
[tex]\dot {KE}\times t=\frac{1}{2} \times 8\times \omega^2[/tex]
[tex]20\times 5=0.5\times 8\times \omega^2[/tex]
[tex]\omega =5\ rad.s^{-1}[/tex] is the angular velocity after 5 seconds.
- Since the disk starts from the rest, its initial angular speed, [tex]\omega_i=0\ rad.s^{-1}[/tex]
Using equation of motion:
[tex]\omega=\omega_i+\alpha.t[/tex]
[tex]5=0+\alpha\times 5[/tex]
[tex]\alpha=1\ rad.s^{-2}[/tex]
Now the angle rotated in the give time:
[tex]\theta=\omega_i.t+\frac{1}{2} .\alpha.t^2[/tex]
[tex]\theta=0+0.5\times 1\times 5^2[/tex]
[tex]\theta=12.5\ rad[/tex]
Explanation:
Formula to calculate energy change is as follows.
dE = [tex]\tau \times \theta[/tex]
[tex]\tau = \frac{dE}{\theta}[/tex]
[tex]\frac{dE}{\theta}[/tex] = 20 J/rad
It is given that mass is 4 kg, and radius is 2 m. Hence, we will calculate the moment of inertia as follows.
I = [tex]\frac{1}{2}mv^{2}[/tex]
= [tex]\frac{1}{2}\times 4 kg \times (2)^{2}[/tex]
= 8 [tex]kg m^{2}[/tex]
Also, [tex]\tau = I \times \alpha[/tex]
20 = [tex]8 \times \alpha[/tex]
or, [tex]\alpha = \frac{20}{8}[/tex]
= 2.5 [tex]rad/s^{2}[/tex]
[tex]w_{o}[/tex] = 0 (rest), t = 4 sec
Also, [tex]\theta = w_{o}t + \frac{1}{2} \alpha t^{2}[/tex]
= [tex]0 + \frac{1}{2} \times \frac{5}{2} \times (5)^{2}[/tex]
= 15.625 rad
or, = [tex]\frac{15.625}{2\pi}[/tex] rev
= 2.48 rev
Therefore, we can conclude that angular displacement is 2.48 rev.
