Respuesta :
To solve this problem it is necessary to apply the kinematic equations of angular motion.
Torque from the rotational movement is defined as
[tex]\tau = I\alpha[/tex]
where
I = Moment of inertia [tex]\rightarrow \frac{1}{2}mr^2[/tex] For a disk
[tex]\alpha =[/tex] Angular acceleration
The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:
[tex]2 \alpha \theta = \omega_f^2-\omega_i^2[/tex]
Where
[tex]\omega_{f,i} =[/tex] Final and Initial Angular velocity
[tex]\alpha =[/tex] Angular acceleration
[tex]\theta =[/tex] Angular displacement
Our values are given as
[tex]\omega_i = 0 rad/s[/tex]
[tex]\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})[/tex]
[tex]\omega_f = 47.12rad/s[/tex]
[tex]\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad[/tex]
[tex]r = 7cm = 7*10^{-2}m[/tex]
[tex]m = 17g = 17*10^{-3}kg[/tex]
Using the expression of angular acceleration we can find the to then find the torque, that is,
[tex]2\alpha\theta=\omega_f^2-\omega_i^2[/tex]
[tex]\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}[/tex]
[tex]\alpha = \frac{47.12^2-0^2}{2*6\pi}[/tex]
[tex]\alpha = 58.89rad/s^2[/tex]
With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so
[tex]\tau = I\alpha[/tex]
[tex]\tau = (\frac{1}{2}mr^2)\alpha[/tex]
[tex]\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)[/tex]
[tex]\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m [/tex]
Therefore the torque exerted on it is [tex]2.45*10^{-3}N\cdot m[/tex]
The torque exerted on the CD is 0.00245N/m
Data;
- final angular acceleration = w2 = 450 rev/min
- angular acceleration = 3.0 rev
- radius = 7.0cm
- mass = 17g
Torque
The torque exerted on the CD can be calculated as
[tex]\tau = Ia[/tex]
- I = moment of inertia
- a = acceleration
The final revolution is
[tex]w_2 = (450 * 2\pi )/60 = 47.12 rad/s[/tex]
3 revs = 18.8495 rads.
a = angular acceleration
g = angle
using a modified equation of motion,
[tex]w_2^2 = w_1^2 + 2a(g_2 - g_1)\\47.12^2 = 0 + 2a (18.84 - 0)\\a = 58.895 rad/s^2[/tex]
Let's use this to find the torque exerted on the CD
[tex]\tau = Ia\\\tau = \frac{1}{2} * (0.017) * 0.07^2 * 58.895\\\tau = 0.00245 N/m[/tex]
The torque exerted on the CD is 0.00245N/m
Learn more on torque and angular acceleration here;
https://brainly.com/question/6860269
