Explanation:
1st- states that when two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction.
2nd- states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it. (most important law)
3rd- states that when two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction. (law of action/reaction)
Time it takes the projectile to hit the ground after being thrown up:
√h/1/2a
√8/(.5)(9.81)
√8/4.905
√1.630988787
= 1.277101714
= 1. 28
hope this helps :)
Some of the most common examples of mechanical waves are water waves, sound waves, and seismic waves. There are three types of mechanical waves: transverse waves, longitudinal waves, and surface waves.
To solve this problem it is necessary to apply the kinematic equations of angular motion.
Torque from the rotational movement is defined as
![\tau = I\alpha](https://tex.z-dn.net/?f=%5Ctau%20%3D%20I%5Calpha)
where
I = Moment of inertia
For a disk
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:
![2 \alpha \theta = \omega_f^2-\omega_i^2](https://tex.z-dn.net/?f=2%20%5Calpha%20%5Ctheta%20%3D%20%5Comega_f%5E2-%5Comega_i%5E2)
Where
Final and Initial Angular velocity
Angular acceleration
Angular displacement
Our values are given as
![\omega_i = 0 rad/s](https://tex.z-dn.net/?f=%5Comega_i%20%3D%200%20rad%2Fs)
![\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})](https://tex.z-dn.net/?f=%5Comega_f%20%3D%20450rev%2Fmin%20%28%5Cfrac%7B1min%7D%7B60s%7D%29%28%5Cfrac%7B2%5Cpi%20rad%7D%7B1rev%7D%29)
![\omega_f = 47.12rad/s](https://tex.z-dn.net/?f=%5Comega_f%20%3D%2047.12rad%2Fs)
![\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad](https://tex.z-dn.net/?f=%5Ctheta%20%3D%203%20rev%20%28%5Cfrac%7B2%5Cpi%20rad%7D%7B1rev%7D%29%20%5Crightarrow%206%5Cpi%20rad)
![r = 7cm = 7*10^{-2}m](https://tex.z-dn.net/?f=r%20%3D%207cm%20%3D%207%2A10%5E%7B-2%7Dm)
![m = 17g = 17*10^{-3}kg](https://tex.z-dn.net/?f=m%20%3D%2017g%20%3D%2017%2A10%5E%7B-3%7Dkg)
Using the expression of angular acceleration we can find the to then find the torque, that is,
![2\alpha\theta=\omega_f^2-\omega_i^2](https://tex.z-dn.net/?f=2%5Calpha%5Ctheta%3D%5Comega_f%5E2-%5Comega_i%5E2)
![\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cfrac%7B%5Comega_f%5E2-%5Comega_i%5E2%7D%7B2%5Ctheta%7D)
![\alpha = \frac{47.12^2-0^2}{2*6\pi}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B47.12%5E2-0%5E2%7D%7B2%2A6%5Cpi%7D)
![\alpha = 58.89rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%2058.89rad%2Fs%5E2)
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
![\tau = I\alpha](https://tex.z-dn.net/?f=%5Ctau%20%3D%20I%5Calpha)
![\tau = (\frac{1}{2}mr^2)\alpha](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%28%5Cfrac%7B1%7D%7B2%7Dmr%5E2%29%5Calpha)
![\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%28%5Cfrac%7B1%7D%7B2%7D%2817%2A10%5E%7B-3%7D%29%287%2A10%5E%7B-2%7D%29%5E2%29%2858.89%29)
![\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m](https://tex.z-dn.net/?f=%5Ctau%20%3D%200.00245N%5Ccdot%20m%20%5Capprox%202.45%2A10%5E%7B-3%7DN%5Ccdot%20m%20)
Therefore the torque exerted on it is ![2.45*10^{-3}N\cdot m](https://tex.z-dn.net/?f=2.45%2A10%5E%7B-3%7DN%5Ccdot%20m)