1) ![1.86\cdot 10^6 rad/s^2](https://tex.z-dn.net/?f=1.86%5Ccdot%2010%5E6%20rad%2Fs%5E2)
2) 2418 rad/s
3) ![27000 m/s^2](https://tex.z-dn.net/?f=27000%20m%2Fs%5E2)
4) 36.3 m/s
Explanation:
1)
The angular acceleration of an object in rotation is the rate of change of angular velocity.
It can be calculated using the following suvat equation for angular motion:
![\theta=\omega_i t +\frac{1}{2}\alpha t^2](https://tex.z-dn.net/?f=%5Ctheta%3D%5Comega_i%20t%20%2B%5Cfrac%7B1%7D%7B2%7D%5Calpha%20t%5E2)
where:
is the angular displacement
is the initial angular velocity
t is the time
is the angular acceleration
In this problem we have:
is the angular displacement
t = 1.3 ms = 0.0013 s is the time elapsed
is the initial angular velocity
Solving for
, we find:
![\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B2%28%5Ctheta-%5Comega_i%20t%29%7D%7Bt%5E2%7D%3D%5Cfrac%7B2%28%5Cpi%2F2%29-0%7D%7B0.0013%7D%3D1.86%5Ccdot%2010%5E6%20rad%2Fs%5E2)
2)
For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:
![\omega_f = \omega_i + \alpha t](https://tex.z-dn.net/?f=%5Comega_f%20%3D%20%5Comega_i%20%2B%20%5Calpha%20t)
where
is the initial angular velocity
t is the time
is the angular acceleration
In this problem we have:
t = 1.3 ms = 0.0013 s is the time elapsed
is the initial angular velocity
is the angular acceleration
Therefore, the final angular speed is:
![\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s](https://tex.z-dn.net/?f=%5Comega_f%20%3D%200%20%2B%20%281.86%5Ccdot%2010%5E6%29%280.0013%29%3D2418%20rad%2Fs)
3)
The tangential acceleration is related to the angular acceleration by the following formula:
![a_t = \alpha r](https://tex.z-dn.net/?f=a_t%20%3D%20%5Calpha%20r)
where
is the tangential acceleration
is the angular acceleration
r is the distance of the point from the centre of rotation
Here we want to find the tangential acceleration of the tip of the claw, so:
is the angular acceleration
r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation
Substituting,
![a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2](https://tex.z-dn.net/?f=a_t%3D%281.86%5Ccdot%2010%5E6%29%280.015%29%3D27900%20m%2Fs%5E2)
4)
Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:
![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
where
u is the initial linear speed
a is the tangential acceleration
t is the time elapsed
Here we have:
(tangential acceleration)
u = 0 m/s (it starts from rest)
t = 1.3 ms = 0.0013 s is the time elapsed
Substituting,
![v=0+(27900)(0.0013)=36.3 m/s](https://tex.z-dn.net/?f=v%3D0%2B%2827900%29%280.0013%29%3D36.3%20m%2Fs)