1) [tex]1.86\cdot 10^6 rad/s^2[/tex]
2) 2418 rad/s
3) [tex]27000 m/s^2[/tex]
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:
[tex]\theta=\omega_i t +\frac{1}{2}\alpha t^2[/tex]
where:
[tex]\theta[/tex] is the angular displacement
[tex]\omega_i[/tex] is the initial angular velocity
t is the time
[tex]\alpha[/tex] is the angular acceleration
In this problem we have:
[tex]\theta=90^{\circ} = \frac{\pi}{2}rad[/tex] is the angular displacement
t = 1.3 ms = 0.0013 s is the time elapsed
[tex]\omega_i = 0[/tex] is the initial angular velocity
Solving for [tex]\alpha[/tex], we find:
[tex]\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2[/tex]
2)
For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:
[tex]\omega_f = \omega_i + \alpha t[/tex]
where
[tex]\omega_i[/tex] is the initial angular velocity
t is the time
[tex]\alpha[/tex] is the angular acceleration
In this problem we have:
t = 1.3 ms = 0.0013 s is the time elapsed
[tex]\omega_i = 0[/tex] is the initial angular velocity
[tex]\alpha = 1.86\cdot 10^6 rad/s[/tex] is the angular acceleration
Therefore, the final angular speed is:
[tex]\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s[/tex]
3)
The tangential acceleration is related to the angular acceleration by the following formula:
[tex]a_t = \alpha r[/tex]
where
[tex]a_t[/tex] is the tangential acceleration
[tex]\alpha[/tex] 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:
[tex]\alpha = 1.86\cdot 10^6 rad/s[/tex] 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,
[tex]a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2[/tex]
4)
Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:
[tex]v=u+at[/tex]
where
u is the initial linear speed
a is the tangential acceleration
t is the time elapsed
Here we have:
[tex]a=27900 m/s^2[/tex] (tangential acceleration)
u = 0 m/s (it starts from rest)
t = 1.3 ms = 0.0013 s is the time elapsed
Substituting,
[tex]v=0+(27900)(0.0013)=36.3 m/s[/tex]
