Answer:
1. [tex]a_{rad}=17545.2\frac{m}{s^{2}} [/tex]
2. [tex] a=4429.45 \frac{m}{s^{2}} [/tex]
Explanation:
Radial acceleration is:
[tex]a_{rad}=\frac{v^2}{r} [/tex] (1)
With r the radius respects the axis of rotation and v the tangential velocity that is related with angular velocity (ω) by:
[tex]v= \omega r[/tex] (2)
By (2) on (1):
[tex]a_{rad}= \frac{(\omega r)^2}{r}= (\omega )^2r=(418,88\frac{rad}{s})^2(0.1m)[/tex]
[tex]a_{rad}=17545.2\frac{m}{s^{2}} [/tex]
To find the acceleration of the tube with the fall, we can use the expression:
[tex]\overrightarrow{J}=\overrightarrow{F}_{avg}(\varDelta t) [/tex] (3)
Due impulse-momentum theorem:
[tex]\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i} [/tex] (4)
with p the momentum and J the impulse. By (4) on (3):
[tex]\overrightarrow{p}_{f}-\overrightarrow{p}_{i}=\overrightarrow{F}_{avg}(\varDelta t) [/tex]
And using Newton's second law (F=ma) and that (P=mv):
[tex]mv_f-mv_i=(ma)(\varDelta t) [/tex] (5)
Final velocity is the velocity just after the encounter with hard floor, and initial momentum us just before that moment so the first one is zero and the second one can be found sing conservation of energy:
[tex]\frac{mv_i}{2}=mgh [/tex]
[tex] v_i=\sqrt{2gh}=\sqrt{2(9.81)(1.0)}=4.43\frac{m}{s}[/tex]
So (5) is:
[tex]-m(4.43)=(ma)(\varDelta t) [/tex]
solving for a:
[tex]a=\frac{4.43}{\varDelta t}=\frac{4.43}{1.0\times10^{-3}}=-4429.45 [/tex]
It’s negative because is opposed to the tube movement.
The acceleration of the test tube when dropped from the given height is 2.0 x 10⁶ m/s² and the centripetal acceleration of the test is 175.5 m/s².
The given parameters;
The angular acceleration at the end of the test tube is calculated as follows;
[tex]a_c = \omega^2 r\\\\a_c = (4000 \ \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1 \ \min}{60 \ s} )^2 \times 0.1\\\\a_c = 175.5 \ m/s^2[/tex]
The linear acceleration of the test tube from the given height is calculated as follows;
h = ut + ¹/₂at²
1 = 0 + 0.5(1 x 10⁻³)²a
1 = 5 x 10⁻⁷a
[tex]a = \frac{1}{5\times 10^{-7}} \\\\a = 2.0 \times 10^6 \ m/s^2[/tex]
Thus, the acceleration of the test tube when dropped from the given height is 2.0 x 10⁶ m/s² and the centripetal acceleration of the test is 175.5 m/s².
Learn more here:https://brainly.com/question/11700262
