What is the mistake in finding circular motion acceleration?

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SUMMARY

The discussion focuses on the calculation of tangential acceleration, actual velocity, angular velocity, and the timing of a body in horizontal circular motion with an angular acceleration defined as α = [-π²/12 * sin(π/6 * t)]. The body, connected to a 2m rod, starts at point A (12 o'clock) and stops at t=3s. The calculations for tangential acceleration yield a = [-π²/6 * sin(π/6 * t)], while the velocity equation results in v(t) = π * cos(π/6 * t). The user concludes that the body will return to point A every 6 seconds but questions whether it can reach point B (6 o'clock), ultimately determining that it cannot due to the limits of the sine function.

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devanlevin
this is my 1st excercise with changing tangent acceleration with circular motion, would appreciate if anyone can tell me where I am going wrong and what to change

a small body connected to a rod with a length of 2m, r=2m, is moving in a horizontal circular motion, with an angular acceleration of
[tex]\alpha[/tex]=[-[tex]\frac{\Pi^{2}}{12}[/tex]sin[tex]\frac{\Pi}{6}[/tex]t]

time is measured in seconds,
at t=0s the body is at point A (at 12 o clock on a clock face) traveling anti clockwise, and at t=3s it is known v=0(the body stops).
find
a) tangent acceleration (a)
b) actual velovity (v)
c) angular velocity ([tex]\omega[/tex])
d) when will the body return to A(12 o clock position)
e) when will the body reach B (6 o clock position)
---------------------------------------------------
a)
a=[tex]\alpha[/tex]*r

a=[-[tex]\frac{\Pi^{2}}{6}[/tex]sin[tex]\frac{\Pi}{6}[/tex]t]
--------------------------------------------------
b)
v=[tex]\int[/tex]adt=[tex]\int[/tex]=[-[tex]\frac{\Pi^{2}}{6}[/tex]sin[tex]\frac{\Pi}{6}[/tex]t]dt=cos[tex]\frac{\Pi}{6}[/tex]t + C
v(t=3)=0=cos[tex]\frac{\Pi}{2}[/tex] + C ===>C=0

v=\Pi*cos[tex]\frac{\Pi}{6}[/tex]t
----------------------------------------------------
c)
[tex]\omega[/tex]=[tex]\frac{v}{r}[/tex]

[tex]\omega[/tex]=[tex]\frac{\Pi}{2}[/tex]cos[tex]\frac{\Pi}{6}[/tex]t
----------------------------------------------------
d)
now i get stuck, i realize that what i need to to is find the time, t, that the angle of the body relative to point A [tex]\vartheta[/tex], is equal to 0 or 2[tex]\Pi[/tex],

thetha=integral(omega)dt=3sin(pi*t/6)

sin(pi*t/6)=0
pi*t/6=pi*K
t=6K

does this mean that it will rach point A every 6 seconds??
-----------------------------------------------------
e)
Point B
similarily with B but comparing thetha with "pi" 180 degrees, but because of the limits of thetha, because it is limites by sinus' limits, it cannot pass 3, therefore cannot reach pi


have i done something wrong, or is this true and the body will never reach B, ie it will start at A, reach C, at 9 o clock, after 3 seconds, stop, return to A, reach D, at 3 o clock, stop, A... in a half circular motion
 
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hope my terminology is correct, translated into english
 

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