# Circular motion, kinematics

1. Nov 13, 2008

### devanlevin

this is my 1st excercise with changing tangent acceleration with circular motion, would appreciate if anyone can tell me where im 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
$$\alpha$$=[-$$\frac{\Pi^{2}}{12}$$sin$$\frac{\Pi}{6}$$t]

time is measured in seconds,
at t=0s the body is at point A (at 12 o clock on a clock face) travelling 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 ($$\omega$$)
e) when will the body reach B (6 o clock position)
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a)
a=$$\alpha$$*r

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

v=\Pi*cos$$\frac{\Pi}{6}$$t
----------------------------------------------------
c)
$$\omega$$=$$\frac{v}{r}$$

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

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

Last edited by a moderator: Nov 13, 2008
2. Nov 13, 2008

### devanlevin

hope my terminology is correct, translated into english