- #1
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
\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) 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 (\omega)
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=\alpha*r
a=[-\frac{\Pi^{2}}{6}sin\frac{\Pi}{6}t]
--------------------------------------------------
b)
v=\intadt=\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 realize 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
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) 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 (\omega)
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=\alpha*r
a=[-\frac{\Pi^{2}}{6}sin\frac{\Pi}{6}t]
--------------------------------------------------
b)
v=\intadt=\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 realize 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
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