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A rod AB with A at the origin rotates such that the angle between the rod...

A rod AB with A at the origin rotates such that the angle between the rod and the x?axis is theta = pi (8t

Solutions

Expert Solution

The collar can be located by a polar coordinate system, which consists of components (r, theta).

$r(t)=50t^2+20t^3$

$\theta (t)=\frac{\pi }{4}(8t-6t^2)$

The position function the gets to be:

$\vec{r}(t)=(50t^2+20t^3)*\vec{u}_r+(\frac{\pi }{4}(8t-6t^2))*\vec{u}_\theta$

Because the unit vectors depend on theta, and theta depends on time, their derivatives are not constants. If we derive the expression as is, we must take that into account. Another way we could use is to choose to convert the position vector to a cartesian coordinate system, as follows:

$\vec{r}(t)=rcos(\theta )*\vec{u}_x+rsin(\theta )*\vec{u}_y$

$\vec{r}(t)=(50t^2+20t^3)cos(\frac{\pi }{4}(8t-6t^2))*\vec{u}_x+(50t^2+20t^3)sin(\frac{\pi }{4}(8t-6t^2))*\vec{u}_y$

The velocity function comes from the derivative of the position function with respect to time, then:

$\vec{v}(t)=\vec{r'}(t)=[r'(t)cos(\theta (t))-r(t)sin(\theta (t))*\theta '(t)]*\vec{u}_x+[r'(t)sin(\theta (t))+r(t)cos(\theta (t))*\theta '(t)]*\vec{u}_y$

You must plug the corresponding functions into the expression above:

$r(t)=50t^2+20t^3$

$r'(t)=100t+60t^2$

$\theta (t)=\frac{\pi }{4}(8t-6t^2)$

$\theta '(t)=\frac{\pi }{4}(8-12t)$

To get the velocity as a function of time:

$\vec{v}(t)=\vec{r'}(t)=[(100t+60t^2)cos(\frac{\pi }{4}(8t-6t^2))-(50t^2+20t^3)sin(\frac{\pi }{4}(8t-6t^2))*(\frac{\pi }{4}(8-12t))]*\vec{u}_x+[(100t+60t^2)sin(\frac{\pi }{4}(8t-6t^2))+(50t^2+20t^3)cos(\frac{\pi }{4}(8t-6t^2))*(\frac{\pi }{4}(8-12t))]*\vec{u}_y$

The acceleration function comes from deriving the velocity function (take into account the use of the chain rule):

$\vec{a}(t)=\vec{v'}(t)=\vec{r''}(t)=[r''cos\theta -r(cos\theta *\theta '+\theta ''*sin\theta )]\vec{u}_x+[r''sin\theta +2r'cos\theta *\theta '+r(-sin\theta *\theta '+\theta ''*cos\theta )]\vec{u}_y$

You must plug the corresponding functions in the expression shown above, and you may get the acceleration with respect to time:

$r(t)=50t^2+20t^3$

$r'(t)=100t+60t^2$

$r''(t)=120t$

$\theta (t)=\frac{\pi }{4}(8t-6t^2)$

$\theta '(t)=\frac{\pi }{4}(8-12t)$

$\theta ''(t)=-3\pi$

Then we finally get the acceleration as follows:

$\vec{a}(t)=\vec{v'}(t)=\vec{r''}(t)=[120t*cos(\frac{\pi }{4}(8t-6t^2)) -(50t^2+20t^3)(cos(\frac{\pi }{4}(8t-6t^2)) *(\frac{\pi }{4}(8-12t))+(-3\pi)*sin(\frac{\pi }{4}(8t-6t^2)) )]\vec{u}_x+[120t*sin(\frac{\pi }{4}(8t-6t^2)) +2(100t+60t^2)cos(\frac{\pi }{4}(8t-6t^2)) *(\frac{\pi }{4}(8-12t))+(50t^2+20t^3)(-sin(\frac{\pi }{4}(8t-6t^2)) *(\frac{\pi }{4}(8-12t))+(-3\pi)*cos(\frac{\pi }{4}(8t-6t^2)) )]\vec{u}_y$

genius_generous answered 1 year ago

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