- #1

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Mg*sin{α} - 1.5m*x(double dot)=0

I am trying to get velocity, and my first thought was to integrate with dt, but I didn't know how to. And I'm not even sure it's possible, anyways, thanks!

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- Thread starter Madtasmo
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- #1

- 3

- 0

Mg*sin{α} - 1.5m*x(double dot)=0

I am trying to get velocity, and my first thought was to integrate with dt, but I didn't know how to. And I'm not even sure it's possible, anyways, thanks!

- #2

osilmag

Gold Member

- 133

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##\int \sin(x) dx = -\cos(x)##

That is the general form for integrating ##\sin(x)##.

That is the general form for integrating ##\sin(x)##.

- #3

osilmag

Gold Member

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##\int d^2x = x dx##

- #4

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Turns out it's possible to do so;

X(dot)=2/3*gt*sin(α)

X(dot)=2/3*gt*sin(α)

- #5

- 321

- 68

Surely you’ve seen ##\ddot x = ## constant.I had gotten to this equation of motion:

Mg*sin{α} - 1.5m*x(double dot)=0

To find ##\dot x## you use ##\ddot x =\frac{d}{dt}\dot x## and separate.

Another common trick worth knowing is from the chain rule:

$$ \frac{d^2x}{dt^2} =\frac{d\dot x}{dt} =\frac{dx}{dt} \frac{d\dot x}{dx} =\dot x \frac{d\dot x}{dx} $$

For example, if you have Newton’s law in the form ##m\frac{d^2x}{dt^2}=F(x)## then the chain rule makes it separable, from which we get the (1-D) work energy theorem (for point masses).

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