MHB Differentiation equation: equation of motion

[markosheehan]

a particle of mass m is projected towards a point O with initial speed √5/3 m/s from a point P where [OP]=3 metres. the particle is repelled from O by a force of magnitude 4m/x³ where x is its distance from O.
(i) show that the equation of motion is v dv=4x^-3 dx
(ii) find how close the particle will get to O
(iii) find its speed when it has traveled half the distance from P to this nearest point
i do not know even how to start this question off could someone show me the step to step guide to answer the qustion

 

[MarkFL]

Re: differentiation equation

First, I would orient an axis of motion with units of meters and with the origin at $O$, and put the particle initially at $x_0=3$, with $$v_0=-\frac{\sqrt{5}}{3}\,\frac{\text{m}}{\text{s}}$$.

Now, according to Newton's 2nd law of motion in 1 dimension, we have acceleration $a$ given by:

$$a=\frac{F}{m}$$

We also know acceleration is the time rate of change of velocity:

$$a=\frac{dv}{dt}$$

And that velocity is the time rate of change of position:

$$v=\frac{dx}{dt}$$

Can you now use these definitions/laws to obtain:

$$v\,dv=4x^{-3}\,dx$$?

 

[markosheehan]

Re: differentiation equation

thanks for helping and i understand everything you say up to the last line how do you get v dv=4x^-3 dx. what does equation of motion mean?

 

[MarkFL]

Re: differentiation equation

thanks for helping and i understand everything you say up to the last line how do you get v dv=4x^-3 dx. what does equation of motion mean?

We know:

$$a=\d{v}{t}=\frac{F}{m}$$

We are given:

$$F=4mx^{-3}$$

and so we obtain:

$$\d{v}{t}=4x^{-3}$$

Can you take the relationship between velocity and position to complete the equation?

What you wind up with is an equation relating the particle's velocity and position, and so it is called an "equation of motion." This equation is an ODE, and since you have initial conditions to go along with it, you have an IVP which can be completely solved to determine the particle's velocity as a function of its position.

 

[markosheehan]

Ok I think I get it thanks

 

[MarkFL]

Okay, now to answer part ii), I recommend switching the dummy variables of integration and using the boundaries as the limits of integration as follows:

$$\int_{v_0}^{v(x)} u\,du=4\int_{x_0}^{x} w^{-3}\,dw$$

Applying the FTOC, what do you get?

Once you have $v(x)$, what do we know about $v$ at the moment when the particle has reached it's nearest point to $O$?

 

[markosheehan]

what does u and w stand for why did u use them at all?. and once i keep going from v dv=4x^-3 dx i get v²/2= x^-2/-2 and then when i apply definte integrals to this i get v²/2-0²/2=x^-2/-2 -0^-2/-2 and this equals v²/2=x^-2/-2 and its final speed is 0 so 0²/2=x^-2/-2 and i can't get passed this i must of done something wrong by the way the answer to part (ii) is 2 metres

 

[markosheehan]

sorry i made a mistake when i was integrating i forgot that i know the initial speed √5/3 this is now the answer √5/3 ²/2 -0²/2=x^-2/-2 -0^-2/-2 however if i go on from this i do not get the right answer do you know what to do?

 

[MarkFL]

When I carry out the integration, I get:

$$v^2(x)=\frac{4}{x_0^2}-\frac{4}{x^2}+v_0^2$$

The variables of integration are "dummy variables" since they get integrated out, and so I replaced them in order to not have the same variables in the integral as I had in the limits.

So, letting $v(x)=0$ and solving for $x$, we obtain:

$$x_{\min}=\frac{2x_0}{\sqrt{4+x_0^2v_0^2}}$$

Plugging in the given values:

$$x_{\min}=\frac{2(3)}{\sqrt{4+(3)^2\left(\dfrac{\sqrt{5}}{3}\right)^2}}=\frac{6}{\sqrt{9}}=2$$