Find the time t at which the angle between two vectors is 45

[chris_0101]

Homework Statement

Suppose a car travels around a circular track of radius 100m. Suppose the speed of the car varies in time as v = (2m/s)t. Find the time t at which the angle between the velocity vector and the acceleration vector is 45 degrees

Homework Equations

Position vector = \vec{r} = \hat{i}bsin(ωt) + \hat{j}bcos(ωt)

The Attempt at a Solution

With the position vector, I found the velocity vector, acceleration vector and their corresponding magnitudes and they are shown below:

\vec{v} = \hat{i}bωcos(ωt) - \hat{j}bωsin(ωt)
|v| = bω

\vec{a} = -\hat{i}bω²sin(ωt) - \hat{j}bω²sin(ωt)
|a| = bω²

v\bulleta = |v||a|cosθ

And basically that is where I am stuck. I do not know how to implement the acceleration/velocity vectors and their magnitudes into the dot product equation since the dot product of v and a will equal 0.

If anyone can help me out with this one, that would be great. Thanks in advance.

 

[Simon Bridge]

Use the velocity vector as your baseline and do everything in it's frame of reference.
v points tangentially to the track.
There are two acceleration vectors - the tangential acceleration a which points along v and the centripetal acceleration a_c which points perpendicularly to v

... the net acceleration will be a_tot=a+a_c ...

So the tangent of the angle that a_tot makes with v is given by what?
When the angle is 45deg, what is the tangent?

 

[chris_0101]

isn't the angle that a_tot makes with v, simply 45 degrees? which is a given value

 

[Simon Bridge]

The angle depends on the speed which depends on time ... you want to know what time gets you 45degrees. What is the formula for centripetal acceleration?

 

[chris_0101]

the formula for centripetal acceleration is a_c = ((v_t)^2)/r

The centripetal acceleration is equal to:
a_c = ((2m/s)^2)/100m
a_c = 0.04 m/s^2

 

[Simon Bridge]

Is the tangential velocity a constant with time then?
I thought you said it was v_t = (2.0m.s^{-2})t ?

 

[chris_0101]

yes, sorry. my mistake. I am extremely confused and lost with this question. So the tangential velocity is the magnitude of the velocity vector, so it would actually be:

a_c = (b(omega))^2 / r

 

[Simon Bridge]

You seem to be getting yourself further and further tangled up for some reason.

Lets have a recap:

in post #1 when you said the speed was varying as: v(t)=(2.0)t

isn't this the equation of the tangential speed? - yes?

 

[chris_0101]

I believe so, yes.

 

[Simon Bridge]

Cool - so if v(t)=(2.0)t, what is the magnitude of the tangential acceleration?

 

[chris_0101]

magnitude of tangential acceleration is simply the derivative of the tangential velocity, which is 2m/s^2

 

[Simon Bridge]

Good - now notice that the tangential acceleration does not change with time - this is important. Also, the tangential acceleration always points the same way as the tangential velocity[1].

The magnitude of the centripetal acceleration was correctly given by you earlier as:

a_c = \frac{v^2}{r}

That v in the numerator is the tangential speed.
Substitute the expression for tangential speed into the above to get the centripetal acceleration as a function of time.

-------------------------------------
[1] ... so long as you remember that slowing down is a negative acceleration

 

[chris_0101]

So I would get:

a_c = ((2t)^2)/R

So I have both tangential acceleration and centripetal acceleration. So I can proceed by substituting it into the equation:

tan(theta) = a_c/a_t

And I think I got it from this point now. a_t and v_t are pointing in the same direction so the above equation will provide the angle between acceleration and velocity.

Thanks, I cannot believe I over analyzed this question. You have helped me a lot, thanks for everything.

 

[Simon Bridge]

Well done.

The real cute bit is that once you have noticed that the two accelerations are the short-sides of a right angled triangle, you just ask yourself what sort of triangle has 45 degree angles... no need for trig.

If that confuses you - the tangent of 45 degrees is 1.BTW: if you must use coordinates - pick then to exploit the symmetry of the situation ... for circular motion, pick polar coordinates. Just sayin.

Happy hacking.