A child slides down the helical water slide AB (Polar Coordinates)

[Alexanddros81]

Homework Statement

13.43 A child slides down the helical water slide AB. The description of motion in cylindrical coordinates is
$R=4m$, $θ=ω^2t^2$ and $z=h[1-(\frac {ω^2t^2} {π})]$, where h=3m and ω=0.75rad/s.
Compute the magnitudes of the velocity vector and acceleration vector when the child is at B.

Homework Equations

The Attempt at a Solution

Is this correct?

 

[NFuller]

Yes, this looks right. You just need to find the value of $t$ at point $B$ to get a number.

 

[Alexanddros81]

Yes, this looks right. You just need to find the value of ttt at point BBB to get a number.

Is this possible to find t with the given data? or is it ok to leave it as it is?

 

[NFuller]

Is this possible to find t with the given data? or is it ok to leave it as it is?

Yes, this is possible since you know the change in $z$.

 

[TSny]

Looks like a maybe a slip when substituting for $v_{\theta}$.

It's often a good idea to work things out in symbols and then plug in numbers at the very end.

 

[Alexanddros81]

Looks like a maybe a slip when substituting for vθ

After correction v becomes 4.62t

Yes, this is possible since you know the change in zzz.

At first I couldn't figure this out but then I realized that the child is sliding down the slide. This means that if we accept as the z positive direction
is upwards then the formula becomes:

$z=h[1- (\frac{ω^2t^2} {π})] => -3=3-0537t^2 => t=3.34s$
So $v=4.62t=15.43 m/s$
and $a=56.66 m/s^2$

Am I correct?

 

[NFuller]

I think so, so long as there aren't any other arithmetic errors I missed.

 

[TSny]

$z=h[1- (\frac{ω^2t^2} {π})] => -3=3-0537t^2 => t=3.34s$

Is the final value of z equal to -3 m?