Mechanics: Angular Velocity Vector Questions

[girlwhoneedsmathhelp]

Homework Statement

The question :

The position vector of a rider on a helter-skelter is given by : r = (2sin t)i + (2cos t)j + (8-1/2 t)k where the units are in metres and seconds. The unit vector k acts vertically upwards.

(I) Find an expression for the velocity of the rider at time t

(ii) Find the speed of the rider at time t

(iii) Find the magnitude and direction of the rider's acceleration when t = pi/4

From the first part I am very confused on how to figure this out. I know that this question has to relate with the formula for v =wr, but I am unsure how to approach it. Even after looking at the answers for I, I still do not know how to do ii and even less for iii.

Relevant Equations

v=wr

Answers are the following :
(i) v=(2cost)i - (2sint)j -(1/2)k
(ii)2.06m/s
(iii)2m/s^2 horizontally towards the vertical axis, making an angle of pi/4 with both the I and j axes.

 

[etotheipi]

You don't need any rotational formulae. You have$$\vec{r} = 2\sin{(t)} \hat{i} + 2\cos{(t)} \hat{j} + (8-\frac{1}{2}t)\hat{k}$$The velocity is $\vec{v} = \frac{d\vec{r}}{dt}$, i.e. the time derivative of the above. Does that help you get started?

 

[Adesh]

Okay, let us know why you think we have to use $v= \omega r$ formula, it's always better to talk. We from basic mechanics know two things:

1. Derivative of position is the velocity.
2. In a circular motion linear velocity, radius and angular velocity are related by $v =\omega r$.

Let me know what are your thoughts on them and their relation to your original question.

 

[etotheipi]

@Adesh It is a good comment but it is heading slightly off topic. I will answer it for the OP: $\vec{v} = \vec{\omega} \times \vec{r}$ pertains to a circular velocity field about a point, e.g. the velocity of a point on a rigid body w.r.t. another point on the rigid body, or the velocity of a particle undergoing circular motion. To digress, it's not totally irrelevant here because we are dealing with helical motion, so we could perform a Galilean boost to a frame where $v_z = 0$ and we would end up with circular motion with $\omega = 1$.

But this question is almost certainly just an exercise in calculus. So let us please first hear back from the OP.

 

[Adesh]

$\vec{v}=\vec{ω}\times \vec{r}$ pertains to a circular velocity field about a point, i.e. the velocity of a point on a rigid body w.r.t. another point on the rigid body, or the velocity of a particle undergoing circular motion.

As the question is about a single body which is considered as a point body so we can just use the formula $v=\omega r$ for any instantaneous circular path.

To digress, it's not totally irrelevant here because we are dealing with helical motion, so we could perform a Galilean boost to a frame where $v_z=0$ and we would end up with circular motion with ω=1.

But how would you get $\omega$? I know we will get a helix of radius $2$ (I mean projection of helix on the $xy$ plane will result in a circle of radius 2), but to get $\omega$ you have to have $v$ which you will get only by differentiating the the expression for position.

 

[etotheipi]

As the question is about a single body which is considered as a point body so we can just use the formula $v=\omega r$ for any instantaneous circular path.

You can do something like this, but it is yet another added degree of complexity to what is really a very simple problem. What you are suggesting is to use an intrinsic coordinate system with basis $\{ \vec{e}_n, \vec{e}_t, \vec{e}_b \}$, and you are using the fact that in intrinsic coordinates$$\vec{v} = v\vec{e}_t$$and$$\vec{a} = \dot{v} \vec{e}_t + \frac{v^2}{\rho} \vec{e}_n$$and it may well be convenient to define a parameter $\omega$, given the cyclical nature of the helical motion, s.t. $v =\rho \omega$, if $\rho$ is the instantaneous radius of curvature (which you also need to go and work out from the equation of the helix!). But I do not suggest this for the OP, it is a billion times harder than necessary.

But how would you get $\omega$? I know we will get a helix of radius $2$ (I mean projection of helix on the $xy$ plane will result in a circle of radius 2), but to get $\omega$ you have to have $v$ which you will get only by differentiating the the expression for position.

If you performed the Galilean boost, $\omega = 1$. That will allow you to find the $x$ and $y$ components of the velocity in the original frame, if you so wished.

But again, added complexity to just $\frac{d}{dt}$ing the original expression. I think it is best now that the OP replies to post #2.

 

[girlwhoneedsmathhelp]

You don't need any rotational formulae. You have$$\vec{r} = 2\sin{(t)} \hat{i} + 2\cos{(t)} \hat{j} + (8-\frac{1}{2}t)\hat{k}$$The velocity is $\vec{v} = \frac{d\vec{r}}{dt}$, i.e. the time derivative of the above. Does that help you get started?

Yes, thanks a lot!