What is the Direction of the Friction Force on a Revolving Car?

[Saitama]

Homework Statement

A car is driven on a large revolving platform which rotates with constant angular velocity $\omega$. At t=0, a driver leaves the origin and follows a line painted radially outward on the platform with constant speed $v_0$. The total weight of car is W and the coefficient of friction between the car and stage is $\mu$.

a. Find the acceleration of the car as a function of time using polar coordinates. Draw a clear vector diagram showing the components of acceleration at some time t>0.

b. Find the time at which the car just starts to skid.

c. Find the direction of the friction force with respect to the instantaneous position vector $\textbf{r}$ just before the car starts to skid. Show your result on a clear diagram.

Homework Equations

The Attempt at a Solution

In polar coordinates, acceleration of car is
\textbf{a}=(\ddot{r}-r\dot{\theta})\hat{r}+(r\ddot{\theta}+2\dot{r}\dot{\theta})\hat{\theta}
Here, $\ddot{r}=0$, $\dot{\theta}=\omega$, $\ddot{\theta}=0$ and $\dot{r}=v_0$. Substituting,
\textbf{a}=-r\omega \, \hat{\textbf{r}}+2v_0\omega \, \hat{\textbf{θ}}
I haven't worked with polar coordinates before so I would like to know if my expression for acceleration vector is correct. Also, how do I draw the vector diagram here? I can show $\hat{\textbf{r}}$ points outward and $\hat{\textbf{θ}}$ is perpendicular to $\hat{\textbf{r}}$ but I don't think the question asks this.

Any help is appreciated. Thanks!

 

[tiny-tim]

Hi Pranav-Arora!

\textbf{a}=(\ddot{r}-r\dot{\theta})\hat{r}+(r\ddot{\theta}+2\dot{r}\dot{\theta})\hat{\theta}

nooo … count the dots!

… how do I draw the vector diagram here? I can show $\hat{\textbf{r}}$ points outward and $\hat{\textbf{θ}}$ is perpendicular to $\hat{\textbf{r}}$ but I don't think the question asks this.

yes that's ok …

to draw pr^ + qθ^, just draw a length p radially and a length q tangentailly

 

[Saitama]

Hi tiny-tim! :)

nooo … count the dots!

Ah yes, I copied down the formula incorrectly. It is $\dot{\theta}^2$ instead of $\dot{\theta}$. Sorry. (But I think the dots are fine. )

The acceleration vector is:
\textbf{a}=-r\omega^2 \, \hat{\textbf{r}}+2v_0\omega \, \hat{\textbf{θ}}

yes that's ok …

to draw pr^ + qθ^, just draw a length p radially and a length q tangentailly

I have attached a sketch, would that work? (The car is at a distance r from the origin)

 

[tiny-tim]

The acceleration vector is:
\textbf{a}=-r\omega^2 \, \hat{\textbf{r}}+2v_0\omega \, \hat{\textbf{θ}}

yes, that's correct …

the first term is the centripetal acceleration, and the second is the Coriolis force (divided by the mass)

I have attached a sketch, would that work?

looks fine!

 

[Saitama]

yes, that's correct …

the first term is the centripetal acceleration, and the second is the Coriolis force (divided by the mass) looks fine!

Thanks for the check tiny-tim! :)

For the b part, I think that as the Coriolis force stays constant during the motion, the car skids when friction is less than the centrifugal force (should I say centrifugal or centripetal? ) . Hence, at the moment it starts slipping, friction force=centrifugal force.
\mu W=r\omega^2
At time $t$, $r=v_0t$. Solving for t,
t=\frac{\mu W}{v_0\omega^2}
Am I doing it right?

 

[tiny-tim]

(should I say centrifugal or centripetal? )

if you're in the rotating frame, you can say centrifugal, because only in that frame does the centrifugal force exist

if you're in the stationary frame, say centripetal acceleration (not force)

… at the moment it starts slipping, friction force=centrifugal force.

no, the only external horizontal force on the car is the friction (in the stationary frame) …

and so that has to be supplying the whole acceleration (and must be in the same direction as it)

 

[haruspex]

if you're in the stationary frame, say centripetal acceleration (not force)

You can refer to centripetal force, provided you treat it as the resultant force required to achieve the centripetal acceleration, not as an actual input force.

 

[Saitama]

no, the only external horizontal force on the car is the friction (in the stationary frame) …

and so that has to be supplying the whole acceleration (and must be in the same direction as it)

Is that not what I have done in my previous post even though I stated it incorrectly?

 

[tiny-tim]

Is that not what I have done in my previous post even though I stated it incorrectly?

no, you said …

… at the moment it starts slipping, friction force=centrifugal force.

… making the friction supplying only the centrifugal force, and not also the coriolis force

 

[Saitama]

no, you said …… making the friction supplying only the centrifugal force, and not also the coriolis force

I should have worded my previous post properly. I did not mean that friction does not supply the Coriolis force. I meant that the Coriolis force i.e $2mv_0\omega$ stays constant. Force of friction acts in two direction, in the direction opposite to that of centrifugal force and in the direction opposite to that of centrifugal force, correct?

I assume that $2mv_0\omega$ is less than the maximum frictional force provided by the surface. Is this assumption wrong? I am completely confused on this. :(

($m$ is the mass of car)

 

[tiny-tim]

Hi Pranav-Arora!

Force of friction acts in two direction, in the direction opposite to that of centrifugal force and in the direction opposite to that of centrifugal force, correct?

that's a strange way of putting it

there is one force of friction, and it supplies the total acceleration

so it is the resultant of the coriolis force and the (mass times) centripetal acceleration

to avoid skidding, the total friction (ie, that resultant force) must be less than µN ​

 

[Saitama]

Hi Pranav-Arora! that's a strange way of putting it

there is one force of friction, and it supplies the total acceleration

so it is the resultant of the coriolis force and the (mass times) centripetal acceleration

to avoid skidding, the total friction (ie, that resultant force) must be less than µN ​

Okay, so something like this:
$$\sqrt{4m^2v_0^2\omega^2+m^2r^2\omega^4}\leq \mu W$$
$$\Rightarrow m\omega\sqrt{4v_0^2+r^2\omega^2} \leq \mu W$$

At the moment when the car starts to skid, $r=v_0t$
$$m\omega\sqrt{4v_0^2+v_0^2t^2\omega^2} = \mu W$$
Solving for t,
$$t=\frac{1}{\omega}\sqrt{\left(\frac{\mu W}{mv_0\omega}\right)^2-4}$$
Looks good?

 

[tiny-tim]

Hi Pranav-Arora!

At the moment when the car starts to skid, $r=v_0t$

r is always v₀t !

(and writing mg instead of W would be easier)

apart from that, looks fine ​

 

[Saitama]

r is always v₀t !

Yes, I placed that after the wrong sentence, sorry.

(and writing mg instead of W would be easier)

The question states W so I used that instead of mg.

I think I can do the c part now. (Sorry for being lazy, I can't draw any sketches at the moment as my mouse isn't working )

Thanks for all the help tiny-tim.

 

[Nick202]

Could you please show the C part?