# Nonuniform circular motion

1. Sep 30, 2008

### subwaybusker

If an object is accelerating tangentially, then how would i go about calculating the radial acceleration since the speed is changing?

2. Sep 30, 2008

### Hootenanny

Staff Emeritus
Obviously the radial acceleration will be a function of time. If the tangential acceleration is constant then you can simply apply the kinematics equations for rotation (which are analagous to the linear equation). If the tangential acceleration is not constant, then the solution is somewhat more complex. Perhaps if you posted the problem in question we could help you out a little more.

3. Sep 30, 2008

### subwaybusker

Bruce is on a bike. He wants to change he speed from 25 to 30km/hr on a curved road, but for safety, the magnitude of his acceleration must not exceed 0.2g. If the radius of the curved road is 5km. what is the minimum time Bruce can change his speed?

Vi=25km/h, so $\omega$i=50 rad/s
Vf=30km/h, so $\omega$f=60 rad/s

|a|^2=|R$\alpha[/tex]|^2+|R[itex]\omega$|^2
(0.2g)^2=|R$\alpha[/tex]|^2+|R[itex]\omega$|^2

a(tang)=dv/dt=(300-250km/h)/t
$\alpha$=(60-50rad/h)/t

then i don't know where to go on to.

4. Sep 30, 2008

anyone?

5. Oct 1, 2008

### tiny-tim

Hi subwaybusker!

(Hootenanny is out for the evening)

(have an omega: ω and a squared: ² )

You don't need ω … centripetal acceleration, ω²r, is also v²/r.

Hint: since v²/r is changing, the maximum dv/dt will also change.

6. Oct 1, 2008

### subwaybusker

okay, i get since that the bike is increasing it's speed, radial ac will increase, but why is the tangential acc changing too? can't dv/dt be a constant? i'm thinking that the tangential is dv(speed)/dt and that you mean the maximum dvelocity/dt which is the total acceleration is changing (correct me if i'm wrong). also, if the speed is always increasing, what do i do with the v²/r since v is never constant?

Last edited: Oct 1, 2008
7. Oct 1, 2008

### tiny-tim

Bruce wants the shortest time, which means the greatest speed, but his acceleration is limited, which means that if the radial acceleration is increasing, then the tangential acceleration must decrease, so that he stays within the safe limit.

8. Oct 1, 2008

### subwaybusker

how can tangential acceleration decrease when it's increasing it's speed from 25 to 30km/h?

9. Oct 1, 2008

### tiny-tim

'cos the acceleration is decreasing, but is still positive.

10. Oct 1, 2008

### subwaybusker

does that mean i'm only find dv/dt and radial acceleration at an instant?

11. Oct 1, 2008

### tiny-tim

At each instant, yes.

12. Oct 1, 2008

### subwaybusker

does that mean i can find the radial acceleration at 30km/hr, use that to find dv/dt and repeat that for 25km/hr and find dv/dt at that point? but somehow i have to relate this to minimum time?

wait, that doesn't sound right. do i have to use some sort of calculus? sorry, but i really don't know how to approach this?

Last edited: Oct 1, 2008
13. Oct 2, 2008

### tiny-tim

Hi subwaybusker!

Yes, you've got the basic idea …

but, as I think you've spotted, you have to do it not only for 25 and 30 , but also for every speed in between :tongue2: …

and then integrate

(and before you ask, no, I'm sorry, there isn't any shorter way! )

have a go!