Angular Velocity of a Car going around a curve

[RobGoodall]

Homework Statement:

A car taking going through a curve of radius 60.0 meters that turns the car through a horizontal ground angle of 90 degrees, if the car goes through the 90 degree curve in a time of 5 seconds, what is the car's Angular Velocity around the curve in radians per second?

Relevant Equations:

ω=dθ/dt

θ=90°= π /2 so the instantaneous angular velocity dθ/dt= lim_{∆ t -> 0} (θ(t + ∆ t)-θ(t))/(∆ t)

When I calculate it out it is π /2 radians per second. Is this correct?

 

[BvU]

Hi, and !

When a 90 degrees turn is done in 5 seconds the $\omega$ can not be $\pi/2$ per second.

How many degrees per second is that ?

 

[RobGoodall]

Hi, and !

When a 90 degrees turn is done in 5 seconds the $\omega$ can not be $\pi/2$ per second.

How many degrees per second is that ?

90, that's why I'm confused.

 

[BvU]

So how many radians per second if $\pi/2$ radians take 5 seconds (and you may asssume constant speed) ?

 

[RobGoodall]

The car is going around a curve so I assumed a constant change in velocity, or would it be constant?
If constant π /10

 

[BvU]

Direction of the velocity vector changes, but its magnitude (what the speedometer indicates) is constant.

 

[RobGoodall]

Direction of the velocity vector changes, but its magnitude (what the speedometer indicates) is constant.

Thank you!

 

[BvU]

You're welcome !

 

[BvU]

So what is your conclusion ?

 

[BvU]

Ah, I missed the $\pi/10$ radians/s in post #5. Well done.

(Don't forget the units !)