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Angular Velocity of a Car going around a curve

  • #1

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?
 

Answers and Replies

  • #2
BvU
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Hi, and :welcome: !

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 ?
 
  • #3
Hi, and :welcome: !

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, thats why I'm confused.
 
  • #4
BvU
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So how many radians per second if ##\pi/2## radians take 5 seconds (and you may asssume constant speed) ?
 
  • #5
The car is going around a curve so I assumed a constant change in velocity, or would it be constant?
If constant π /10
 
  • #6
BvU
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Direction of the velocity vector changes, but its magnitude (what the speedometer indicates) is constant.
 
  • #7
Direction of the velocity vector changes, but its magnitude (what the speedometer indicates) is constant.
Thank you!
 
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  • #8
BvU
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You're welcome !
 
  • #9
BvU
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So what is your conclusion ?
 
  • #10
BvU
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Ah, I missed the ##\pi/10## radians/s in post #5. Well done.

(Don't forget the units !)
 

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