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Circular MotionA race car starts from rest on a circular trackof radius 430 m. Its...

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jennyjoo123 | Student | (Level 3) eNoter

Posted November 20, 2010 at 10:24 AM via web

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Circular Motion

A race car starts from rest on a circular track
of radius 430 m. Its speed increases at the
constant rate of 0.3 m/s2. At the point where
the magnitudes of the radial and tangential
accelerations are equal, determine the speed
of the race car.
Answer in units of m/s.

 

Is it V^2/r.

 

1 Answer | Add Yours

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krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted November 20, 2010 at 11:49 AM (Answer #1)

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The radial acceleration of the car moving in a circular path is given by:

Radial acceleration = (v^2)/r

Where:

v = tangential speed (velocity)

r = radius of circular path = 430 m (Given)

It is given:

Tangential acceleration = 0.3 m/s^2

When:

Radial acceleration = Tangential acceleration:

==> (v^2)/r = 0.3

==> (v^2)/430 = 0.3

==> v^2 = 0.3*430 = 129

Therefore:

v = 129^(1/2) = 11.3578 m/s

Answer:

Speed of the car = 11.3578 m/s

(Here you can see that the expression V^2/r given in the question refers to the radial acceleration, although the way I have expressed it is slightly different.)

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