Rotational bodies

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November 12th 2009, 01:24 PM
jddery

Rotational bodies

A car turns a corner with a radius of curvature of 18.2 m while braking to reduce its speed. If the brakes generate an angular deceleration of 0.7 rad/s2 what is the magnitude of the acceleration of the car half way through the corner when the car's linear speed is 4.2 m/s?

[Specify units of 'm/s2' in your answer.]

http://moodle.phys.ualberta.ca/file....n/car-bend.jpg
November 12th 2009, 03:45 PM
skeeter

Quote:

Originally Posted by jddery

A car turns a corner with a radius of curvature of 18.2 m while braking to reduce its speed. If the brakes generate an angular deceleration of 0.7 rad/s2 what is the magnitude of the acceleration of the car half way through the corner when the car's linear speed is 4.2 m/s?

[Specify units of 'm/s2' in your answer.]

http://moodle.phys.ualberta.ca/file....n/car-bend.jpg

$|a| = \sqrt{a_T^2 + a_c^2}$

where tangential acceleration is $a_T = r\alpha$

and centripetal acceleration is $a_c = \frac{v^2}{r}$
November 12th 2009, 05:23 PM
jddery

hey thanks that worked

now how did you come up with that equation? is that a defined equation in a text book?
or did you simply know that adding the tangential acceleration with the centripetal acceleration by means of Pythagorus would result in acceleration?
November 12th 2009, 08:18 PM
mathaddict

Quote:

Originally Posted by jddery

hey thanks that worked

now how did you come up with that equation? is that a defined equation in a text book?
or did you simply know that adding the tangential acceleration with the centripetal acceleration by means of Pythagorus would result in acceleration?

THe car experiences 2 types of acceleration ie tangential a and centripetal a (towards the centre) . So its resultant acceleration would be the vector sum of the 2 accelerations . Phythagoras would give you the result since the 2 accelrations are 90 to each other .