GR0177 #23



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Comments 
secretempire1 20120828 08:07:17  It's rather easy to visualize this problem.
You have a particle moving tangentially to the orbit at 10m/s.
Simply being constrained to this orbit gives it a centripetal acceleration equal to , that points inward radially.
The problem also states that the particle's speed is increasing (tangentially) at 10 m/sē. This gives us two acceleration vectors, one pointing radially inward with a magnitude of 10, and one tangentially (making each vector at 90 degrees to each other) with a magnitude of 10. Two equal vectors at 90 degrees to each other yields a vector at 45 degrees. Therefore, the angle between the tangent and the 45 degree acceleration vector is just 9045 = 45.   f4hy 20090402 15:05:26  Looks like there is a typo. It should read . You squared the velocity but left the squared symbol.   senatez 20061102 12:36:28  The accellaration vector will be a vector with equal magnitude along the tangent and normal direction, so it will be along the 45 degree angle. I dont think a dot product is necessary to "see" the angle.   tachyon 20051111 08:56:30  I think you meant to say "Vectorially add the normal and tangential ACCELERATION", not the velocity.
I really appreciate the website, since I'm taking the test tomorrow!!(if I hadn't said so in previous posts)
yosun 20051111 14:17:59 
tachyon: yes, that's what i meant; i was conjuring up solutions on the fly, typing as fast as i could, from the illegible stuff i jotted down when i took the test for practice. thanks for the typoalert; it's been corrected.

 

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