GR0177 #23
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Comments |
secretempire1 2012-08-28 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/s². 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 90-45 = 45. | | f4hy 2009-04-02 15:05:26 | Looks like there is a typo. It should read . You squared the velocity but left the squared symbol. | | senatez 2006-11-02 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 2005-11-11 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 2005-11-11 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 typo-alert; it's been corrected.
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