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GR0177 #23
Problem
 GREPhysics.NET Official Solution Alternate Solutions
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Mechanics}Vectors

If a particle is moving in a circle, then its acceleration points towards the center---even if it is moving at a constant (magnitude) velocity. This is the normal component of acceleration.

Now, if its tangential velocity is increasing, then it has a tangential acceleration. The tangential acceleration is given as just .

The normal acceleration is given by the centripetal acceleration formula .

Vectorially add the normal and tangential acceleration and dot it with the tangential velocity to find that the angle between them is just 45 degrees.  Alternate Solutions
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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)
 yosun2005-11-11 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.      LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$