GREPhysics.NET: GR0177 #23

GR0177 #23

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

Mechanics $\Rightarrow$ }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 $a_t=10 m/s^2$ .

The normal acceleration is given by the centripetal acceleration formula $a_n=v^2/r = 100^2/10=10$ .

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.

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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 $\frac{v^2}{r} = \frac{10^2}{10}=10 \frac{m}{s^2}$ , 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.

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f4hy
2009-04-02 15:05:26

Looks like there is a typo. It should read $\frac{v^2}{r}=\frac{10^2}{10} = 10$ . You squared the velocity but left the squared symbol.

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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.

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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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Looks like there is a typo. It should read $\frac{v^2}{r}=\frac{10^2}{10} = 10$ . You squared the velocity but left the squared symbol.

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