Someone please help: Circular Tangetial Motion (1 Viewer)

123ryoma12 · Oct 16, 2016

Why is the tangential accelration mgsin(theta)
Shouldn't it be
sin(theta) = mg/tangential acceleration
please help i'm going crazy

InteGrand · Oct 16, 2016

123ryoma12 said:
View attachment 33505
Why is the tangential accelration mgsin(theta)
Shouldn't it be
sin(theta) = mg/tangential acceleration
please help i'm going crazy

$\noindent The acute angle in the picture between the red weight vector and the tangent is $90^\circ - \theta$, since the angle between the radius and tangent is $90^\circ$ and using co-interior angles in parallel lines. So the component of the weight force in the tangential direction (pointing towards the left side of the tangential line on the picture) will be $mg\cos \left(90^\circ - \theta \right) = mg \sin \theta$. (This isn't the tangential acceleration by the way. That would just be $g\sin \theta$.)$

123ryoma12 · Oct 16, 2016

isnt that still mg/tangentialforce?
cos(90-theta) = adjacent/hypo = mg/tangetialforce
so tangential force = mg/sin(theta)
im sure im wrong but this is how im seeing it right now

InteGrand · Oct 16, 2016

123ryoma12 said:
isnt that still mg/tangentialforce?
cos(90-theta) = adjacent/hypo = mg/tangetialforce
so tangential force = mg/sin(theta)
im sure im wrong but this is how im seeing it right now

You flipped the two around. The "hypotenuse" is the mg and the tangential force is the "adjacent". Basically this (zoom in if it's a bit unclear initially): http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0011/pendulum_FBD.gif .

123ryoma12 · Oct 16, 2016

oh thanks a lot for this photo lol
I've been doing it the same method with banked tracks.

Someone please help: Circular Tangetial Motion (1 Viewer)

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