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Acceleration of a curved trajectory 
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#1
Nov1212, 05:46 AM

P: 8

Feynman_Lectures_on_Physics_Volume_1_Chapter_11
In paragraph 116 he says that the tangent acceleration is the change of speed v but if I look at fig. 118 the change in speed is slightly smaller than the change in tangent velocity. (I drew a circle with the radius of the speed of v_I that has it's middle in the origin of the vectors mentally.) I assume he just neglects that difference but I don't really understand the idea of neglecting certain things. What things can you neglect and what not? He also says that the acceleration at right angles to the curve is the magnitude of the velocity times the change in angle. The magnitude of which velocity? v_I or v_II? How can you just multiply speed with an angle? I know you can if you use sine, cosine or something like that but multiply it with an angle directly? 


#2
Nov1212, 08:25 AM

Mentor
P: 11,900

You can neglect things which go to 0 quicker than your relevant effect, if the timestep goes to 0. In other words, everything where effectsize/timestep goes to 0 as timestep goes to 0.
In general, your post looks confusing to me, therefore the very general answer. 


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