Angle of acceleration in non-uniform circular motion

AI Thread Summary
The discussion revolves around calculating the angle of acceleration in non-uniform circular motion, specifically addressing the confusion regarding the use of the tangent function. Participants clarify that the angle can be expressed using either ##\tan^{-1}(x)## or arctan(x), and they discuss the correct ratio of radial to tangential acceleration. The angle in question is between the radial acceleration and the total acceleration, with participants noting the use of the gamma symbol for degrees in the textbook. A diagram is referenced to illustrate the relationship between the accelerations and the angles involved, leading to a better understanding of the problem. The conversation concludes with a resolution of the initial confusion regarding the angle measurement.
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Homework Statement
Pls see below
Relevant Equations
Pls see below
For (c),
1676794421345.png

Solution is
1676794442239.png

Can someone please explain how they calculated that angle? I thought they would do ##arc\tan (\frac {32}{3.35})##

Many thanks!
 
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Callumnc1 said:
View attachment 322504
Can someone please explain how they calculated that angle? I thought they would do ##arc\tan (\frac {32}{3.35})##

Many thanks!
##\tan^{-1}## is one way of writing ##\arctan##.
Or is it the 3.35/32 instead of the other way up that bothers you?
 
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haruspex said:
##\tan^{-1}## is one way of writing ##\arctan##.
Or is it the 3.35/32 instead of the other way up that bothers you?
The OP seems to be unaware that the trig function inverses can be written in two ways: e.g., ##\tan^{-1}(x)## is the same as arctan(x), and similar for the other circular trig functions.

On another note, does anyone recognize the symbol that follows 5.98 in the solution? Evidently it means "degrees" since ##\tan^{-1}(\frac{3.35}{32.0}) \approx 5.98## (degrees), but the symbol used looks like ##\gamma## to me, which I've never seen used to signify degrees .
 
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haruspex said:
##\tan^{-1}## is one way of writing ##\arctan##.
Or is it the 3.35/32 instead of the other way up that bothers you?
Thank you for your reply @haruspex!

Sorry I should been more accurate - it is the 3.35/32 that is bothering me

Many thanks!
 
Mark44 said:
The OP seems to be unaware that the trig function inverses can be written in two ways: e.g., ##\tan^{-1}(x)## is the same as arctan(x), and similar for the other circular trig functions.

On another note, does anyone recognize the symbol that follows 5.98 in the solution? Evidently it means "degrees" since ##\tan^{-1}(\frac{3.35}{32.0}) \approx 5.98## (degrees), but the symbol used looks like ##\gamma## to me, which I've never seen used to signify degrees .
Thank you for your reply @Mark44!

Sorry I was actually aware that ##tan^{-1} = arctan##. However, I was not sure why they wrote 3.35\32 instead 32\3.35 inside the tan function.

Also yeah that textbook solutions dose use the gamma symbol for degrees for some reason.

Many thanks!
 
Callumnc1 said:
Thank you for your reply @haruspex!

Sorry I should been more accurate - it is the 3.35/32 that is bothering me

Many thanks!
The 32 acceleration is radial. The angle quoted is that between the radial acceleration and the total acceleration. The cosine of that angle would be radial/total and its tangent is tangential/radial.
 
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haruspex said:
The 32 acceleration is radial. The angle quoted is that between the radial acceleration and the total acceleration. The cosine of that angle would be radial/total and its tangent is tangential/radial.
Thank you for reply @haruspex !

When I draw a diagram of the accelerations acting on the mass when ##\theta = 20~degrees##
1676838074204.png

Then draw a vector addition diagram to get the total acceleration,
1676838284246.png

I see that ##\tan\theta = \frac{a_c}{a_t} = \frac{32}{3.35}##

I know this is bad practice to change symbols when solving a problem, but if I add an angle phi (use to represent tangent/radial acceleration) ,
1676838541321.png

Then I can see that ##\tan\phi = \frac{a_t}{a_c}##. I think I'm now not sure how they got acceleration to be below the cord at 5.89 degrees.
1676838800945.png

Many thanks!
 
Callumnc1 said:
View attachment 322528
Many thanks!
The angle of a line "below the cord" means the angle between the cord and the line, measured on the underside of the cord. In your last diagram above, that is the angle between the pale brown cord and the black total acceleration vector, measured anticlockwise from the cord.
 
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haruspex said:
The angle of a line "below the cord" means the angle between the cord and the line, measured on the underside of the cord. In your last diagram above, that is the angle between the pale brown cord and the black total acceleration vector, measured anticlockwise from the cord.
Oh thank you for your help @haruspex! I see it now :)
1676846089139.png
 

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