Angle between normal force and radial line for cylindrical coordinates

• simphys
In summary: $$becuase ## \dot r =...$$$$\frac{dr}{d \theta} = -r\frac{d\theta}{d r}$$ becuase theta is the angle between the radial line and the normal force and r is the distance from the center of the radial line to the point of interest.
simphys
Homework Statement
The 0.75-kg smooth can is guided along the circular
path using the arm guide. If the arm has an angular velocity ##\theta^. = 2rad/s## and an angular acceleartion of 0.4 rad/s^2 at the instant ##\theta = 30degrees##, determine the force of the guide on the can. Motion occurs in the horizontal plane.
Relevant Equations
cylindrical stuff

so I was wondering. there is this normal force on the can from the path. And there's this formula to find the angle between the radial line and the tangent or also between the normal force and either the radial or theta axis. the formula is ##\psi = r/dr/d\theta##. The thing is that here they have just used ##\theta = 30degrees## for some reason and I can't quite see why they use 30 degrees as being the angle between the axis and the normal force instead of the psi angle. Could someone clarify this please?

for more clarity:

here, 30 degrees is used, but why is that?

simphys said:
for more clarity:
View attachment 319772
here, 30 degrees is used, but why is that?
because the net force you have labeled as ##N## (without friction) always acts toward the paths center.

simphys
erobz said:
because the net force you have labeled as ##N## (without friction) always acts toward the paths center.
what do you mean? the path's center is not in the pin of the arm guide right? and the force F is the force exerted by the arm guide on the can/ball or whatever.

simphys said:
what do you mean? the path's center is not in the pin of the arm guide right? and the force F is the force exerted by the arm guide on the can/ball or whatever.
for the circular arc the normal force from the guide path is directed toward the center of the path ( ie the center of the circle - not the center of the pin for the arm ). It must be normal to the tangent at the point of interest.

simphys
erobz said:
for the circular arc the normal force from the guide path is directed toward the center of the path ( ie the center of the circle - not the center of the pin for the arm ). It must be normal to the tangent at the point of interest.
yes yes of coursre that I know, and that is why I am asking why I can't use the formula to determine the angle between the radial line r shown and the normal force. Instead they use 30 degrees here, but how do you determine that?

simphys said:
yes yes of coursre that I know, and that is why I am asking why I can't use the formula to determine the angle between the radial line r shown and the normal force. Instead they use 30 degrees here, but how do you determine that?
I'm sorry I don't think I understand your question. They are telling you its 30 deg ( ## \theta = 30^{\circ} ## ) at the instant shown. Because there is no friction ( and its an isosceles triangle), that angle between ##r## and ##N## must be ##\theta##.

erobz said:
I'm sorry I don't think I understand your question. They are telling you its 30 deg ( ## \theta = 30^{\circ} ## ) at the instant shown. Because there is no friction ( and its an isosolies triangle), that angle between ##r## and ##N## must be ##\theta##.
well, so if you look at this example the angle psi, the formula I mentioned above is used for the normal force angle.

simphys said:
well, so if you look at this example the angle psi, the formula I mentioned above is used for the normal force angle.
View attachment 319773

I believe you can use it?

simphys
erobz said:
I believe you can use it?
actually you might be right.. I got something in the 40's, but I had the wrong equation for r=f(##\theta##) one sec let me try again.

simphys said:
actually you might be right.. I got something in the 40's, but I had the wrong equation for r=f(##\theta##) one sec let me try again.
nope... I am getting -40.89

erobz said:
Its says this is in the horizontal plane, so this is a top view (gravity is acting into the page)
oh, that is what they mean.., okay thanks a lot. I thought that it was downwards or something.

simphys said:
nope... I am getting -40.89

simphys
erobz said:
##r = cos(\theta)## and for the given, r = 0.866 and for rdot --> ##rdot = sin(\theta) * \theta(dot)## = -1

simphys said:
##r = cos(\theta)## and for the given, r = 0.866 and for rdot --> ##rdot = sin(\theta) * \theta(dot)## = -1
$$\frac{dr}{d \theta} \neq \dot r$$

simphys
erobz said:
$$\frac{dr}{d \theta} \neq \dot r$$

how come is it not?

oh dear......

simphys said:
how come is it not?
becuase ## \dot r = \frac{dr}{dt}##

I get - 60 degrees which indeed makes it 30 degrees now..... thanks a lot!! Need to watch out on those two mistakes I just made here.

erobz

1. What is the normal force in cylindrical coordinates?

The normal force in cylindrical coordinates is the component of the force that is perpendicular to the surface of the cylinder at a given point. It is represented by the symbol "N" and is measured in units of force, such as Newtons (N).

2. How is the normal force related to the radial line in cylindrical coordinates?

The normal force is always perpendicular to the radial line in cylindrical coordinates. This means that the angle between the normal force and the radial line is always 90 degrees.

3. Can the angle between the normal force and the radial line change in cylindrical coordinates?

Yes, the angle between the normal force and the radial line can change depending on the orientation of the surface at a given point. For example, if the surface is curved, the angle will vary at different points along the surface.

4. How is the angle between the normal force and the radial line calculated in cylindrical coordinates?

The angle between the normal force and the radial line can be calculated using trigonometric functions, specifically the cosine function. The formula is: cosθ = N/R, where θ is the angle, N is the normal force, and R is the radial line.

5. What is the significance of the angle between the normal force and the radial line in cylindrical coordinates?

The angle between the normal force and the radial line is important in determining the direction and magnitude of the force on an object in cylindrical coordinates. It is also used in various engineering and physics calculations involving cylindrical coordinates.

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