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Determine the magnitude of its radial acceleration for ball

  • #1
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< Mentor Note -- Thread moved from the General Physics forum to HH >

A ball swings counterclockwise in a vertical circle at the end of a rope 1.50 m long. When the ball is 36.9 degrees past the lowest point on its way up, its total acceleration is (-22.5i + 20.2j) m/s^2. For that instant, (a) sketch a vector diagram showing the components of its acceleration, (b) determine the magnitude of its radial acceleration\


Attempt at understanding:

I am able to visually picture the system and has been shown in the following link (question #2):
http://facultyfiles.deanza.edu/gems/lunaeduardo/Winter2013Exam1Solutions.PDF

What I don't understand is why the components are being added to find the radial acceleration. Why isn't one of the components simply used to find the radial component. For example, if 22.5 is the horizontal component with an angle for 36.9 degrees with the rope, then why isn't the radial acceleration simply acceleration (radial) = 22.5*cos(36.9).

Unless I am missing some concepts here, please clarify this.
 
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Answers and Replies

  • #2
haruspex
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if 22.5 is the horizontal component with an angle for 36.9 degrees with the rope, then why isn't the radial acceleration simply acceleration (radial) = 22.5*cos(36.9)
That is clearly not right since by the same logic you could say the radial acceleration is the radial component of the vertical acceleration, but that will in general give a different number.
Acceleration is the second derivative of displacement. It might be easier to think about it in terms of displacements.
If you move the object left by distance x (from the given position) then you have moved it closer to the centre (reduced the radius) by ##x \sin(\theta)##. If you then move it up by y then you have moved it another ##y \cos(\theta)## closer to the centre. The two add up.
 
  • #3
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Hi haruspex,

Thank you for your reply. I guess I am thinking of right angle triangles where with one component, you can get the hypotenuse or adjacent. I am guessing I am trying to apply the same principles here but I still do not understand why you just add the component of the "i" and "j" along the rope. Can you please explain this a little more in detail. Thanks again!
 
  • #4
haruspex
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Hi haruspex,

Thank you for your reply. I guess I am thinking of right angle triangles where with one component, you can get the hypotenuse or adjacent. I am guessing I am trying to apply the same principles here but I still do not understand why you just add the component of the "i" and "j" along the rope. Can you please explain this a little more in detail. Thanks again!
First, let me correct what I wrote before. I should have specified small displacements:
If you move the object left by a small distance dx (from the given position) then you have moved it closer to the centre (reduced the radius) by ##dx \sin(\theta)##. If you then move it up by dy then you have moved it another ##dy \cos(\theta)## closer to the centre. The two add up.​
Let me try to put it in terms of your right angled triangle, a, b, c. Overall, you wish to travel along c, the hypotenuse direction. You start off along a, at angle theta to the hypotenuse. Having reached the corner, in terms of movement in the hypotenuse direction you have gone distance ##a \cos \theta=c \cos^2 (\theta)##. You now go along b. This takes you distance ##b \sin(\theta) =c \sin^2 (\theta)## in the original hypotenuse direction, for a total movement of ##c \cos^2 (\theta)+c \sin^2 (\theta) = c## in that direction.
That was in terms of displacements, and performing them sequentially. But can you see that this carries over to velocities, and thus to accelerations? And if in each short time dt you make one movement, then the other, it's the same as doing both movements at once in time dt.
 
  • #5
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Thanks again for your informative post. I am curious, if you have a force vector at an angle of 30 degrees with the horizontal, and if given the x component of the force, can't you simply find the force vector using triangles (Fx*cos 30)? I am so sorry but I seem to completely missing something. Or do you actually have to add the x vector and y vector in order to obtain the magnitude vector? I thought you would only add the x and y component vectors when you don't know the angle. Please clarify for me.

Thanks a lot for your help again!
 
  • #6
haruspex
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if you have a force vector at an angle of 30 degrees with the horizontal, and if given the x component of the force, can't you simply find the force vector using triangles (Fx*cos 30)?
No, it's the other way around, and I often see this mistake made.
If F acts at 30 degrees to the horizontal and you know the horizontal component of F has to balance some other force G then Fx = -G, but by definition Fx = F cos(30), so F = Fx sec(30) =-G sec(30).
 
  • #7
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No, it's the other way around, and I often see this mistake made.
If F acts at 30 degrees to the horizontal and you know the horizontal component of F has to balance some other force G then Fx = -G, but by definition Fx = F cos(30), so F = Fx sec(30) =-G sec(30).
Sorry, that was a mistake on my part as I didn't think before submitting. I understand that part actually. Now, would this same principle not apply for obtaining the radial acceleration using the x or y component of the total acceleration from the initial question.
 
  • #8
haruspex
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Sorry, that was a mistake on my part as I didn't think before submitting. I understand that part actually. Now, would this same principle not apply for obtaining the radial acceleration using the x or y component of the total acceleration from the initial question.
No. In the example in post #6, you know that the total resultant from the two components is F: ##\vec F_x+\vec F_y = \vec F##. In the OP question, the x and y accelerations have a resultant in an unknown direction, of which we are extracting the radial component: ##\vec F_x+\vec F_y = \vec F_{rad} + \vec F_{tang}##.
 

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