Ball rotated on a string vector diagram

  • Thread starter Ab17
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  • #1
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Homework Statement



is 1222.5i 1 20.2j2 m/s . For that instant, (a) sketch a vector diagram showing the components of its acceler- ation, (b) determine the magnitude of its radial accel- eration, and (c) determine the speed and velocity of the ball.
42. A ball swings counterclockwise in a vertical circle at
the end of a rope 1.50 m long. When the ball is 36.9°
past the lowest point on its way up, its total acceleration is is -22.5i +20.2j m/s2 . For that instant, (a) sketch a vector diagram showing the components of its acceler- ation, (b) determine the magnitude of its radial accel- eration, and (c) determine the speed and velocity of the ball.


Homework Equations


a= acì + atj

The Attempt at a Solution


If the total acceleration is given by -22
5ì + 20.2j m/s2 then isnt 22.5 is the radial acceleration and 20.2m/s2 is the tangential acceleration. Im confused can you please show me the vector diagram too
 

Answers and Replies

  • #2
BvU
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Hi Ab,

Where did you find that relevant equation ?

Why don't you start drawing a picture of the situation. And work it around to a free body diagram of the ball.
 
  • #3
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The equation is in vector form
 
  • #4
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There is tension and gravity
 
  • #5
BvU
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The equation is in vector form
In the sense that it gives an x component and a y component.
There is tension and gravity
Yes. Please make the sketch.
 
  • #7
BvU
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That's a start. There is a bit more information: 36.9##^\circ##, 1.5 m and ##\vec a = -22.5 \,\hat\imath + 20.2\, \hat\jmath ## m/s2. (In fact I find these values a bit strange: I would expect a negative vertical component. Brings me to the question: what are ##\hat\imath## and ## \hat\jmath## ? I don't see them in the drawing).

Now you need a relevant equation to link the forces and the acceleration. Any idea ?
 
Last edited:
  • #8
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F= ma
The tension is causing the centripetal acc
 
  • #9
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So the centripetal acc will be in direction of tension and towards the center
 
  • #10
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I and j are unit vectors
 
  • #11
BvU
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I and j are unit vectors
Yes, that's why they have a hat. But so far you didn't reveal which way they point :smile:.
So the centripetal acc will be in direction of tension and towards the center
Correct. It is a given that the ball describes a circular trajectory. Not with a constant speed (why not?).
Now it's time for another equation. And perhaps you can already draw the figure that part (a) asks for ?

[edit]
I find these values a bit strange: I would expect a negative vertical component
This was a bit premature. Please ignore.
 
  • #13
BvU
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A ball swings counterclockwise in a vertical circle at
the end of a rope 1.50 m long. When the ball is 36.9° past the lowest point on its way up
I'm not good at approval stamping. But I am good at asking nasty questions, so:
  • What makes you think the ##\hat \imath## and ##\hat \jmath## are radial and tangential and not simply horizontal and vertical ?
  • Where is that 36.9° angle sitting ?
 
  • #14
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Because If a is the total acceleration then its components have to be I and J
 
  • #15
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36.9 is below the x axis
 
  • #16
BvU
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Because If a is the total acceleration then its components have to be I and J
Agree. But this goes for horizontal and vertical just as well.
36.9 is below the x axis
It cleary says something else altogether.
 
  • #17
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It tells us the location of the ball isnt
 
  • #18
BvU
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When the ball is 36.9° past the lowest point on its way up
to me means the angle between the negative vertical axis (the ##\ -\bf \hat\jmath\ ## direction) is 36.9° .
So the complement of the angle you draw.
 
  • #19
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So i should break centripetal acc into components or somthn like that
 
  • #20
BvU
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Sort of. You know the ball describes a circular trajectory. There is an equation needed to link acceleration and speed. At the moment we don't have any relevant equations at all. I hope by now ee do agree that only two forces work on the ball. And we know the resultant acceleration. To link ##\vec T + m\vec g## to the given resultant acceleration we also need a relevant equation.
Set up a plan of approach and carry it out !
 
  • #22
BvU
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Good plan? Does it work and give you the desired answers ? I doubt it.

However, the ##|\vec F_{\rm centripetal} | = {mv^2\over r}\ ## is an essential equation indeed!

I see two different unknown a with the same symbol. Recipe for disaster :smile:.

And I don't see the known ##\ \vec a = -22.5 \hat\imath + 20.2 \hat\jmath\ ## in your plan ?

Fortunately you didn't start carrying out the plan yet :mad:.

For your next post I hope to see a plan + execution :rolleyes: (or at least attempt at ..)

But: we're getting under way !
 
  • #23
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ac = v2/r
 
  • #25
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ac = v2/r
V = sqrt(ac.r)
V = 3.43 m-s1
 

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