# Ball rotated on a string vector diagram

1. May 23, 2016

### Ab17

1. The problem statement, all variables and given/known data

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.

2. Relevant equations
a= acì + atj

3. 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

2. May 23, 2016

### BvU

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. May 23, 2016

### Ab17

The equation is in vector form

4. May 23, 2016

### Ab17

There is tension and gravity

5. May 23, 2016

### BvU

In the sense that it gives an x component and a y component.

6. May 23, 2016

### Ab17

• FBD

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7. May 23, 2016

### BvU

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: May 26, 2016
8. May 23, 2016

### Ab17

F= ma
The tension is causing the centripetal acc

9. May 23, 2016

### Ab17

So the centripetal acc will be in direction of tension and towards the center

10. May 23, 2016

### Ab17

I and j are unit vectors

11. May 23, 2016

### BvU

Yes, that's why they have a hat. But so far you didn't reveal which way they point .
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 ?

This was a bit premature. Please ignore.

12. May 24, 2016

### Ab17

Is this ok

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13. May 24, 2016

### BvU

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. May 24, 2016

### Ab17

Because If a is the total acceleration then its components have to be I and J

15. May 24, 2016

### Ab17

36.9 is below the x axis

16. May 24, 2016

### BvU

Agree. But this goes for horizontal and vertical just as well.
It cleary says something else altogether.

17. May 24, 2016

### Ab17

It tells us the location of the ball isnt

18. May 24, 2016

### BvU

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. May 25, 2016

### Ab17

So i should break centripetal acc into components or somthn like that

20. May 26, 2016

### BvU

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 !