# Ball rotated on a string vector diagram

## 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.

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

BvU
Homework Helper
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.

The equation is in vector form

There is tension and gravity

BvU
Homework Helper
The equation is in vector form
In the sense that it gives an x component and a y component.
There is tension and gravity

• FBD

#### Attachments

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BvU
Homework Helper
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:
F= ma
The tension is causing the centripetal acc

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

I and j are unit vectors

BvU
Homework Helper
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 .
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 ?

I find these values a bit strange: I would expect a negative vertical component
This was a bit premature. Please ignore.

BvU
Homework Helper
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 ?

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

36.9 is below the x axis

BvU
Homework Helper
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.

It tells us the location of the ball isnt

BvU
Homework Helper
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.

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

BvU
Homework Helper
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 !

Break mg into components and then t-mgcos@ = mv2/r and the other component is responsible for tangential accleration so mgsin@ = ma
gsin@ = a

BvU
Homework Helper
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 .

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 .

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

But: we're getting under way !

ac = v2/r

Mgcos@ = Mac
gcos@ = ac
ac = 7.84m.s2

ac = v2/r
V = sqrt(ac.r)
V = 3.43 m-s1

BvU
Homework Helper
ac = v2/r
You mean ac = v2/r
##g\cos\alpha= a_c##
##a_c = ## 7.84 m.s2
So T is not contributing and we can forget the given ##
\ \vec a = -22.5 \hat\imath + 20.2 \hat\jmath\
## ?
Furthermore it is m/s2, not m.s2
For your next post I hope to see a plan + execution (or at least attempt at ..)
Perhaps try to keep them separate.