Small question about speed of the ball

[masterchiefo]

Homework Statement

Hello everyone,

I have this very quick and simple question:
This is the drawing of the problem and I have to find the speed of the ball.

Homework Equations

The Attempt at a Solution

I found a_c to be 0.83m/s²
and R = 1.2855

do I simply use this formula to find the speed? a_c = v²/R
?
thank you

 

[Simon Bridge]

The relationship between the (magnitude of) tangential velocity and the centripetal acceleration is $a_c=v_t^2/r$ ... you know this already (it should be in your notes next to the equation you wrote). So the only remaining issue must be whether or not the velocity you are asked to find is the tangential velocity.
How could you go about figuring that out?

 

[masterchiefo]

The relationship between the (magnitude of) tangential velocity and the centripetal acceleration is $a_c=v_t^2/r$ ... you know this already (it should be in your notes next to the equation you wrote). So the only remaining issue must be whether or not the velocity you are asked to find is the tangential velocity.
How could you go about figuring that out?

Well it only says to find the speed of the ball, nothing else. I assumed its the tangential velocity. what other velocity is there to be calculated ?

 

[Simon Bridge]

Exactly ... so you have answered your own question.

 

[masterchiefo]

Exactly ... so you have answered your own question.

But I don't understand because I don't get the correct answer.
answer is 3.25 and I get
0.83 = v2/1.2855
v=1.03294

 

[Simon Bridge]

From your value of $a_c$ I get their answer - so it is down to how you did the math.
What did you use for your value of "r"?

 

[ehild]

But I don't understand because I don't get the correct answer.
answer is 3.25 and I get
0.83 = v2/1.2855
v=1.03294

Your ac is wrong. How did you get it?

 

[masterchiefo]

But I don't understand because I don't get the correct answer.
answer is 3.25 and I get 1.03294

From your value of $a_c$ I get their answer - so it is down to how you did the math.
What did you use for your value of "r"?

1.2855 = sin(40)*2

Your ac is wrong. How did you get it?

Forces in X = m*ac = -T*sin(40)
1kg*ac = -T*sin(40)
Forces in Y = 0 = -w+T*cos(40)
0 = -1kg*9.81+T*cos(40)

then when I do a solve with both equations I get T = 12.8 N and ac = 0.839

 

[ehild]

1.2855 = sin(40)*2

Forces in X = m*ac = -T*sin(40)
1kg*ac = -T*sin(40)
Forces in Y = 0 = -w+T*cos(40)
0 = -1kg*9.81+T*cos(40)

then when I do a solve with both equations I get T = 12.8 N and ac = 0.839

ac=T sin(40), how is 12.8 sin(40) equal to 0.839?

 

[masterchiefo]

Your ac is wrong. How did you get it?

oh wow I really feel stupid now. I misstyped it in my calculator, I assumed it was correct since I was getting the correct value for T.

thank you very much :P

 

[ehild]

oh wow I really feel stupid now. I misstyped it in my calculator, I assumed it was correct since I was getting the correct value for T.

thank you very much :P

You are welcome.
The tension and weight as vectors ad up to the centripetal force and from the right triangle you can see at once that Fcp=W tan(theta)

 

[masterchiefo]

From your value of $a_c$ I get their answer - so it is down to how you did the math.
What did you use for your value of "r"?

thank you very much for helping me, appreciate it ;P