Circular Motion of a ball Problem

[cse63146]

Homework Statement

A 60 g ball is tied to the end of a 50-cm-long string and swung in a vertical circle. The center of the circle, as shown in Figure, is 150 cm above the floor. The ball is swung at the minimum speed necessary to make it over the top without the string going slack.

If the string is released at the instant the ball is at the top of the loop, where does the ball hit the ground?

http://session.masteringphysics.com/problemAsset/1000600/5/knight_Figure_07_62.jpg

Homework Equations

Critical velocity = V{c} = \sqrt{r*g}
\Deltad=V{i}*t + (1/2)(a)(t^2)

The Attempt at a Solution

I attempted this problem as a projectile motion problem, and divided the calculations into two parts: Horizontal and Verticle

Verticle Part

So I used this equation: \Deltad=V{i}*t + (1/2)(a)(t^2) and I know the verticle initial is 0 so I can eliminate V{i}*t of the equation and get: \Deltad=(1/2)(a)(t^2)

2 = (1/2)*(9.8)*(t^2)
2 = 4.9*(t^2)
t^2 = 2 / 49
t = 0.64 seconds

Horizontal Part

The horizontal parts only has 3 variables: V, d, and t (which I know from the verticle part). and V = V{c} = \sqrt{r*g} = 0.59 m/s.

and d = v*t = 0.59*0.64 = 0.3776 m

This doesn't look right. I was wondering if someone can point me in the right direction. Any help is greatly appreciated. Thanks.

 

[PhanthomJay]

Looks ok except for a math error in your critical speed calc.

 

[cse63146]

yea I see what you mean. I accidently used the mass instead of the radius.

so it would be v_{c}=\sqrt{r*g} = \sqrt{0.5*9.8} = 6.9

therefore:

d = v*t = (6.9) * (0.64) = 4.4 m/s?

Does this seem right?

Edit: Still not the right answer. Help?

 

[Doc Al]

What are you asked to find? (I don't see a question in the problem statement.)

 

[cse63146]

oops.Sorry about that.

If the string is released at the instant the ball is at the top of the loop, where does the ball hit the ground?

 

[PhanthomJay]

yea I see what you mean. I accidently used the mass instead of the radius.

so it would be v_{c}=\sqrt{r*g} = \sqrt{0.5*9.8} = 6.9

therefore:

d = v*t = (6.9) * (0.64) = 4.4 m/s?

Does this seem right?

Edit: Still not the right answer. Help?

you've made anoteher math error,
\sqrt{r*g} = \sqrt{0.5*9.8}= \sqrt4.9 = 2.2m/s And then calculate the distance d which wil be in meters.

 

[cse63146]

yep

d = v*t = 2.2*0.64 = 1.4 m

Thanks a lot!

still can't figure out why I put down 6.9 instead of 4.9...