# Uniform Circular Motion and Projectile Motion Help

## Homework Statement

A 60 g ball is tied to the end of a 40-cm-long string and swung in a vertical circle. The center of the circle 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, how far to the right does the ball hit the ground? ## Homework Equations

vy^2 = voy^2 + 2a(y-yo)
vy = voy + at
ac=vx^2/r
(x-xo) = voxt + (1/2)axt^2

## The Attempt at a Solution

I used vy^2 = voy^2 + 2a(y-yo) to find the final velocity in the y direction :
vy^2 = (0)^2 + 2(-9.8)(-0.19)
vy = 1.93 m/s

Then, I used that final velocity and found the time it took to hit the ground :
vy = voy + at
(1.93) = (0) + (9.8)t
t = 0.1969 s

After that, I used the ac=vx^2/r equation to find the velocity in the x direction :
(9.8) = (vx^2)/(0.4)
vx = 1.98 m/s

Then, I used my velocity in the x direction, my time, and my zero acceleration in the x direction to find my delta x :
(x-xo) = voxt + (1/2)axt^2
(x-xo) = (1.98)(0.1969) + (1/2)(0)(0.1969)^2
(x-xo) = 0.3899 m OR 38.99 cm

The online program told me that it was wrong and all of my friends at my department don't know where I went wrong, can someone please help me?

#### Attachments

Last edited:

Chestermiller
Mentor
You did not provide the entire problem statement. What exactly are you supposed to calculate?

How did you calculate the velocity at the top of the arc?
You are aware that the acceleration of the ball is not constant, right?

You did not provide the entire problem statement. What exactly are you supposed to calculate?

How did you calculate the velocity at the top of the arc?
You are aware that the acceleration of the ball is not constant, right?
I edited the forum now, and I thought the acceleration of the ball would be zero, therefore having a constant velocity, after it was released.

berkeman
Mentor
I asked her to go back and edit her post with the last sentence in the Problem Statement. Now it finally makes sense.

@Jacquelyn your equations do not look right to me. If the string is released at the top of the circle, there is no Vy at that instant. The velocity is horizontal initially.

The time it takes to hit the ground is the same as if it were dropped from that height, right? (Why?)

So you need to figure out what the Vx is at the top of the circle to barely keep the string taut, and use that for your horizontal motion equation and the time you got from the falling motion to figure out how far it goes to the side before hitting the ground. Does that make sense?

• Jacquelyn and CWatters
I asked her to go back and edit her post with the last sentence. Now it finally makes sense.

@Jacquelyn your equations do not look right to me. If the string is released at the top of the circle, there is no Vy at that instant. The velocity is horizontal initially.

The time it takes to hit the ground is the same as if it were dropped from that height, right? (Why?)

So you need to figure out what the Vx is at the top of the circle to barely keep the string taut, and use that for your horizontal motion equation and the time you got from the falling motion to figure out how far it goes to the side before hitting the ground. Does that make sense?
That does all make sense, that's what I tried to do (and thought I did). It takes the same time to hit the ground because there are no forces in the y direction (except for gravity) that would make the ball it the ground faster and the x component does not affect the y.

I tried to use the ac = v^2/r to find my initial velocity in the x direction because I know that the centripetal acceleration inwards and the tangential velocity would be perpendicular, and at the top of the circle the acceleration would be 9.8 because of gravity so I used : (9.8) = (vx^2)/(0.4) --> vx = 1.98 m/s

Chestermiller
Mentor
That does all make sense, that's what I tried to do (and thought I did). It takes the same time to hit the ground because there are no forces in the y direction (except for gravity) that would make the ball it the ground faster and the x component does not affect the y.

I tried to use the ac = v^2/r to find my initial velocity in the x direction because I know that the centripetal acceleration inwards and the tangential velocity would be perpendicular, and at the top of the circle the acceleration would be 9.8 because of gravity so I used : (9.8) = (vx^2)/(0.4) --> vx = 1.98 m/s
This result for the initial velocity in the x direction is correct.

How long does it take for the ball to fall to the ground if it starts out with zero y velocity at an elevation of 0.19 m?

• Jacquelyn
This result for the initial velocity in the x direction is correct.

How long does it take for the ball to fall to the ground if it starts out with zero y velocity at an elevation of 0.19 m?
For this I used the equation :
(y-yo) = voyt + (1/2)at^2
(-0.19) = (0)t + (1/2)(-9.8)t^2
t = 0.1969 s

Because I am given the acceleration due to gravity, the distance in the y component, and the initial velocity in the y direction being zero.

Chestermiller
Mentor
For this I used the equation :
(y-yo) = voyt + (1/2)at^2
(-0.19) = (0)t + (1/2)(-9.8)t^2
t = 0.1969 s

Because I am given the acceleration due to gravity, the distance in the y component, and the initial velocity in the y direction being zero.
Very nice. Now, in this amount of time, how far does it travel horizontally?

• Jacquelyn
Very nice. Now, in this amount of time, how far does it travel horizontally?
Now, because it is a projectile in free fall and there are no forces acting on it in the x direction, I am assuming that my acceleration in the x direction is zero. Therefore I am using the equation :
(x-xo) = voxt + (1/2)axt^2
(x-xo) = (1.98)(0.1969) + (1/2)(0)(0.1969)^2
(x-xo) = 0.3899 m OR 38.99 cm

Chestermiller
Mentor
Now, because it is a projectile in free fall and there are no forces acting on it in the x direction, I am assuming that my acceleration in the x direction is zero. Therefore I am using the equation :
(x-xo) = voxt + (1/2)axt^2
(x-xo) = (1.98)(0.1969) + (1/2)(0)(0.1969)^2
(x-xo) = 0.3899 m OR 38.99 cm
Correct. Maybe you reported the result with too many significant figures. In my judgment, you are only entitled to 2 significant figures. So, 39 cm.

• Jacquelyn
Correct.
The online homework program I am using (Pearson) says that the answer is wrong.

Chestermiller
Mentor
The online homework program I am using (Pearson) says that the answer is wrong.
Like a said. Maybe it's a question of significant figures. Try 39 cm or 0.39 m.

• Jacquelyn
Like a said. Maybe it's a question of significant figures. Try 39 cm or 0.39 m.
Both of those did not work. I am just glad that I understand what I am doing because when the program told me I was wrong I got very confused. I will try to email my professor, thank you for your help!

• berkeman and Chestermiller
gneill
Mentor
How long does it take for the ball to fall to the ground if it starts out with zero y velocity at an elevation of 0.19 m?
Shouldn't that be 1.9 m?

• Jacquelyn
Chestermiller
Mentor
Shouldn't that be 1.9 m?
Yikes. You're right. Senior moment.