# Circular motion to projectile motion

• ChetBarkley
In summary, the conversation discusses the calculation of the velocity and distance of a ball being swung in a vertical circle. The calculated velocity at the bottom of the swing is 4.9 m/s in the x-direction. Using projectile motion formulas, the time is found to be 0.2 seconds, and the resulting distance is 0.98m. There is some confusion about the problem statement and the desired distance, but the overall work is deemed correct and concise.
ChetBarkley
Homework Statement
A 100g ball on a 60 cm string is swung in a vertical circle about a point 200 cm above the ground. The tension in the string when the ball is at the very bottom of the circle is 5.0N. A very sharp knife is suddenly inserted when the ball is parallel to the ground to cut the string. What is the distance the ball will land after the string is cut.
Relevant Equations
T - mg = m(v^2/r)
delta y = v_y(t) + 1/2at^2
delta x = v_x(t)
So first I found the velocity of the ball at the bottom of the swing from the force equations, which I got to be 4.9 m/s and this is only in the x-direction. Then using the projectile motion for delta y I found time, which is 0.2s. Then using that time I found the delta x to be 0.98m.
I just want to make sure my work is correct and concise.

0.2 seconds?

Just wonder how you calculated that

ChetBarkley said:
Homework Statement:: A 100g ball on a 60 cm string is swung in a vertical circle about a point 200 cm above the ground. The tension in the string when the ball is at the very bottom of the circle is 5.0N. A very sharp knife is suddenly inserted when the ball is parallel to the ground to cut the string. What is the distance the ball will land after the string is cut.
Relevant Equations:: T - mg = m(v^2/r)
delta y = v_y(t) + 1/2at^2
delta x = v_x(t)

So first I found the velocity of the ball at the bottom of the swing from the force equations, which I got to be 4.9 m/s and this is only in the x-direction. Then using the projectile motion for delta y I found time, which is 0.2s. Then using that time I found the delta x to be 0.98m.
I just want to make sure my work is correct and concise.
That's not what I got. I didn't, however, do any intermediate calculations. The ##0.2s## looks wrong. The ##4.9 \ m/s## looks right.

BvU said:
0.2 seconds?

Just wonder how you calculated that
So delta y = 0.2 and there is not velocity in the y direction once the string is cut, meaning v naught y is zero. So my equation looks like 0.2 = 0t+.5(9.8)t^2. Solving for t: t=sqrt(.2/4.9).

PeroK said:
That's not what I got. I didn't, however, do any intermediate calculations. The ##0.2s## looks wrong. The ##4.9 \ m/s## looks right.
So delta y = 0.2 and there is not velocity in the y direction once the string is cut, meaning v naught y is zero. So my equation looks like 0.2 = 0t+.5(9.8)t^2. Solving for t: t=sqrt(.2/4.9)

ChetBarkley said:
So delta y = 0.2
How do you get that?

ChetBarkley said:
So delta y = 0.2 and there is not velocity in the y direction once the string is cut, meaning v naught y is zero. So my equation looks like 0.2 = 0t+.5(9.8)t^2. Solving for t: t=sqrt(.2/4.9).
This statement makes no sense: "when the ball is parallel to the ground". Does it mean when the string is parallel to the ground (giving ##\Delta y=-2##) or when the ball is moving parallel to the ground (giving ##v_{0 y}=0##)? But it anyway cannot give both.

Last edited:
nasu
And there are two points in the rotation when the ball moves parallel to the ground.
Badly stated problem, prof gets a C -.

PeroK said:
How do you get that?
It's given in the problem

haruspex said:
This statement makes no sense: "when the ball is parallel to the ground". Does it mean when the string is parallel to the ground (giving ##\Delta y=-0.2##) or when the ball is moving parallel to the ground (giving ##v_{0 y}=0##)? But it anyway cannot give both.
It's when the ball is moving when parallel to the ground, giving ##v_{0 y}=0##.

ChetBarkley said:
It's when the ball is moving when parallel to the ground, giving ##v_{0 y}=0##.
But not ##\Delta y=-0.2##. Reread the question.

ChetBarkley said:
So delta y = 0.2
Make a sketch. With a center at 2 m and a radius of 0.6 m, the lowest point is at 1.4 m. If it lands on the ground (also not stated in the prolem), how can you possibly 'calculate' a ##\Delta y## of 0.2 m ?

##\ ##

BvU said:
the lowest point is at 1.4 m
The highest is at 2.6m. Which one is desired? C-

BvU said:
Make a sketch. With a center at 2 m and a radius of 0.6 m, the lowest point is at 1.4 m. If it lands on the ground (also not stated in the prolem), how can you possibly 'calculate' a ##\Delta y## of 0.2 m ?
I imagine the idea was that ##200 \ cm = 0.2 \ m##.

PeroK said:
I imagine the idea was that ##200 \ cm = 0.2 \ m##.
But not such a good idea as 200cm=2m.

PeroK
I must admit I always have to do a mental check to avoid confusing ##cm## and ##mm##.

Tom.G
PeroK said:
I must admit I always have to do a mental check to avoid confusing ##cm## and ##mm##.
Try furlongs and inches ...

## \ ##

hutchphd
BvU said:
Try furlongs and inches ...

## \ ##
That's no problem because they are all different! Inches, feet, yards and miles are all distinct.

BvU
jbriggs444

## 1. What is circular motion?

Circular motion is a type of motion in which an object moves in a circular path around a fixed point or axis. This type of motion is characterized by a constant speed and a continuously changing direction.

## 2. How is circular motion related to projectile motion?

Projectile motion is a type of motion in which an object is launched into the air and moves along a curved path under the influence of gravity. This type of motion can be thought of as a combination of horizontal and vertical circular motions.

## 3. What are the key differences between circular motion and projectile motion?

The main difference between circular motion and projectile motion is that in circular motion, the object moves in a continuous circular path, while in projectile motion, the object follows a curved path and eventually returns to the ground due to the influence of gravity.

## 4. How are the equations for circular motion and projectile motion related?

The equations for circular motion and projectile motion are related through the concepts of centripetal force and acceleration. In circular motion, the centripetal force acts towards the center of the circle, while in projectile motion, the force of gravity acts downwards, causing a curved path.

## 5. Can circular motion and projectile motion occur simultaneously?

Yes, circular motion and projectile motion can occur simultaneously. This can happen when an object is launched at an angle from a circular path, resulting in a combination of circular and projectile motion. An example of this is a roller coaster, where the cars move in a circular path while also experiencing projectile motion as they go up and down hills.

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