# 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.f

#### 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

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.

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).

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)

So delta y = 0.2
How do you get 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).
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 -.

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

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##.

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.

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 ?

##\ ## the lowest point is at 1.4 m
The highest is at 2.6m. Which one is desired? C-

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##.

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
I must admit I always have to do a mental check to avoid confusing ##cm## and ##mm##.
Try furlongs and inches ... ## \ ##

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

• BvU
• jbriggs444