Circular motion to projectile motion

AI Thread Summary
The discussion centers on calculating the projectile motion of a ball after its string is cut while swinging in a vertical circle. The initial velocity at the bottom of the swing was determined to be 4.9 m/s, and the time of flight calculated was 0.2 seconds, which some participants questioned as potentially incorrect. There was confusion regarding the vertical displacement (delta y) and the interpretation of the problem statement about the ball's position when the string was cut. Participants emphasized the importance of clarifying the problem's details and ensuring accurate unit conversions, particularly between centimeters and meters. The conversation highlights the need for careful calculations and clear communication in physics problems.
ChetBarkley
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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.
 
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0.2 seconds?

:welcome:

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?

:welcome:

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.
 
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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
 
  • #10
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##.
 
  • #11
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.
 
  • #12
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 ?

##\ ##
 
  • #13
:wink:
BvU said:
the lowest point is at 1.4 m
The highest is at 2.6m. Which one is desired? C-
 
  • #14
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##.
 
  • #15
PeroK said:
I imagine the idea was that ##200 \ cm = 0.2 \ m##.
But not such a good idea as 200cm=2m.
 
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  • #16
I must admit I always have to do a mental check to avoid confusing ##cm## and ##mm##.
 
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  • #17
PeroK said:
I must admit I always have to do a mental check to avoid confusing ##cm## and ##mm##.
Try furlongs and inches ...?:)

## \ ##
 
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  • #18
BvU said:
Try furlongs and inches ...?:)

## \ ##
That's no problem because they are all different! Inches, feet, yards and miles are all distinct.
 
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  • #19
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