Conservation of Energy + Projectile Motion

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SUMMARY

The discussion focuses on calculating the height from which a ball must roll down a ramp to land 100 cm away in a hole 0.05 cm deep, launching at a 15-degree angle. The conservation of energy principle is applied, leading to the equation mgh = 1/2mv^2 + 1/2Iw^2, where I is the moment of inertia for a solid sphere. The user initially calculates a negative height of -70.9 cm, indicating a miscalculation likely due to unit inconsistencies and algebraic errors. The overall approach using conservation of energy and projectile motion equations is correct, but requires careful attention to unit conversions and algebraic clarity.

PREREQUISITES
  • Understanding of conservation of energy principles
  • Familiarity with projectile motion equations
  • Knowledge of unit conversions between centimeters and meters
  • Basic algebraic manipulation skills
NEXT STEPS
  • Review the conservation of energy in the context of rolling objects
  • Study projectile motion equations, specifically the components of velocity
  • Practice unit conversion techniques between metric systems
  • Examine common algebraic pitfalls in physics calculations
USEFUL FOR

Students in physics courses, educators teaching mechanics, and anyone involved in experimental physics or engineering applications related to projectile motion and energy conservation.

andy343
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Homework Statement


So I have a ball rolling down a ramp structure from rest and up a curve,to launch out at 15 degrees. The ball lands 100cm away in a hole 0.05cm deep.
The height in which it comes out of the ramp is 0.05cm high
angle at which ball launches out = 15 degrees above horiontal
dx=100cm
dy=-0.05+-0.05= -0.1cm
*also the lowest point of the ramp is on the ground

find the height in which the ball needs to be rolled down the ramp in order to land 100cm away and into a hole 0.05cm deep.

Homework Equations


Conservation of energy
projectile motion equations

The Attempt at a Solution


Use conservation of energy equation to find velocity and than input that into dy=vyit+1/2ayt^2
Ei=Ef
Ei=KEt+KEr
mgh=1/2mv^2+1/2Iw^2
I=2/5mr^2
w=v/r
mgh=7/10mv^2
solve for v, v=sqrt(10/7gh)---= launching velocity
Use formula dy=vyit+1/2ayt^2
t=dx/vx
vx=cos(15)vi
vyi=sin(15)vi
-0.1=sin15vi(100/cos15vi)+1/2(-9.81)(1/(cos15)(vi))^2
vi cancels vi=launching velocity = sqrt(10/7gh)
-0.1=sin15(100/cos15)+1/2(-9.81)(100^2/(cos15)^2(10/7(9.81)h)
9.81 cancels
-0.1=sin15(100/cos15)+1/2(-1)(100^2/(cos15)^2(10/7h)
solve for h
h=-70.9cm
Umm, I don't get why my height becomes negative..so I think I did something wrong but
I don't know where. ---This is actually from my lab and not a problem so I might be missing something.
 
Last edited:
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I'm slightly confused about what the 100/cos15 is. Is that your time?
 
Um oh sorry about the confusion...I accidently inputed some wrong numbers, give me a second
and yes that is my time
 
Last edited:
Where does "100" come from?
 
So it lands in a hole 100cm* away in 0.05cm* deep hole.
 
Last edited:
You can't mix meters and centimeters in one formula. "9.81" is in meters.
 
lol, okay I think I know why my calculations are wrong now...my bad
TYVM
One more thing, so am I approaching the calculation/problem correctly? by using conservation of energy to find launch velocity and than input that into the dy=vyit+1/2ayt^2?
 
The overall approach seems OK. There might be errors in algebra - the formulas are very hard to read.
 

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