Posted by jeff on Monday, December 22, 2008 at 12:55am.
A frictionless track with a loop of radius 36 cm sits on a table 0.9 meters above the ground. If a 0.15 kg object just makes the loop, how high above the table did the object start and how far from the table does it land?
Can you check my work
Mgh  Mg(2R) = 0.5MV^2
gh  2gR = 0.5V^2
gh2gR = 0.5Rg
h = 2.5R = 2.5(.36m)= 0.9
for the second part to figure out how far from the table it lands:
Y = Voyt + .5gt^2
0.9 = 4.9t^2
t = 0.428571s
x = Voxt
x =((Rg)^0.5)*t

physics  Damon, Monday, December 22, 2008 at 6:34am
Your method is correct. I did not check the arithmetic but do not see any obvious errors.

physics  drwls, Monday, December 22, 2008 at 6:47am
If it just makes the loop without losing contact, V^2/R = g at the top of the loop (y = 2R). The highest elevation must be such that
m g (h2R) = (1/2)m V^2 = (1/2)m g R
h = (5/2) R is correct
I don't understand what you did after that.
At the bottom of the loop, I assume that the velocity (V') starts out horizontal and satisfies the equation
(1/2)mV'^2 = 2.5 m R g
V' = sqrt (5Rg) = 4.2 m/s = Vox
The time t that it spends falling satisfies
(1/2) g t^2 = 0.9 m
t = sqrt(1.8/9.8) = 0.43 s
Vox*t = 1.8 m
It is possible that the object is "launched" at some angle other than horizontal. That would depend upon how the loop was constructed. No sketch was provided.

physics  drwls, Monday, December 22, 2008 at 6:51am
In your second part, I agree with your t calculation but not your Vox
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