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
gh-2gR = 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
3 answers
Your method is correct. I did not check the arithmetic but do not see any obvious errors.
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 (h-2R) = (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.
m g (h-2R) = (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.
In your second part, I agree with your t calculation but not your Vox