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
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.
In your second part, I agree with your t calculation but not your Vox
Let's check your work step by step.
First, let's look at the equation you've written:
Mgh - Mg(2R) = 0.5MV^2
This equation represents the conservation of mechanical energy, where M is the mass of the object, h is the initial height above the table, R is the radius of the loop, and V is the velocity of the object at the top of the loop.
Next, you simplified the equation by canceling out the mass and multiplying 2R by g:
gh - 2gR = 0.5V^2
This is correct so far.
Then, you further simplified the equation by dividing both sides by Rg:
h - 2R = 0.5R
At this step, it seems like there was a mistake in the calculation. When you divided both sides by Rg, the 2R term on the left side should also be divided by R:
h/R - 2 = 0.5
Now, let's solve for h:
h/R = 0.5 + 2
h/R = 2.5
To find h, multiply both sides by R:
h = 2.5R = 2.5 * 0.36 m = 0.9 m
So the object started 0.9 meters above the table.
Now, let's move on to the second part of the question, where you need to find how far from the table the object lands.
You correctly set up the equation using the vertical motion equation:
Y = Voyt + 0.5gt^2
Here, Y represents the vertical displacement, Voy is the initial vertical velocity, t is the time of flight, and g is the acceleration due to gravity.
You plugged in the values and got the equation:
0.9 = 4.9t^2
This equation correctly relates the initial height to the time of flight.
To solve for t, divide both sides by 4.9:
t^2 = 0.9 / 4.9
t^2 ≈ 0.183673
Taking the square root of both sides gives:
t ≈ 0.428571 s
So the time of flight is approximately 0.428571 seconds.
To find how far from the table the object lands, you correctly used the equation:
x = Voxt
Here, x represents the horizontal distance, Vox represents the initial horizontal velocity, and t is the time of flight.
To determine Vox, we need to first find the vertical velocity Voy at the top of the loop. We can use the equation:
Voy = (Rg)^0.5
Plugging in the values:
Voy = (0.36 m * 9.8 m/s^2)^0.5
Voy ≈ 1.92 m/s
Since the loop is frictionless, the horizontal and vertical velocities are constant throughout the motion. Therefore, Vox = Voy.
Using the time t ≈ 0.428571 s and the horizontal velocity Vox ≈ 1.92 m/s:
x = Vox * t
x ≈ 1.92 m/s * 0.428571 s
x ≈ 0.82 m
So the object lands approximately 0.82 meters from the table.
In summary, your initial height calculation is correct. However, there was a mistake in dividing 2R by R, resulting in an incorrect equation. Other than that, your calculations for the time of flight and horizontal distance are correct. The object starts 0.9 meters above the table and lands approximately 0.82 meters from the table.