An object is released from rest and falls in free fall motion. The speed v of the object after it has fallen a distance y is given by
v2 = 2gy.
In an experiment, v and y are measured and the measured values are used to calculate g. If the percent uncertainty in the measured value of v is 3.34% and the percent uncertainty in the measured value of y is 5.00%, what is the percent uncertainty in the calculated value of g? (Do not enter units for this answer.)
To determine the percent uncertainty in the calculated value of g, we need to find the fractional uncertainty in g and then convert it to a percentage.
Given:
Percent uncertainty in v = 3.34%
Percent uncertainty in y = 5.00%
Using the formula for v^2 = 2gy, we can rearrange it to solve for g:
g = v^2 / (2y)
Let's calculate the fractional uncertainties first:
Fractional uncertainty in v (δv) = Percent uncertainty in v / 100 = 3.34 / 100 = 0.0334
Fractional uncertainty in y (δy) = Percent uncertainty in y / 100 = 5.00 / 100 = 0.0500
Now, we can calculate the fractional uncertainty in g (δg). Since g is calculated using the division of v^2 and y, we need to use the rule for multiplication and division of measured quantities:
δg / g = sqrt((2δv/v)^2 + (δy/y)^2)
Substituting the values we have:
δg / g = sqrt((2 * 0.0334)^2 + (0.0500)^2)
δg / g = sqrt(0.0028 + 0.0025)
δg / g = sqrt(0.0053)
δg / g = 0.0727
Finally, to get the percent uncertainty in g, we multiply the fractional uncertainty by 100%:
Percent uncertainty in g = δg / g * 100
Percent uncertainty in g = 0.0727 * 100
Percent uncertainty in g = 7.27%
Therefore, the percent uncertainty in the calculated value of g is approximately 7.27%.
To calculate the percent uncertainty in the calculated value of g, we need to determine how the percent uncertainties in the measured values of v and y affect the percent uncertainty in the calculated value of g.
The equation given is v^2 = 2gy.
Taking the natural logarithm of both sides of the equation, we get ln(v^2) = ln(2gy).
Now, differentiating both sides of the equation with respect to t, we have:
(1/v^2) * (dv/dt) = (d/dt) ln(2gy).
Using the chain rule, we can rewrite this as:
(1/v^2) * (dv/dy) * (dy/dt) = (1/y) * (d/dt) ln(2gy).
Since the object is falling in free fall motion, its acceleration can be approximated as g. Therefore, dv/dt = g and dy/dt = v.
Substituting these values into the equation, we have:
(1/v^2) * g * v = (1/y) * (d/dt) ln(2gy).
Simplifying this equation, we get:
g/v = (1/y) * (d/dt) ln(2gy).
Since v^2 = 2gy, we can substitute this into the equation:
g/v = (1/y) * (d/dt) ln(v^2).
Integrating both sides of the equation with respect to t, we get:
∫ g/v dt = ∫ (1/y) * (d/dt) ln(v^2) dt.
Integrating, we have:
g ∫ dt/v = ∫ (1/y) * (2v/v^2) dt.
Simplifying further, we get:
g * t/v = ∫ (2/y) dt.
Since the object is falling from rest, t = y/v. Substituting this into the equation, we have:
g * (y/v) / v = ∫ (2/y) dt.
Now, solving for g, we get:
g = ∫ (2/y) dt / (y/v) / v.
Simplifying this further, we have:
g = 2v / y^2.
Now, we can calculate the percent uncertainty in the calculated value of g using the percent uncertainties in the measured values of v and y.
The percent uncertainty in v, denoted as Δv/v, is given as 3.34%.
The percent uncertainty in y, denoted as Δy/y, is given as 5.00%.
To find the percent uncertainty in g, we can use the formula:
Δg/g = √[(Δv/v)^2 + (2 * Δy/y)^2].
Substituting the given values, we have:
Δg/g = √[(0.0334)^2 + 2 * (0.0500)^2].
Evaluating this expression, we get:
Δg/g = √[0.00111556 + 2 * 0.0025].
Δg/g = √[0.00611556].
Finally, calculating the square root, we get:
Δg/g ≈ 0.07826.
So, the percent uncertainty in the calculated value of g is approximately 7.826%.
v is sq
v is squared, doubling the uncertainty
the value is then divided by y, which adds its uncertainty
(2 * 3.34%) + 5.00%