I am having trouble with GR8677 #6 and was wondering if anyone could help me out. The problem seems quite simple and I know i could eliminate all but 2 of the answers by looking at the units but for some reason the solution eludes me .
-Agaliarept
GR8677 #6
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(GR8677 #6) Solution
If you wanted to solve for the tangential acceleration of the particle without looking at the answer choices then you could reduce the problem to the standard “block sliding down a frictionless incline plane”. If you take theta to be the angle the plane makes from the horizontal then you are probably familiar with the result that the acceleration of the block tangent to the plane is a_t = g*sin(theta). If you wanted to express the acceleration in terms of the slope of the block (s = rise/run) then you would get a_t = g*s/sqrt(s^2+1).
Armed with this standard result, problem #6 becomes quite easy since you know the slope of the of the function y = x^2/4 is s = dy/dx = x/2. Plugging this into the result above you get:
a_t = g*(x/2)/sqrt((x/2)^2+1) = g*x/sqrt(x^2+4)
Note: Your use of units and of knowing limits 0 < a < g are really good techniques on multiple choice problems. Another quick way to solve this one is let x go to infinity.
Armed with this standard result, problem #6 becomes quite easy since you know the slope of the of the function y = x^2/4 is s = dy/dx = x/2. Plugging this into the result above you get:
a_t = g*(x/2)/sqrt((x/2)^2+1) = g*x/sqrt(x^2+4)
Note: Your use of units and of knowing limits 0 < a < g are really good techniques on multiple choice problems. Another quick way to solve this one is let x go to infinity.
method gives the right solution, but if the question were trickier, it'd be wrong. the problem gives "dimensionless units"... simple dimensional analysis won't quite work here.
danty wrote:You do not need to calculate the tangential acceleration. It is obvious that is greater than 0 and lower than g, so focus on the other three choices.
The only choice that gives dimension of acceleration is D: gx/(x^2 + 4)^1/2