Solving a Rolling Bead's Motion in 2D Space

[aak452]

Homework Statement

a If the bead starts at the origin at time t=0, how long does it take for it to reach the
end of the track (30m away in the figure)? Provide the answer in terms of v$_{}$. b How does the parameter α depend on time, ie. what is α(t)?
d What is the velocity vector as a function of t (in symbolic form)?
e What is the acceleration vector as a function of t (in symbolic form)?
f Based on your results for the velocity and acceleration, what is the radius of the "kissing circle" at the top of the track? Recall, the kissing circle at a given point goes through the point, has the same tangent as the curve, and a radius that reproduces the perpendicular component of the acceleration, i.e.:
R = |→v|^2/a?
f Will Rbottom the radius of the kissing circle at the bottom of the track
be <, > or = to R$_{}$?
h How fast would the rod need to move for the bead to leave the track? Provide an answer in symbolic form and a numerical value for the speed.

The Attempt at a Solution

a) I tried approximating the length of the track through simply turning each curve into a line. Then divide the total distance by v$_{}$?
b)α is just the x-position of the bead, so it would increase depending on time. This answer, however, seems WAY too simple...perhaps I am interpreting the question wrong?
c) I tried taking the derivative of the given parametric equation...but am not sure how to derive a parametric equation, as it has two parts.
d)I suppose I would the derive the equation obtained from deriving the original parametric equation?
After this point, I am just hopelessly lost. I think I am having trouble applying the easy concepts I learn from the textbook to more conceptual problems. Any help would be MUCH appreciated!

 

[anathema]

Hello!

Thank you for your post. Let's break down each part of the problem and work through them together.

a) To find the time it takes for the bead to reach the end of the track, we can use the equation d = vt, where d is the distance (30m in this case), v is the velocity, and t is the time. However, we need to find the velocity in terms of v_{} first. Since the velocity is constantly changing, we need to use calculus to find it. We can use the parametric equations given in the problem to find the velocity vector as a function of t. This is shown in part d of the problem. Once we have the velocity vector, we can find the magnitude of the velocity (speed) by taking the square root of the sum of the squares of the x and y components of the velocity vector. We can then plug this value into the equation d = vt and solve for t.

b) You are correct that α is the x-position of the bead, but it is not increasing with time. It is a constant value that represents the starting position of the bead at t=0.

c) To derive the parametric equation, we can use the chain rule. Let x(t) and y(t) be the x and y components of the parametric equation, respectively. Then, dx/dt = x'(t) and dy/dt = y'(t). The velocity vector is given by dx/dt i + dy/dt j. We can then use the chain rule to find the acceleration vector, which is shown in part e of the problem.

d) As mentioned above, the velocity vector is given by dx/dt i + dy/dt j. We can use the chain rule to find dx/dt and dy/dt, and then plug those values into the velocity vector.

e) The acceleration vector is given by d^2x/dt^2 i + d^2y/dt^2 j. We can use the chain rule to find d^2x/dt^2 and d^2y/dt^2, and then plug those values into the acceleration vector.

f) To find the radius of the kissing circle at the top of the track, we can use the formula R = |v|^2/a, where v is the velocity and a is the acceleration. We can use the velocity and acceleration vectors found in parts d and e