- #1
Bakeman
- 4
- 0
A block of mass m is positioned at the top of a quarter-circle ramp with radius R and friction coefficient μ. To find the velocity v at the bottom, I did the force summation equations in the n- and t-directions and came up with:
n: N - mgsinθ = mv^2/R
t: mgcosθ - μN = ma
I then solve for N in the n-direction equation, then subsitute into the t-direction equation. Solving for a, I get:
a = g(cosθ - μsinθ) - μv^2/R
I realize that I need to use this in the kinematic equation: vdv = ads, or in this case, vdv = aRdθ. Doing this, I get:
vdv = (Rg(cosθ - μsinθ) - μv^2)dθ
This can supposedly be rearranged into an easily solvable differential equation in the form of dy/dx + f(x)y = g(x), but I'm not seeing it. I'm assuming that v = y and θ = x, which would make the equation dv/dθ + f(θ)y = g(θ).
What am I missing?
n: N - mgsinθ = mv^2/R
t: mgcosθ - μN = ma
I then solve for N in the n-direction equation, then subsitute into the t-direction equation. Solving for a, I get:
a = g(cosθ - μsinθ) - μv^2/R
I realize that I need to use this in the kinematic equation: vdv = ads, or in this case, vdv = aRdθ. Doing this, I get:
vdv = (Rg(cosθ - μsinθ) - μv^2)dθ
This can supposedly be rearranged into an easily solvable differential equation in the form of dy/dx + f(x)y = g(x), but I'm not seeing it. I'm assuming that v = y and θ = x, which would make the equation dv/dθ + f(θ)y = g(θ).
What am I missing?