Saved Bookmarks
| 1. |
A particle moves along the plane trajectory y(x) with velocity v whose modulus is constant. Find the acceleration of the particle at the point x=0 and the curvature radius of the trajectory at that point if the trajectory has the form (a) of a parabola y=ax^2, (b) of an ellipse (x//a)^2+(y//b)^2=1, a and b are constants here. |
|
Answer» Solution :Let us DIFFERENCIATE twice the path equation `y(x)` with respect to time. `(dy)/(dt)=2ax(dx)/(dt)`, `(d^2y)/(dt^2)=2a[((dx)/(dt))^2+x(d^2x)/(dt^2)]` Since the PARTICLE moves UNIFORMLY, its acceleration at all points of the path is normal and at the point `x=0` it coincides with the direction of derivative `d^2y//dt^2`. Keepping in mind that at the point `x=0`, `|(dx)/(dt)|=v`, We get`w=|(d^2y)/(dt^2)|_(x=0)=2av^2=w_n` So, `w_n=2av^2=v^2/R`, or `R=(1)/(2a)` Note that we can also calculate it from the formula of problem `(1.35b)` (b) Differentiating the equation of the trajectory with respect to time we see that `b^2x(dx)/(dt)+a^2y(dy)/(dt)=0` (1) which implies that the vector `(b^2xveci+a^2yvecj)` is normal to the velocity vector `vecv=(dx)/(dt)veci+(dy)/(dt)vecj` which, of course, is along the tangent. Thus the FORMER vactor is along the normal and the normal component of acceleration is clearly `w_n=(b^2x(d^2x)/(dt^2)+a^2y(d^2y)/(dt^2))/((b^4x^2+a^4y^2)^(1//2))` on using `w_n=vecw.vecn//|vecn|`. At `x=0`, `y!=b` and so at `x=0` `w_n=+-(d^2y)/(dt^2):|_(x=0)` Differenciating (1) `b^2((dx)/(dt))^2+b^2x((d^2x)/(dt^2))+a^2((dy)/(dt))^2+a^2y((d^2y)/(dt^2))=0` Also from (1) `(dy)/(dt)=0` at `x=0` So `((dx)/(dt))=+-v` (since tangential velocity is constant `=v`) Thus `((d^2y)/(dt^2))=+-(b)/(a^2)v^2` and `|w_n|=(bv^2)/(a^2)=v^2/R` This gives `R=a^2//b`. |
|