Sorry , it was just a mistake that I didn't saw , indeed it's radial accleration , and it's directed toward the center , on the top as I sad before , there is only gravity and friction (static friction) .
because on each point on the loop , we can break the accleration into two components , one pointing to the center of the loop , and the other tangential to the path , it's like an orbital motion .
ok , on the top of the loop we have :
the vertical accleration which is tangential to the loop : a_{x} = \frac{v^{2}}{R}
and the normal accelration which points downward : a_{y} = m.g
is it right ?
By using conservation of mechanical energy energy :
m.g.h=w(f_{r})+\frac{1}{2}m.v^{2}+\frac{1}{2}I.\omega ^{2}=-f_{r}.\frac{h}{\sin(\alpha)} + \frac{1}{2}m.v^2+\frac{1}{3}.m.v^{2}=-f_{r}.\frac{h}{\sin(\alpha)}+\frac{5}{6}m.v^{2}
g.h = \frac{5}{6}.v^{2}-f_{r}.\frac{h}{\sin(\alpha)}
on the top of the loop there is only gravity and friction , they said that we remiss the reaction force , so we have (by applying the sum of the forces , and the sum of the torques) ;
m.g-f_{r} = m.a
f_{r}=\frac{2}{3}m.r^{3}.a
so : a = \frac{g}{1+\frac{2}{3}r^{3}}
and then ?
when the ball is on the loop , there is gravitational force friction , and reaction force , they have also sad that the reaction force was zero on the top of the loop , in addition to that , they give a hint which says that we have to use Newton's law to find out the relation between the...
Homework Statement
An empty ball , of mass m moment of inertia I = (2m.r²)/3, is rolling across the path shown below :
there is friction fr from A to C .
r is the radius of the ball , and R is the radius of the circular part within the path .
what would be the minimal height h , so...
I was trying to solve a problem involving work , as we know :
w = \int_{a}^{b} \vec{f}.d\vec{s}
but in my problem the path was cyrcular , so how to evaluate this kind of integral ?
the Normal force isn't restricted on the beginning point as you're saying , but it follows the object until the end of the path , and so do the gravity of course.