this question seems small
But
it is very Big
let us solve it by finding piece by piece
[MATH]|\vec{a} (t)|^2 = {a_T}^2 + {a_N}^2[/MATH]
where [MATH]\vec{a} (t)[/MATH] is the acceleration vector, [MATH]a_T[/MATH] is the tangential acceleration, and [MATH]a_N[/MATH] is the normal acceleration
then
[MATH]a_N = \sqrt{|\vec{a} (t)|^2 - {a_T}^2}[/MATH]
[MATH]\vec{v} (t) = \frac{d \vec{r} (t)}{dt} = \ <-100\pi \sin \pi t, 100\pi \cos \pi t>[/MATH]
[MATH]|\vec{v} (t)| = 100\pi[/MATH]
[MATH]\vec{a} (t) = \frac{\vec{v} (t)}{dt} = \ <-100{\pi}^2 \cos \pi t, -100{\pi}^2 \sin \pi t>[/MATH]
[MATH]|\vec{a} (t)| = 100{\pi}^2[/MATH]
[MATH]a_T = \frac{d \ |\vec{v} (t)|}{dt} = 0[/MATH]
then
[MATH]a_N = \sqrt{(100{\pi}^2)^2 - {0}^2} = 100{\pi}^2[/MATH]
[MATH]\vec{N} (t) = \frac{\frac{d \ \vec{T} (t)}{dt}}{\left |\frac{d \ \vec{T} (t)}{dt}\right |}[/MATH]
where [MATH]\vec{N} (t)[/MATH] is the unit normal vector and [MATH] \vec{T} (t)[/MATH] is the unit tangent vector
[MATH]\vec{T} (t) = \frac{\vec{v} (t)}{|\vec{v} (t)|} = \frac{<-100\pi \sin \pi t, 100\pi \cos \pi t>}{100\pi} = \ <-\sin \pi t, \cos \pi t>[/MATH]
[MATH]\frac{d \ \vec{T} (t)}{dt} = \ <-\pi \cos \pi t, -\pi \sin \pi t>[/MATH]
then
[MATH]\vec{N} (t) = \frac{<-\pi \cos \pi t, -\pi \sin \pi t>}{\pi} = \ <-\cos \pi t, -\sin \pi t>[/MATH]
And finally
[MATH]\vec{F} (t) = m \cdot a_N \cdot \vec{N} (t) = 100 \cdot 100{\pi}^2 \cdot <-\cos \pi t, -\sin \pi t> \ = \ <-10000{\pi}^2\cos \pi t, -10000{\pi}^2\sin \pi t>[/MATH] lbf