A head slide without friction around a loop the (figure). The bead slide is released is from rest at a height `h = 3.50R`. How large is the normal force on the bead at point `(A)` if its mass is `50 g` ?

Options

A. `0.10 N` downward
B. `0.10 N` upward
C. `1.0 N` downward
D. `1.0 N` upward

Answer

C

Explanation

The speed at the top can be found from the conservatiobn of energy for the bead-track -Earth system .
We define the bottom of the loop as the zero level for the gravational potential energy.
Since `v_(1) = 0, E_(i) = K_(i) = U_(i) = 0 + mgh = mg (3.50 R)`
The total energy of the bead at point , `(A)` can be
written as `E_(A) - K_(A) + U_(A) = (1)/(2) mv_(A)^(2) + mg(2R)`
Since mechainical energy is conserved `E_(i) + E_(A)`, we get
`mg(3.50 R) = (1)/(2) mv_(A)^(2) + mg(2R)`
simplifying `v_(A)^(2) = 3.0 gR implies v_(A) = sqrt(3.0 gR)`
To find the force at the top , we constant a force diagram as shown.

Here we assume that `N` is downward, like `mg`
`Sigma F_(y) = ma_(y) : N+ mg = (mv^(2))/( r)`
`N = m[(v^(2))/( R) - g] = m [(3.0gR)/(R) - g] = 2.0 mg`
`N = 2.0 (50 xx 10^(-3) kg) (10.0 m//s^(2))`
`= 1.0 N` downward.