Question

a bead on a wire. for the bead on the wire, put the possible fbs in order of decreasing velocity a) iv,v,i,iii,ii b) ii,iii,v,iv,i c) ii,iii,iv,v,i d) i,ii,iii,iv,v e) ii,iii,i,v,iv

Explanation:

Step1: Analyze centripetal - force condition

For a bead moving in a vertical circular path, the net - force towards the center provides the centripetal force $F_c=\frac{mv^2}{R}$. At any point, $F_c = n\pm mg$ (where $n$ is the normal force and $mg$ is the gravitational force. The sign of $mg$ depends on the direction of the normal force and gravity relative to the center of the circle).

Step2: Consider the top - most point

At the top - most point of the vertical circle, the centripetal force is $F_c=n + mg$. When $n = 0$ (the minimum speed condition to complete the circle), $v_{min}=\sqrt{gR}$. As the normal force $n$ increases at the top - most point, the speed $v$ increases.

Step3: Analyze each case

In case (i), $mg>n$, so $F_c=mg - n$. In case (ii), $mg=n$, so $F_c = 0$. In case (iii), $n>mg$, so $F_c=n - mg$. In case (iv), $n$ and $mg$ act in opposite directions and $n>mg$, providing a larger centripetal force compared to case (iii). In case (v), $n = 0$, and $F_c=mg$.
The larger the centripetal force, the larger the speed $v=\sqrt{\frac{F_cR}{m}}$.
The order of centripetal forces from largest to smallest is (iv) > (v) > (i) > (iii) > (ii). Since $v=\sqrt{\frac{F_cR}{m}}$, the order of velocities from largest to smallest is (iv), (v), (i), (iii), (ii).

Answer:

A. iv,v,i,iii,ii