A thin circular loop of radius R rotates about its vertical diameter with an angular frequency w. Show that a small bead on the wire loop remains at its lowermost point for ω ≤ √ (g / R) . What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω = √(2g / R )? Neglect friction.
The free body diagram of a thin circular loop showing all the forces acting on a small bead is shown below:
The equation to balance the horizontal forces acting on the beads is given as,
m r ω 2 = N sin θ m ( R sin θ ) ω 2 = N sin θ m R ω 2 = N
The equation to balance the vertical forces acting on the beads is given as,
m g = N cos θ m g = m R ω 2 cos θ cos θ = g R ω 2 (1)
We know that,
cos θ ≤ 1
By substituting the given values in the above equation, we get
g R ω 2 ≤ 1 ω ≤ g R
The vertical downward force is ω = 2 g R .
By substituting the given values in equation (1), we get
cos θ = g R ( 2 g R ) 2 cos θ = 1 2 θ = cos − 1 ( 1 2 ) = 60 °
Thus, the angle made by the radius vector is 60 ° .
The free body diagram of a thin circular loop showing all the forces acting on a small bead is shown below:
The equation to balance the horizontal forces acting on the beads is given as,
The equation to balance the vertical forces acting on the beads is given as,
We know that,
By substituting the given values in the above equation, we get
The vertical downward force is
By substituting the given values in equation (1), we get
Thus, the angle made by the radius vector is
