 #1
Adesh
 735
 188
 Homework Statement:
 Is centripetal force the force acting towards the Center of rotation or towards the axis of the rotation?
 Relevant Equations:

Centripetal Acceleration ##a =
\frac{v^2}{r}##.
Centripetal force is defined as the force causing the body to follow a curved path, acting towards the center and always orthogonal to the direction of motion. For uniform circular motion the formula for centripetal acceleration is $$a_c = \frac{v^2}{r}$$.
But my understanding of centripetal acceleration began to shake when I confronted this problem:
A smooth wire of length ##2 \pi r## is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed ##\omega## about the vertical diamteter ##AB##, as shown in figure, the bead is at rest with respect to the circular ring at position P as shown. Find the value of ##\omega ^2##..
I attempted to solve it this way: at point P there is gravitational acceleration ##a_g =  g ~\hat{j}##. Now, I thought for point P to be in static situation (w.r.t. to the circle) this downward acceleration must be cancelled.
But if centripetal acceleration’s direction of point P were to act horizontally from point towards AB then we won’t get ##a_g## cancelled. How about having a centripetal acceleration acting from P to the point O? In this way we surely going to get a vertical component of centripetal acceleration.
By some simple geometry:
We would get ## \theta = \frac{\pi}{6}##. Now, the vertical component of ##a_c## is ##a_c \cos \theta = \frac{v^2}{r} \frac{\sqrt 3}{2} = \frac{\sqrt 3 \omega^2 ~r}{2} ##. Now for equilibrium we must have $$ a_c \cos \theta = g \\
\frac{\sqrt 3~\omega^2 ~r}{2 } =g \\
\omega ^2 = \frac{2g}{\sqrt 3~r}$$.
Although, I have reached to the correct answer but I suspect the step when I said the centripetal acceleration would act from P towards O.
But my understanding of centripetal acceleration began to shake when I confronted this problem:
A smooth wire of length ##2 \pi r## is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed ##\omega## about the vertical diamteter ##AB##, as shown in figure, the bead is at rest with respect to the circular ring at position P as shown. Find the value of ##\omega ^2##..
I attempted to solve it this way: at point P there is gravitational acceleration ##a_g =  g ~\hat{j}##. Now, I thought for point P to be in static situation (w.r.t. to the circle) this downward acceleration must be cancelled.
But if centripetal acceleration’s direction of point P were to act horizontally from point towards AB then we won’t get ##a_g## cancelled. How about having a centripetal acceleration acting from P to the point O? In this way we surely going to get a vertical component of centripetal acceleration.
By some simple geometry:
We would get ## \theta = \frac{\pi}{6}##. Now, the vertical component of ##a_c## is ##a_c \cos \theta = \frac{v^2}{r} \frac{\sqrt 3}{2} = \frac{\sqrt 3 \omega^2 ~r}{2} ##. Now for equilibrium we must have $$ a_c \cos \theta = g \\
\frac{\sqrt 3~\omega^2 ~r}{2 } =g \\
\omega ^2 = \frac{2g}{\sqrt 3~r}$$.
Although, I have reached to the correct answer but I suspect the step when I said the centripetal acceleration would act from P towards O.