How Does the Angle Affect Bead Stationarity in Rotating Circular Motion?

[kidia]

A small bead with a mass of 100grams slides along a semicircular wire with a radius of 10 centimeters that rotates about a vertical axis at a rate of 2 revolutions per second.Find the value of angle p for which the bead will remain stationary with respect to rotating wire.

Any idea on this?

 

[ramollari]

The whole problem is to find the angle which the line joining the bead and the centre of the semicircle makes with the horizontal.
The solution is to solve a resulting equation either for \theta or for r. Consider vertical equilibrium first to deduce the vertical component of the normal force. Then, consider the horizontal forces and acceleration at a given time to solve for \theta. Remember that you can express r in terms of R_{Circ} and \theta.

 

[thepatientmental]

To find the value of angle p, we can use the concept of centripetal force in circular motion. The centripetal force is the force that keeps an object moving in a circular path and is directed towards the center of the circle.

In this scenario, the centripetal force is provided by the tension in the wire, which acts towards the center of the circle. The weight of the bead, which acts downwards, is balanced by the normal force from the wire.

To find the angle p, we can use the equation for centripetal force:

F = mω²r

Where F is the centripetal force, m is the mass of the bead, ω is the angular velocity (2πf in this case, where f is the frequency of rotation) and r is the radius of the circle.

We know the values of m, ω, and r, so we can rearrange the equation to solve for the angle p:

p = arccos(F/mω²r)

Substituting the values, we get:

p = arccos(T/(0.1kg)(2π(2 rev/s))²(0.1m))

Where T is the tension in the wire. Since we want the bead to remain stationary, the net force on the bead must be zero. This means that the tension in the wire must be equal to the weight of the bead:

T = mg

Substituting this into the equation for p, we get:

p = arccos((0.1kg)(9.8m/s²)/(0.1kg)(2π(2 rev/s))²(0.1m))

Simplifying, we get:

p = arccos(0.5)

Using a calculator, we can find that the angle p is approximately 60 degrees.

Therefore, if the wire is rotated at a rate of 2 revolutions per second, the bead will remain stationary when the wire is at an angle of 60 degrees from the vertical axis.