# Circular Motion Part 3

1. A conical funnel rotates about its own axis with angular speed . A particle remain stationary on the inner surface of the funnel and it describes a horizontal circular motion of radius r as shown in figure. The coefficient of friction between particle and inner surface of the funnel is . What is the minimum value of , so that the particle remain stationary on the inner surface of the funnel.

Â As the funnel rotates with angular speed , then it is considered as accelerated frame. Therefore, the particle experiences a pseudo force due to it. Let us consider, R is the normal reaction between particle and funnel. The frictional force between the particle and funnel acts along the inner surface of funnel in the upward direction to hold it at that position.

Therefore, Rcos Â + fsin Â = mg

Or, Rcos Â + Rsin Â = mg —– (i)

And Rsin Â – fcos Â =

Or, Rsin Â – Rcos Â = —- (ii) [when we observe from ground i.e. inertial frame of reference, the particle rotates due to the centripetal force.]

Dividing equation (ii) from equation (i) we get, Â = = Â

=.

2. A mass m is released from the top of a hemisphere (point P) of radius r with horizontal speed v. Calculate the angle with respect to the vertical where it leaves the contact from the hemisphere.

Let us consider, the mass will leave the contact from the hemisphere at point Q. If R is the reaction force acts on the mass by the hemisphere, then mgcos Â â€“ R provides the required centripetal force to the mass during its circular motion. If at point Q, the speed of the mass is u, then mgcos â€“ R = .

When the mass leaves the contact from the hemisphere, then R = 0, so, mgcos = .

cos = Â —– (i)

We draw a perpendicular QS on PO. Therefore, SO = h = rcos .

Using conservation of energy at point P and Q we get, + mg(PS) =

Or, Â + mg(r- rcos) =

Or, Â = + 2gr(1- cos)

u =

Putting the value of u in equation (i) we get, cos = Â

Or, cos = + 2 – 2cos

Or, 3cos Â = Â + 2 Â

= .

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