Pendulum on a Merry Go Round

Question

Consider a pendulum mounted on the edge of a merry go round. When the merry go round is at rest, the angle which the rope makes with the vertical is zero. When the merry go round is rotating, to an observer on ground, the pendulum will make an angle in the anticlockwise direction from the vertical(the pendulum will move out when it is mounted, on a rotating merry go round).

Obviously, the tension in the rope provides a centripetal force, which is directed inward, but the bob moves outwards, why is this?

Is it because of the "fictitious" centrifugal force?

Now consider a rod and a ball, the ball is attached to the rod, in such a way that it can slide over it without friction.

When the rod is rotated 2D, still the centripetal force will act toward the center,yet the ball moves outwards.

These are my queries:

1. In the second case and the first case what makes the bob move outward?

2. In the second case, the only force being applied is the tangential force, that makes the rod spin, so where does the centripetal force come from? In the first case, the tension in the string provided the centripetal force, what provides the centripetal force here in the second case?

Answer 1

Think of both problems in terms of inertia (rather than in terms of centripetal/centrifugal force). In both cases the bob "wants" to travel on a straight line. In this case the deviation from a straight line (due to any external force) is the "fictious centrifugal force".

1. In both cases inertia "makes the bob move outward". (It doesn't really "move outward", It tries to move in a straight line.
2. In tescond case there is no centripetal force so the bob will fly off the end of the rod, al. la hai ali.

Answer 2

Obviously, the tension in the rope provides a centripetal force, which is directed inward, but the bob moves outwards, why is this?

Due to its inertia (its wanting to move in a straight line instead of a circle per Newton's first law).

Is it because of the "fictitious" centrifugal force?

Yes, in the non inertial (rotating) frame of the merry go round. The centrifugal force is an "inertial" force. It is fictitious because the outward movement of the pendulum bob is not due any physical contact pushing the bob out. Being thrown to the side of a hard turning vehicle is likewise a centrifugal force. Being thrown forward on hard braking of a car is likewise an inertial (fictitious) force, but in this case it is not a centrifugal force (force associated with circular motion).

Now consider a rod and a ball, the ball is attached to the rod, in such a way that it can slide over it without friction.

When the rod is rotated 2D, still the centripetal force will act toward the center, yet the ball moves outwards.

If there is no friction between the ball and rod, there will be no centripetal force. It is static friction between the ball and rod that provides the centripetal force. If there is no static friction, or if the maximum possible static friction force is exceeded, the ball will slide outward. It's similar to a hard cornering car. Static friction between the tires and the road provides the centripetal force for turning. Without static friction there would be no centripetal force and the the car will continue in a straight line to an observer on the ground per Newton's first law, or skid outwards to an observer in the car.

1. In the second case and the first case what makes the bob move outward?

Inertia. Per Newton's first law a body at rest or moving at constant speed in a straight line will remain a rest or continue to move at constant speed in a straight line unless acted upon by a net external force. For the frictionless rod there is no external force to keep the ball from moving out.

2. In the second case there is no centripetal force so the bob will fly off the end of the rod, al. la hai ali.

Yes it will fly off the rod. I don't know what you mean by "al. la hai ali".

Hope this helps.