Irodov Problem 1.105

a) Since the force F rotates with a constant angular velocity and is localized to a plane without any loss of generalization we can express it in vector form as . This means,













b)The particle will stop when |v| = 0, ie. at times when





The distance traveled during any two consecutive intervals will be the same since the way |v|
changes is identical. This if given by,

Irodov Problem 1.104

The body will first decelerate in its upwards journey and come to a complete stop in the air. It will then begin to fall down and experience air drag. We will consider both these situations separately.

Upward journey:
There are two forces acting on the body, i) the force of gravity mg and ii) the air drag which acts downwards in order to oppose the body's upward motion. In other words both gravity and air drag oppose the body's upwards motion. So we have,














The maximum height achieved by the body is given by (1).

Downward journey
After achieving its maximum height the body begins its decent. During decent, gravity accelerates the body downwards, but the air drag acts upwards trying to oppose the body's downward motion. Thus we have,














From (1) and (2) we have,

Irodov Problem 1.103

The force diagram is depicted in the Figure beside. Initially the force will be too weak to overcome the friction force f offered by the surface. As time goes by the force at will exceed the maximum possible value of f that the surface has to offer namely Nk. At this time the mass will start to accelerate. Let us call this time when the body starts to move as t0.

Forces in the vertical direction
There are two forces acting in the vertical direction, i) the force of gravity mg and ii) the Normal reaction from the surface N. There is no component of acceleration for the body in the vertical direction. Thus we have,

N - mg = 0 (1)

Forces in the horizontal direction
There are two forces acting on the body in the horizontal direction, i) the force F=at and ii) the force of friction f. Let the acceleration of the body be w. Thus we have,

at - f = mw (2)

At the point when the body begins to move f = Nk, F=at0 and w=0. Thus, from (1) and (2) we have,




After time t0 the mass begins to accelerate. During this time the friction force f is fixed at mgk . From (1) and (2) we have,