Irodov Problem 1.95

Basically this is the equation of a particle moving along an ellipse with major and minor axes of length 2a and 2b. The force acts radially inwards for this particle.

Irodov Problem 1.94

Viewed from the point of a stationary observer, the mass turns along the cylinder because of the Normal reaction N offered by the surface of the cylinder. This normal reaction is the same force that pushes the cylinder outwards as well. We shall solve this problem in two ways i) using the knowledge of centripetal acceleration and ii) mathematically. This will help understand the origin of centripetal acceleration.

Method 1 : Using Centripetal Acceleration The body experiences centripetal acceleration radially inwards as it rotates. The component of the tangential velocity in the direction of the normal reaction is and thus the centripetal acceleration is . There is no other acceleration along this direction. Thus from newton's laws we have,





Method 2 : Mathematically
If there were no gravity, the position the particle vectorially is given by,








This is basically the centripetal acceleration directed radially inwards. The normal reaction N also is directed radially inwards as sown in the Figure. Even if there were gravity acting on the particle, there would be no component of gravity along the direction of centripetal acceleration which is directed radially inwards along the direction of the Normal reaction N, and hence would not change the answer. Now the rest of the solution is simply as in Method 1.

Irodov Problem 1.93

The very basic difference between this problem and all other that while in all previous problems there was no friction between the pulley and the string, here there is some friction between the pulley and the thread. So in all the previous problems the thread would simply slip over the pulley and there was no force of friction hindering this slippage. In this problem we have to consider the effect of force of friction between the pulley and the string. One big difference when there is some friction between the string and the pulley is that the tension in different parts of the string will be different and not the same everywhere. The reason is that the tension has to provide for the friction force in addition to the gravity acting on the masses. To understand how the force of friction acts on the string consider an infinitesimally small segment of the string as shown in the figure.

We consider an infinitesimally small part of the string that subtends an angle at the center of the pulley. The section is shown magnified in the figure and depicts the various forces acting on the piece of the string. There four forces acting on the piece of string , i) the force T with which the rest of the string on the right pulls this string piece, ii) the force T+dT with which the rest of the string on the left, iii) the force of friction dF acting on the string that opposes its slippage and iv) the normal reaction dN that acts normal to the pulley. The tension will keep continuously increasing along the direction of mass m2 since friction always acts in a direction so as to oppose the slippage as shown in the figure. Let us consider forces acting in the tangential and radial directions.


Along the normal direction we have three forces acting, i) the components of tension pulling this string piece along the radial direction and as shown in the figure ( for small values of , can be approximated by see here ) and ii) the normal reaction dN acting upwards. Since the piece of string does not have any component of acceleration in the radial direction we have,






As the limits of all the variables dN, dT and tend to 0, the term will tend to 0 and equation (1) will be reduced to,


Now let us consider forces acting on the string in the tangential direction. There are three forces acting along this direction, i) the component of tension acting on the left side of the string
, ii) the component acting on the right of the string and iii) the force of friction dF as shown in the figure. At the slipping point, dF = kdN . The string has no mass and the system has just begun to accelerate. Thus we have,




From (3) and (2) we can write,









a) Now let us consider the forces acting on the masses. On mass m1 there are two forces, i) the gravitational force m1g and the tension T1. The masses are not accelerating just before the slipping condition so,


By the same reasoning,


But we already know that at the point of slipping. In other words from (5),(6), (7) we can write,



(b) In this part the masses are accelerating at a rate w and so we can rewrite equations (6) and (7) as,



Since the string is massless, equations (1)-(5) will not change (0.w=0), in other words
even though for this part .

From (8), (9) and (10) we have,