Irodov Solution 1.89

At any radius r the maximum providable force of friction is given by mgk. With respect to a stationary observer, the rotating cycle is subject to a centripetal acceleration of directed towards the center of the circle (in the rotating frame of the cycle this will be felt as the centrifugal force acting radially outwards) . Thus, at any radius the at the maximum sustainable velocity v we have,



Now the radius corresponding to the maximum sustainable velocity can be determined by maximizing the square of the expression for the square of the velocity as well and making sure that its second derivative is -ive.











At r=R/2 thus, the cyclist can maintain the maximum possible velocity without slipping and its value can be determined by substituting in the expression for velocity determined above as .

Irodov Problem 1.90

From the point of view of a stationary observer, as the car rotates at a radius R, he is subject to centripetal acceleration direction radially inwards given by at a time t. Further, because the car is accelerating it has a tangential acceleration of . Both these acceleration components are provided by the components of friction force acting in the two directions (tangential) and (radial) . Thus based on newton's laws,



The net force of friction is the Euclidean norm of the two components,





But the maximum force of friction that can be offered is mgk thus, at slipping point,







The distance traveled by the car will be,

Irodov Problem 1.88

Let the original length of the spring be l and suppose that because of the rotation the spring elongates by a length ( is the percentage elongation of the spring). Thus, the total length of the spring will be . The angle formed by the mass to the pivot (as shown int the figure) is given by,


From the point of view of a stationary observer there is only one force acting on the mass - the force of the spring that tries to pull the mass back given by . The mass because of its rotation is subject to a centripetal acceleration that is directed radially inwards towards the pivot (this will become centrifugal force directed outwards with respect to a rotating observer on the mass). The component of this centripetal acceleration along the direction of the spring is given by,



Now writing the F=ma due to Newton we have,






No matter what the direction of rotation the centripetal acceleration will always be directed in the same direction and will have the same magnitude. In other words, the rotation direction will not matter - the elongation will be the same.

Irodov Problem 1.87

As shown in the figure, at the time when the mass is at an angle it has descended a height of and thus has lost a potential energy of . This potential energy will be converted into the mass's kinetic energy and thus we have,



The Figure beside shows the forces acting on the mass at some given angle. There are two forces acting on the mass in the radial direction, i) the normal reaction N offered by the surface and ii) the component of force of gravity . The mass also has the centripetal acceleration (note that this becomes the centrifugal force in the reference frame of the observer on the mass and will act radially outwards) given by in the radial direction directed towards the center of the circle.
Thus, we have,



When the body is about to leave the surface, the normal reaction becomes N=0. In other words we have,







At this angle the velocity of the mass can be calculated using (1) and substituting the obtained value of angle as, .