Irodov Problem 1.271

There are three forces acting on the cube during its motion, i) the force of gravity mg acting downwards, ii) the normal reaction N from the surface acting upwards and iii) the force of friction f opposing the motion of the cube. Let the acceleration of the cube in the horizontal direction be a. Then we have,



The maximum possible force of friction f offered by the surface is given by Nk. Since the cube is moving, the maximum force of friction is acting on the cube. In other words, f = Nk. So, from (2) we have,



From (1) and (3) it is now clear that,


In (5), vo is the initial velocity of the cube.

Now let us consider the torques acting on the cube about and axis AA' that passes through the cube CG, is perpendicular to the velocity vector and lies in a plane parallel to the surface. The force of gravity mg acts through the CG (along the line OO') and hence does not induce any torque about AA'. The force of friction acts at the surface uniformly at all points of contact and hence acts at a distance a/2 from AA' and induces a torque of mgka/2. This must mean that the force friction should have a tendency to rotate the cube in the clockwise direction about the CG. However, the cube does not rotate. This is because the normal reaction N, shifts away from the axis passing through the CG (OO') to another parallel axis BB' and induces a counter-torque to that induced by the force of friction and exactly cancels it.

Suppose that the axis BB' is at a distance from the axis OO'. Then, the torque induced by the normal reaction about the axis AA' is given by . Hence, for the cube not to rotate about the axis AA' we have,






Now let us consider the question of dissapearance of angualr momentum about an axis CC' that lies in the surface and is perpendicular to the direction of motion of the cube.




Suppose at some instant of time, the CG of the cube is at a distance x from the axis CC'. The force of friction passes through this axis and hence induces no torque whatsoever. There are however, two other torques acting on the cube i) that due to gravity given by mgx and ii) that due to the normal reaction -N(x+ak/2) = -mg(x+ak/2).

Thus, there is a net torque of -mgak/2 acting on the cube at any given point of time w.r.t to the axis CC'. The initial angular momentum of the cube w.r.t to the axis CC' is mvoa/2 is thus, reduced as time passes given by,








This clearly explains how angular momentum is lost w.r.t to the axis CC'.

Irodov Problem 1.270

Consider an infinitesimally small piece of the rod of length dx at a distance x from O'. If the linear density (mass per unit length) of the rod is p then the mass of this piece if pdx. The piece rotates about the axis OO' at a radius of and so the centrifugal force acting on the rod (as seen in the rotating frame) is given by . The component of the centrifugal force that is perpendicular to the rod is responsible for inducing a torque on the rod and is given by,
as shown in the figure. The moment arm of the force is x and so the torque acting on the piece of rod is given by,




The net torque acting on the rod due to the centrifugal force is given by,








The force of gravity mg acts at the CG (center) of the rod at a distance l/2 from O' and generates a counter-torque to the torque due to the centrifugal force. The component of gravity in the direction perpendicular to the rod is given by and thus the torque generated by gravity if given by . In order for the rod to rotate at a stable angle, the torque due to gravity must exactly counter-balance that due to the centrifugal force. Thus, we have,

Irodov Problem 1.269

In this problem as the rod rotates, each point along the rod rotates at a different radius and thus experiences a different centrifugal force. The centrifugal force acts in the radially outward direction and generates a torque on the rod trying to make the rod perpendicular to the axis of rotation.

Consider an infinitesimally small length of the rod of length dx that is at a distance x from the center of the rod O. Let the linear density (mass per unit length) of the rod be p. The mass of this infinitesimally small piece of the rod is pdl. The radius of rotation of this piece of rod, as shown in the figure is . The piece thus, experiences a centrifugal force acting radially outwards of magnitude,



The component of this force perpendicular to the length of the rod is responsible for inducing the torque on the rod and is given by,



The moment arm of this force on the infinitesimally small piece of rod is x, and so the torque is given by,





The net torque acting on the part BC of the rod is exactly equal to that acting on the part AC. The net torque can be found by integrating dT for all x over this part of the rod given by,

Irodov Problem 1.268

Let be a certain radius vector of a point rotating with an angular velocity vector. Let the unit vector in the direction of be. Let us choose and x,y,z Cartesian coordinate system such that the z-axis aligns with the angular velocity vector and the point rotates in the x,y plane. Then we have,







In Eqn (1) w is the magnitude of the angular velocity vector. Let r be the magnitude of , then we have,









Differentiating once again,















When seen from the rotating frame both these acceleration appear as pseudo forces with a negative sign. Now consider a infinitesimally small piece of the rotating body of mass dm. The centrifugal force and the Coriolis force experienced by this mass are respectively given by,