Question

A uniform cube with an edge a rests on a horizontal plane whose friction coefficient equals k. The cube is set in motion with an initial velocity, travels some distance over the plane and comes to a standstill. Explain the disappearance of the angular momentum of the cube relative to the axis lying in the plane at right angles to the cube's motion direction. Find the distance between the resultants of gravitational forces and the reaction forces exerted by the supporting plane.
A. $\Delta x = \dfrac{{ka}}{4}$
B. $\Delta x = \dfrac{{3ka}}{2}$
C. $\Delta x = \dfrac{{ka}}{2}$
D. None of these

Answer 1

As the cube is in rotational equilibrium, so torque about centre of mass of all the forces must be zero. In this question, the cube is not moving in the horizontal direction. So, N=mg
Here N is the normal reaction, m is the mass of the cube and g is the acceleration due to gravity. When we take torque to be equal to zero then by making an equation, we can find the distance between the point of application of normal reaction and the Centre of gravity of the cube.

Complete step by step solution:
Given:- edge of cube rest on horizontal plane.
Coefficient of friction between cube and plane = $k$
When the cube is given an underlying speed or velocity on a table towards any direction it procures a rakish energy about a pivot that is angular momentum about an axis on the table which is normal to the initial velocity and just lower to the centre of gravity.
When the cube stops moving, the angular momentum acquired by the cube will vanish or fade away which is the effect of torque on the cube.
Frictional forces are unable to behave like the angular momentum because the frictional forces are in the plane with the axis.
But if force of the normal reaction is infrequent, the torque will make angular momentum fade away.
We can evaluate distance between point of application of normal reaction and the Centre of gravity of the cube =$\Delta x$:
Take the moment about the centre of gravity of all the forces. This must disappear because the cube cannot turn on the table.
Then if the force of friction is:
$fr\dfrac{a}{2} = N.\Delta x$………………………………………………………………………………… (I)
Where $N$= force acting in the perpendicular direction
Hence, force $N{\text{ }} = {\text{ }}mg$
Where $m$= mass and
$g$= acceleration due to gravity
So, equation (I) becomes:
$\therefore fr\dfrac{a}{2} = mg.\Delta x$………………………………………………………………………... (II)
Frictional force is force generated due to motion between two surfaces.
$fr = k \times m \times g$
So, equation (II) becomes:

$$k \times m \times g \times \dfrac{a}{2} = m \times g \times \Delta x \\\ \therefore k \times \dfrac{a}{2} = \Delta x \\\ \therefore \Delta x = K \times \dfrac{a}{2} \\\$$

Therefore the distance between point of application of normal reaction and the Centre of gravity of the cube = $\Delta x = \dfrac{{ka}}{2}$
Here angular momentum is continuously decreasing about the centre.

Hence, option (C) is the correct answer.

Note: We can represent clockwise angular momentum by positive sign, and counter clockwise angular momentum by negative sign. We must keep in mind that as angular momentum is a vector quantity the proper direction must be taken for an exact answer.