We have read about the dynamics of particle motion before where we discovered that they were capable of undergoing translational motion only, but as we know, rigid bodies can undergo translational as well as rotational motion. So, in such cases, both the linear and the angular velocity need to be analyzed.

In order to simplify these problems, we define the translational and rotational motion of the body separately. The rotational motion of the object is referred to as the rotational motion of an object about a fixed axis. In this section, we will discuss the dynamics of rotational motion of an object about a fixed axis.

A rotating body, as can be seen in the figure above, will have a point that has zero velocity, about which the object undergoes rotational motion. This point can be on the body or at any point away from it. Since the axis of rotation is fixed, we consider only those components of the torques applicable on the object that is along this axis as only these components cause rotation in the body. The perpendicular component of the torque will tend to turn the axis of rotation for the object from its position.

These results in the emergence of some necessary forces of constraint which finally tends to cancel the effect of these perpendicular components, thus restricting the motion of the axis from its fixed position, rendering its position to be maintained. Since the perpendicular components cause no effect; these components are not considered during the calculation. For any rigid body undergoing a rotational motion about a fixed axis, we need to consider only the forces that lie in planes perpendicular to the axis.

Forces which are parallel to the axis will give torques perpendicular to the axis and need not be taken into account. Also, only the components of position vector that are perpendicular to the axis are considered. Components of position vectors along the axis result in torques perpendicular to the axis and thus are not to be taken into account.

A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs of spring constant k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in horizontal plane.

The unstretched length of each spring is L.

The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall.

The disk rolls without slipping with velocity Vo.

Coefficient of friction is µ.

Suppose we need to find the force when the disc is displaced by a distance ‘x’.

It is said that the disc moves without slipping. So, the frictional force is static and hence we cannot say f = μ N, instead we consider it to be ‘f’.

Also, a = α R —– (1)

From free body diagram, 2kx – f = ma —— (2)

f.R = Iα —— (3)

Where, I = mR22 Using (1), (2) and (3)

we get, Net force = 4kx3

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