Combined Rotation and Translation , Rolling Motion

Combined Rotation & Translation:

In this type of motion, the axis of rotation is not stationary. If a body rotates about an axis with angular velocity ω then the linear velocity of any particle located at a distance r from the axis of rotation, measured with respect to the axis of rotation

$ \displaystyle \vec{v} = \vec{\omega}\times \vec{r} $

If the axis of rotation also moves with velocity $\vec{v_o}$ with respect to a frame of reference, then the velocity of the particle relative to that frame will be
$ \displaystyle \vec{v} = \vec{\omega}\times \vec{r} + \vec{v_o}$

Rolling Motion :

If velocity of the surface on which a body is rolling is $\vec{v_s}$ (with respect to a frame) and R is the radius of the body, which is rotating with an angular velocity ω around an axis, then the condition for pure rolling is,

$ \displaystyle \vec{\omega}\times \vec{r} + \vec{v_o} = \vec{v_s} $

where $\vec{v_o}$ is the velocity of the axis with respect to the frame.

If $\vec{v_s}= 0 $ then the condition for rolling is vo = ω R

Illustration : A cylindrical drum, pushed along by a board rolls forward on the ground. There is no slipping at any contact. Find the distance moved by the man who is pushing the board, when axis of the cylinder covers a distance L

Solution : Let vo be the linear speed of the axis of the cylinder and ω be its angular speed about the axis.

As it does not slip on the ground hence , vo = ω R .

Where R is the radius of the cylinder.

Speed of the topmost point is

v = vo + ω R

v = vo + vo = 2 vo

Since time taken by the axis to move a distance L is equal to t = L/vo

In the same interval of time distance moved by the topmost point is

$\large s = 2v_o \times \frac{L}{v_o} = 2 L$

As there is no slipping between any point of contact hence distance moved by the man is 2L.

Energy of a Rolling body

The translational K.E. of a rolling body

$ \displaystyle K.E_t = \frac{1}{2}m v^2 $

and, the rotational K.E. of a rolling body

$ \displaystyle K.E_r = \frac{1}{2}I_o \omega^2 $

Since we know that Io = mk2 and for rolling ω = v/r

$ \displaystyle K.E_r = \frac{1}{2}m k^2(\frac{v^2}{r^2}) = \frac{1}{2}\frac{k^2}{r^2}mv^2 $

Therefore, the total K.E. of the rolling body is

K.E. = K.Et + K.Er

Putting v = r ω for rolling, we obtain the total K.E. in terms of ω as

$ \displaystyle K.E = \frac{1}{2} m (r\omega)^2 (1 + \frac{k^2}{r^2}) $

$ \displaystyle K.E = \frac{1}{2} (mr^2 + mk^2)\omega^2 $

Since mk2 = M.I. of a body about its center of mass O = Io

mk2 + mr2 = Io + mr2 = Ip

Where Ip = M.I of the body about P

Therefore we conclude that, the combined effects of translation of center of mass of a body and its rotation about an axis passing through the c.m. are equivalent to its pure rotation about an axis passing through the point of contact P, of the rolling body.

% energy of translation $ \displaystyle = \frac{K. E_t}{K.E} \times 100 $

$ \displaystyle = \frac{r^2}{k^2 + r^2} \times 100 $

% energy of rotation $ \displaystyle = \frac{k^2}{k^2 + r^2} \times 100 $

Illustration : A thin hollow sphere of mass m = 2kg, radius = 1/2 m rolling on a horizontal surface with a constant frequency n = 60 rpm. Find the total mechanical energy of the hollow sphere.

Solution : Total energy $ \displaystyle E = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $

For rolling v = r ω & I = (2/3)mr2

$ \displaystyle E = \frac{1}{2}m(r\omega)^2 + \frac{1}{2}(\frac{2}{3}mr^2)\omega^2 $

$ \displaystyle E = \frac{1}{2}mr^2\omega^2 (1+\frac{2}{3}) = \frac{5}{6}mr^2\omega^2 $

on putting the given values , E = 16.35 J.

Exercise : Referring to the previous illustration,

(a) what fraction of energy is translational?

(b) what is the ratio of translation & rotational K.E.?

Role of friction in Rolling Motion

If a body rolls without slipping such that the velocity of its centre of mass does not change then no frictional force acts on the body.

However, If a force acts on the body, the velocity of its centre of mass or its angular velocity about the centre changes and, there is a tendency of the body to slide at its point of contact. Thus, friction force acts on the body to oppose this tendency to slide.

Rolling on inclined plane :

METHOD – I (Force- torque method) :

When a body rolls without slipping along an inclined plane, there are three forces acting on the body

(1) mg – weight

(2) N – Normal reaction

(3) f – static friction

Force Equation :

Suppose that the body accelerates down the plane with an acceleration a.

The net force parallel to the plane

Fx = mg Sinθ – f = ma …(i)

Since the body does not move perpendicular to the plane, the net force acting on it along the perpendicular is zero.

Fy = N – mg Cosθ = 0 …(ii)

Torque Equation:

Since mg-> and N-> pass through the centre O of the rolling body, torque about O due to these is zero. The torque due to friction, clockwise about O is

τ = r f Sin90° = r f

We know that τ = Icm α = Ioα

r f = Ioα …(iii)

$\displaystyle f = \frac{I_o \alpha}{r}$

$\displaystyle f= \frac{I_o a}{r^2}$

From (i)

$\displaystyle mg sin\theta – \frac{I_o a}{r^2}= m a$

$\displaystyle mg sin\theta = (m+\frac{I_o}{r^2})a$

$\displaystyle a = \frac{mg sin\theta}{m+\frac{I_o}{r^2}}$

Also Read :

∗ Motion of Rigid body & Rotational Kinematics
∗ K.E of Rotation & Moment of Inertia
∗ Moment of inertia of a thin ring & Radius of Gyration
∗ Parallel & Perpendicular axes Theorem
∗ Torque & Pseudo Torque
∗ Angular momentum & its Conservation
∗ Relation b/w torque & angular acceleration

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