02 Centre of Rotation

02 Centre of Rotation#

Aim#

To show that a free rotating body rotates around its centre of mass.

Subjects#

  • 1D40 (Motion of the Center of Mass)

  • 1Q60 (Rotational Stability)

Diagram#

../../../../../_images/figure_03.png

Fig. 8 .#

Equipment#

  • Two different wooden spheres \(\left(\mathrm{d}_{1}=5 \mathrm{~cm}\right.\) and \(\left.\mathrm{d}_{2}=4 \mathrm{~cm}\right)\), connected by a light stick.

  • Two different wooden spheres ( \(\mathrm{d}_{1}=5 \mathrm{~cm}\) and \(\mathrm{d}_{2}=4 \mathrm{~cm}\) ), connected by a rubber band (length unwound around \(50 \mathrm{~cm}\) ).

  • Board with square grid \(\left(10 \times 10 \mathrm{~cm}^{2}\right)\).

  • Video camera on tripod (see Diagram).

  • Projector to project image of square grid to the audience.

Presentation#

The two spheres connected to each other by means of the light stick are balanced (see Diagram to show where the centre of mass (\(\mathrm{CM}\)) is located. As can be observed, the location of the \(\mathrm{CM}\) divides the distance ( \(d\) ) between the two centres of the spheres in roughly \(\frac{1}{3} d\) and \(\frac{2}{3} d\).

Then the system of the two spheres connected by the rubber band is taken. The rubber band is twisted (see Diagram and Remarks). The system is placed on the board with the grid and then left by itself. The combination starts to rotate while unwinding the twisted rubber band. As can be observed, the combination rotates around a fixed point. This can be recognized as the \(\mathrm{CM}\) shown in the first part of the demonstration. During the rotation the distance between the spheres increases, but its centre of rotation keeps the ratio \(\frac{1}{3}:\frac{2}{3}\).

Explanation#

\(\underline{1 .}\)

The total mass is located at the \(\mathrm{CM}\). No external forces are acting, so the \(\mathrm{CM}\) has to remain at its position on the board as Newton’s first law tells us. (When the system should rotate around any other point but the \(\mathrm{CM}\), the \(\mathrm{CM}\) performs a rotation and an external torque should be needed for that.)

../../../../../_images/figure_12.png

Fig. 9 .#

\(\underline{2 .}\)

When the body rotates it does so around an axis perpendicular to \(d\). When no external forces are acting, the angular momentum vector \(\vec{L}\) remains constant. The body in our demonstration consists of two masses: \(m_{1}\) and \(m_{2}\) (see Figure 9a).

\(\vec{L}=\vec{L}_{1}+\vec{L}_{2} ; \vec{L}=\vec{R}_{1} \times m_{1} \vec{v}_{1}+\vec{R}_{2} \times m_{2} \vec{v}_{2}\).

The axis of rotation characterizes itself by the fact that, at any position on this axis, \(\vec{L}\) will have the same magnitude and direction (at \(O^{'}\) the horizontal components of \(L_{1 a}\) and \(L_{2 a}\) cancel, and their vertical components add up to \(\vec{L}\): see Sources. This means that only one axis of rotation is possible. When, for instance, an axis of rotation is chosen passing through the centre of \(\mathrm{m}_{1}\), then the total angular momentum adds up to \(L_{2 b}\) (see Figure 9 b). And so, being constant in magnitude, its direction constantly changes (describing a cone). Such a situation needs an external torque.

In our demonstration, there is no external torque and the sphere-system rotates around an axis somewhere between the two spheres (Figure 10). At \(O\), \(\vec{L}=\vec{R}_{1} \times m_{1} \vec{v}_{1}+\vec{R}_{2} \times m_{2} \vec{v}_{2}\), directed along the axis of rotation. At \(O^{'}\), the horizontal components of \(L_{1}\) and \(L_{2}\) need to cancel in order to keep \(L\) along the axis of rotation.

\(\left|\vec{L}_{1}\right|=\frac{R_{1} m_{1} v_{1}}{\cos \alpha}\)

\(L_{1 h}=\frac{R_{1} m_{1} v_{1}}{\cos \alpha} \sin \alpha=R_{1} m_{1} v_{1} \tan \alpha=m_{1} v_{1} y\)

In the same way: \(L_{2 h}=m_{2} v_{2} y, v=\omega R\), so \(L_{1 h}=m_{1} \omega R_{1} y L_{1 h}=m_{1} \omega R_{1} y\) and

\(L_{2 h}=m_{2} \omega R_{2} y L_{2 h}=m_{2} \omega R_{2} y\). These two are equal when \(m_{1} R_{1}=m_{2} R_{2}\), and this is so when the axis of rotation passes through the \(\mathrm{CM}\).

../../../../../_images/figure_2.png

Fig. 10 .#

Remarks#

  • The easiest way to wind the rubber band is done by holding the small sphere in your hand and whirling the large sphere round over the ground.

  • The wooden spheres having diameters of \(4 -\) and \(5 \mathrm{~cm}\) will have a mass ratio of \(1\) to \(1.95\) , so very close to a factor \(2\).

  • See also the demonstration [Dumb-Bell](/book/1 mechanics/1Q rot dyn/1Q40 Cons of Angular Momentum/1Q4007 Dumb Bell/1Q4007.md).

Sources#

  • Mansfield, M and O’Sullivan, C., Understanding physics, pag. 139-141

  • PSSC, College Physics, pag. 352-354

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