The vector sum of a system of non-collinear forces acting on a rigid body is given to be non-zero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero, does this mean that it is necessarily zero about any arbitrary point?
No, not necessarily.
Given, $$\quad \sum_\limits i F_i \neq 0$$
$$\text { The sum of torques about a certain point } O, \sum_\limits i \mathbf{r}_i \times \mathbf{F}_i=0$$
$$\begin{aligned} &\text { The sum of torques about any other point } \mathrm{O}^{\prime}\\ &\sum_i\left(\mathbf{r}_i-\mathbf{a}\right) \times \mathbf{F}_i=\sum_i \mathrm{r}_i \times \mathbf{F}_i-\mathbf{a} \times \sum_i \mathrm{~F}_i \end{aligned}$$
Here, the second term need not vanish.
Therefore, sum of all the torques about any arbitrary point need not be zero necessarily.
A wheel in uniform motion about an axis passing through its centre and perpendicular to its plane is considered to be in mechanical (translational plus rotational) equilibrium because no net external force or torque is required to sustain its motion. However, the particles that constitute the wheel do experience a centripetal the acceleration directed towards the centre. How do you reconcile this fact with the wheel being in equilibrium?
How would you set a half wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane? Will you require external forces to sustain the motion?
Wheel is a rigid body. The particles that constitute the wheel do experience a centripetal acceleration directed towards the centre. This acceleration arises due to internal elastic forces, which cancel out in pairs. In a half wheel, the distribution of mass about its centre of mass (through which axis of rotation passes) is not symmetrical. Therefore, the direction of angular momentum of the wheel does not coincide with the direction of its angular velocity. Hence, an external torque is required to maintain the motion of the half wheel.
A door is hinged at one end and is free to rotate about a vertical axis (figure). Does its weight cause any torque about this axis? Give reason for your answer.
Consider the diagram, where weight of the door acts along negative y-axis.
A force can produce torque only along a direction normal to itself as $$\tau=\mathbf{r} \times \mathbf{F}$$. So, when the door is in the $$x y$$-plane, the torque produced by gravity can only be along $$\pm z$$ direction, never about an axis passing through $$y$$-direction. Hence, the weight will not produce any torque about $$y$$-axis.
$$(n-1)$$ equal point masses each of mass $m$ are placed at the vertices of a regular n-polygon. The vacant vertex has a position vector a with respect to the centre of the polygon. Find the position vector of centre of mass.
Let $$\mathbf{b}$$ be the position vector of the centre of mass of a regular $$n$$-polygon. $(n-1)$ equal point masses are placed at $$(n-1)$$ vertices of the regular $$n$$-polygon, therefore, for its centre of mass
$$\begin{gathered} r_{\mathrm{CM}}=\frac{(n-1) m b+m a}{(n-1) m+m}=0 \quad(\because \text { Centre of mass lies at centre }) \\ \Rightarrow \quad (n-1) m b+m a=0 \\ \Rightarrow \quad b=-\frac{a}{(n-1)} \end{gathered}$$
Find the centre of mass of a uniform (a) half-disc, (b) quarter-disc.
$$\begin{aligned} &\text { Let } M \text { and } R \text { be the mass and radius of the half-disc, mass per unit area of the half-disc }\\ &m=\frac{M}{\frac{1}{2} \pi R^2}=\frac{2 M}{\pi R^2} \end{aligned}$$
(a) The half-disc can be supposed to be consists of a large number of semicircular rings of mass $$d m$$ and thickness $$d r$$ and radii ranging from $$r=0$$ to $$r=R$$.
Surface area of semicircular ring of radius $$r$$ and of thickness $$d r=\frac{1}{2} 2 \pi r \times d r=\pi r d r$$
$$\therefore$$ Mass of this elementary ring, $$d m=\pi r d r \times \frac{2 M}{\pi R^2}$$
$$d m=\frac{2 M}{R^2} r d r$$
If $$(x, y)$$ are coordinates of centre of mass of this element,
then, $$(x, y)=\left(0, \frac{2 r}{\pi}\right)$$
Therefore, $$x=0 \text { and } y=\frac{2 r}{\pi}$$
Let $$x_{C M}$$ and $$y_{C M}$$ be the coordinates of the centre of mass of the semicircular disc.
$$\begin{aligned} &\begin{aligned} \text { Then }\quad x_{\mathrm{CM}} & =\frac{1}{M} \int_0^R x d m=\frac{1}{M} \int_0^R 0 d m=0 \\ y_{\mathrm{CM}} & =\frac{1}{M} \int_0^R y d m=\frac{1}{M} \int_0^R \frac{2 r}{\pi} \times\left(\frac{2 M}{R^2} r d r\right) \\ & =\frac{4}{\pi R^2} \int_0^R r^2 d r=\frac{4}{\pi R^2}\left[\frac{r^3}{3}\right]_0^R \\ & =\frac{4}{\pi R^2} \times\left(\frac{R^3}{3}-0\right)=\frac{4 R}{3 \pi} \end{aligned} \end{aligned}$$
$$\therefore \text { Centre of mass of the semicircular disc }=\left(0, \frac{4 R}{3 \pi}\right)$$
(b) Centre of mass of a uniform quarter disc. Mass per unit area of the quarter disc $$=\frac{M}{\frac{\pi R^2}{4}}=\frac{4 M}{\pi R^2}$$
Using symmetry
For a half-disc along $$y$$-axis centre of mass will be at $$x=\frac{4 R}{3 \pi}$$
For a half-disc along $$x$$-axis centre of mass will be at $$x=\frac{4 R}{3 \pi}$$
Hence, for the quarter disc centre of mass $$=\left(\frac{4 R}{3 \pi}, \frac{4 R}{3 \pi}\right)$$
