$$(n-1)$$ 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 $$=\left(\frac{4 R}{3 \pi}, \frac{4 R}{3 \pi}\right)$$
Two discs of moments of inertia $$I_1$$ and $$I_2$$ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed $$\omega_1$$ and $$\omega_2$$ are brought into contact face to face with their axes of rotation coincident.
(a) Does the law of conservation of angular momentum apply to the situation? Why?
(b) Find the angular speed of the two discs system.
(c) Calculate the loss in kinetic energy of the system in the process.
(d) Account for this loss.
Consider the diagram below
Let the common angular velocity of the system is $$\omega$$.
(a) Yes, the law of conservation of angular momentum can be applied. Because, there is no net external torque on the system of the two discs.
External forces, gravitation and normal reaction, act through the axis of rotation, hence, produce no torque.
(b) By conservation of angular momentum
$$\begin{aligned} & L_f=L_i \\ \Rightarrow \quad & I \omega=I_1 \omega_1+I_2 \omega_2 \\ \Rightarrow \quad & \omega=\frac{I_1 \omega_1+I_2 \omega_2}{I}=\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2} \quad\left(\because I=I_1+I_2\right) \end{aligned}$$
$$\begin{aligned} \text{(c)}\quad & K_f=\frac{1}{2}\left(I_1+I_2\right) \frac{\left(I_1 \omega_1+I_2 \omega_2\right)^2}{\left(I_1+I_2\right)^2}=\frac{1}{2} \frac{\left(I_1 \omega_1+I_2 \omega_2\right)^2}{\left(I_1+I_2\right)} \\ & K_i=\frac{1}{2}\left(I_1 \omega_1^2+I_2 \omega_2^2\right) \\ & \Delta K=K_f-K_i=-\frac{I_1 I_2}{2\left(I_1+I_2\right)}\left(\omega_1-\omega_2\right)^2<0 \end{aligned}$$
(d) Hence, there is loss in KE of the system. The loss in kinetic energy is mainly due to the work against the friction between the two discs.
A disc of radius $$R$$ is rotating with an angular $$\omega_0$$ about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is $$\mu_\kappa$$.
(a) What was the velocity of its centre of mass before being brought in contact with the table?
(b) What happens to the linear velocity of a point on its rim when placed in contact with the table?
(c) What happens to the linear speed of the centre of mass when disc is placed in contact with the table?
(d) Which force is responsible for the effects in (b) and (c)?
(e) What condition should be satisfied for rolling to begin?
(f) Calculate the time taken for the rolling to begin.
(a) Before being brought in contact with the table the disc was in pure rotational motion hence, $$v_{C M}=0,$$.
(b) When the disc is placed in contact with the table due to friction velocity of a point on the rim decreases.
(c) When the rotating disc is placed in contact with the table due to friction centre of mass acquires some linear velocity.
(d) Friction is responsible for the effects in (b) and (c) .
(e) When rolling starts $$v_{C M}=\omega R$$.
where $$\omega$$ is angular speed of the disc when rolling just starts.
$$\begin{aligned} &\text { (f) Acceleration produced in centre of mass due to friction }\\ &a_{\mathrm{CM}}=\frac{F}{m}=\frac{\mu_k m g}{m}=\mu_k g . \end{aligned}$$
$$\begin{aligned} &\text { Angular retardation produced by the torque due to friction. }\\ &\alpha=\frac{\tau}{I}=\frac{\mu_{\mathrm{k}} m g R}{I} \quad\left[\because \tau=\left(\mu_k N\right) R=\mu_k m g R\right] \end{aligned}$$
$$\begin{array}{lll} \therefore & v_{\mathrm{CM}} & =u_{\mathrm{CM}}+a_{\mathrm{CM}} t \\ \\ \Rightarrow & v_{\mathrm{CM}} & =\mu_{\mathrm{k}} g t \\ \text { and } & \omega & =\omega_0+\alpha t \\ \Rightarrow & \omega & =\omega_0-\frac{\mu_k m g R}{I} t \end{array} \quad\left(\because u_{\mathrm{CM}}=0\right)$$
$$\begin{aligned} &\text { For rolling without slipping, } \frac{v_{\mathrm{CM}}}{R}=\omega\\ &\begin{aligned} \Rightarrow \quad \frac{v_{\mathrm{CM}}}{R} & =\omega_0-\frac{\mu_k m g R}{I} t \\ \frac{\mu_k g t}{R} & =\omega_0-\frac{\mu_k m g R}{I} t \\ t & =\frac{R \omega_0}{\mu_k g\left(1+\frac{m R^2}{l}\right)} \end{aligned} \end{aligned}$$
Two cylindrical hollow drums of radii $$R$$ and $$2 R$$, and of a common height $$h$$, are rotating with angular velocities $$\omega$$ (anti-clockwise) and $$\omega$$ (clockwise), respectively. Their axes, fixed are parallel and in a horizontal plane separated by $$3 R+\delta$$. They are now brought in contact $$(\delta \rightarrow 0)$$
(a) Show the frictional forces just after contact.
(b) Identify forces and torques external to the system just after contact.
(c) What would be the ratio of final angular velocities when friction ceases?
(a) Consider the situation shown below, we have shown the frictional forces.
(b) $$F^{\prime}=F=F^{\prime}$$ where $$F^{\prime}$$ and $$F^{\prime}$$ are external forces through support.
$$\Rightarrow \quad F_{\text {net }}=0 \quad \text {(one each cylinder)}$$
External torque = $$F \times 3 R,(\text { anti-clockwise })$$
(c) Let $$\omega_1$$ and $$\omega_2$$ be final angular velocities of smaller and bigger drum respectively (anti-clockwise and clockwise respectively)
Finally, there will be no friction.
Hence, $$\quad R \omega_1=2 R \omega_2 \Rightarrow \frac{\omega_1}{\omega_2}=2$$