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, $$\quad R \omega_1=2 R \omega_2 \Rightarrow \frac{\omega_1}{\omega_2}=2$$
A uniform square plate $$S$$ (side $$c$$) and a uniform rectangular plate $$R$$ (sides $$b, a$$) have identical areas and masses.
Show that
(a) $$I_{x R} / I_{x S}<1$$
(b) $$I_{y R} / I_{y s}>1$$
(c) $I_{z R} / I_{z S}>1$
By given question
Area of square $$=$$ Area of rectangular plate
$$\Rightarrow \quad c^2=a \times b \Rightarrow c^2=a b$$
Now by definition
(a) $$\frac{I_{x R}}{I_{x S}}=\frac{b^2}{c^2} \quad\left[\because I \propto(\text { area })^2\right]$$
From the diagram $$b< c$$
$$\Rightarrow \quad \frac{I_{x R}}{I_{x S}}=\left(\frac{b}{c}\right)^2 < 1 \Rightarrow I_{x R} < I_{x S}$$
$$\begin{aligned} &\text { (b) }\\ &\frac{I_{y_R}}{I_{y_S}}=\frac{a^2}{c^2} \end{aligned}$$
$$\begin{aligned} &\text { as }\\ &\begin{aligned} & a>c \\ & \frac{I_{y_R}}{I_{y_S}}=\left(\frac{a}{c}\right)^2>1 \end{aligned} \end{aligned}$$
$$\begin{aligned} & \text { (c) } I_{z R}-I_{z S} \propto\left(a^2+b^2-2 c^2\right)=a^2+b^2-2 a b=(a-b)^2 \quad\left[\because c^2=a b\right] \\ & \Rightarrow \quad\left(I_{z R}-I_{z S}\right)>0 \Rightarrow \frac{I_{z R}}{I_{z S}}>1 \end{aligned}$$
A uniform disc of radius $$R$$, is resting on a table on its rim. The coefficient of friction between disc and table is $$\mu$$ (figure). Now, the disc is pulled with a force $$F$$ as shown in the figure. What is the maximum value of $$F$$ for which the disc rolls without slipping?
Consider the diagram below
Frictional force $$(f)$$ is acting in the opposite direction of $$F$$.
Let the acceleration of centre of mass of disc be a then
$$F-f=M a\quad \text{.... (i)}$$
where $$M$$ is mass of the disc
The angular acceleration of the disc is
$$\alpha=a / R\quad \text{(for pure rolling)}$$
from
$$\tau=l \alpha$$
$$f R=\left(\frac{1}{2} M R^2\right) \alpha \Rightarrow f R=\left(\frac{1}{2} M R^2\right)\left(\frac{a}{R}\right)$$
$$M a=2 f\quad \text{.... (ii)}$$
From Eqs. (i) and (ii), we get
$$\begin{array}{cc} & f=F / 3[\because N=M g] \\ \because & f \leq \mu N=\mu M g \\ \Rightarrow & \frac{F}{3} \leq \mu M g \Rightarrow F \leq 3 \mu M g \\ \Rightarrow & F_{\max }=3 \mu M g \end{array}$$