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
$$\begin{array}{cc} & f=F / 3[\because N=M g] \\ \because & f \leq \mu N=\mu M g \\ \Rightarrow & \frac{F}{3} \leq \mu M g \Rightarrow F \leq 3 \mu M g \\ \Rightarrow & F_{\max }=3 \mu M g \end{array}$$