Single Choice

The moment of inertia of the system about the center of ring will be

A$$ 20 kg m^2 $$
Correct Answer
B$$ 40 kg m^2 $$
C$$ 10 kg m^2 $$
D$$ 60 kg m^2 $$

Solution

To solve this problem let's first calculate the moment of inertia of this ring about its center. To calculate the moment of inertia, we will calculate the moment of inertia of the rod and multiply it by four and then add the moment of inertia of the ring to it.
i.e $$I=[\frac{M(R\sqrt{2})^{2}}{12}+M(\frac{R}{\sqrt{2}})^2]*4+mR^2$$
Using $$M=6 kg, R=1m$$ we get,
$$I=20kgm^2$$
Hence, option $$(A)$$ is the correct answer.

To calculate the minimum value of the coefficient of static friction we procced as follows.
Let's now write the balanced force equation for this body. We have,
$$(4M+m)gsin\theta-F=(4M+m)a$$,
where $$F$$ is the force due to friction.
Solving this equation we have,
$$FR=I(\frac{a}{R})$$
solving, we have $$a=\frac{7g}{24}$$
For the system to not slide we must have the friction force greater than the sliding force.
i.e $$F=20a\le\mu N$$
or, $$20\le\mu (4M+m)gcos 30$$
Given $$M=6kg, m=4kg$$
Hence, $$\mu\ge \frac{5}{12\sqrt{3}}$$
$$\therefore \mu_{min}=\frac{5}{12\sqrt{3}}$$

SIMILAR QUESTIONS

Rotational Dynamics

Fig. shows a uniform solid block of mass $$M$$ and edge lengths $$a$$, $$b $$ and $$ c $$ . Its M.I about an axis passing through one corner and perpendicular (as shown) to the large face of the block is :

Rotational Dynamics

The M.I of solid cylinder of mass $$'M'$$, length $$'l'$$ and radius $$'R'$$ about a transverse axis passing through its centre is :

Rotational Dynamics

The surface density ($$mass/area$$) of a circular disc of radius $$a$$ depends on the distance from the center of $$ \rho (r) = A + Br$$. Find its moment of inertia about the line perpendicular to the plane of the disc through its centre.

Rotational Dynamics

The moment of inertia of a uniform solid right circular cone of mass $$10\ kg$$, height $$2\ m$$ and vertical angle $$90^o$$ above a diameter of its base is:

Rotational Dynamics

A uniform disc of radius $$R$$ has a round disc of radius $$R/3$$ cut as shown in fig. The mass of the remaining (shaded) portion of the disc equals $$M$$. Find the moment of inertia of such a disc relative to the axis passing through geometrical centre of original disc and perpendicular to the plane of the disc.

Rotational Dynamics

Two loops P and Q are made from a same uniform wire. The radii of P and Q are $$r_1$$ and $$r_2$$ respectively and their moments of inertia are $$l_1$$ and $$l_2$$ respectively. If $$\dfrac{l_2}{l_1}$$ = 4, then $$\dfrac{r_2}{r_1}$$ is equal to:

Rotational Dynamics

A disc of radius $$R$$, is cut from a disc of radius $$R_{2}$$ as shown in figure. if the disc left has mass $$M$$ then find the moment of inertia of the disc left about an axis passing through its centre and perpendicular to its plane

Rotational Dynamics

If the mass of the hydrogen atom is $$1.7\times 10^{-24}\ g$$ and interatomic distance in a molecule of hydrogen is $$4\times 10^{-8}\ cm$$, then the moment of inertia [in $$kg-m^{2}$$] of a molecule of hydrogen about the axis passing through the centre of mass and perpendicular to the line joining the atoms will be:-

Rotational Dynamics

A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R, as shown in the figure. The moment of inertia of this lamina about axes passing through O and P is $$l_o $$ and $$l_p$$,respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $$\dfrac {l_p}{l_o}$$ to the nearest integer is :

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