A disc is a spinning with an angular velocity $$\omega $$ rad/s about the axis of spin. The couple applied to the disc causing precession will be (where $$I$$ = Mass moment of inertia of the disc and $$\omega {\text{P}}$$ = Angular velocity of precession of the axis of spin)

A disc is spinning with an angular velocity? rad/s about the axis of spin. The couple applied to the disc causing precession will be

A

1/2Iω2

B

Iω2

C

1/2 Iω ωp

D

Iω ωp

A disc is a spinning with an angular velocity ω rad/s about the axis of spin. The couple applied to the disc causing precession will be

A

1/2 Iω2

B

Iω2

C

1/2 Iωωp

D

Iωωp

A disc spinning on its axis at 20 rad/s will undergo precession when a torque 100 N-m is applied about an axis normal to it at an angular speed, if mass moment of inertia of the disc is the 1 kg-m2

A

2 rad/s

B

5 rad/s

C

10 rad/s

D

20 rad/s

Two monochromatic waves having frequencies $$\omega $$ and $$\omega + \Delta \omega \left( {\Delta \omega \ll \omega } \right)$$ and corresponding wavelengths $$\lambda $$ and $$\lambda - \Delta \lambda \left( {\Delta \lambda \ll \lambda } \right)$$ of same polarization, travelling along X-axis are superimposed on each other. The phase velocity and group velocity of the resultant wave are respectively given by

A

$$\frac{{\omega \lambda }}{{2\pi }},\,\frac{{\Delta \omega {\lambda ^2}}}{{2\pi \Delta \lambda }}$$

B

$$\omega \lambda ,\,\frac{{\Delta \omega {\lambda ^2}}}{{\Delta \lambda }}$$

C

$$\frac{{\omega \Delta \lambda }}{{2\pi }},\,\frac{{\Delta \omega \Delta \lambda }}{{2\pi }}$$

D

$$\omega \Delta \lambda ,\,\omega \Delta \lambda $$

The moment of inertia of a uniform sphere of radius, r about an axis passing through its centre is given by $$\frac{2}{5}\left( {\frac{{4\pi }}{3}{r^5}\rho } \right).$$ A rigid sphere of uniform mass density $$\rho $$ and radius R has two smaller spheres of radii $$\frac{R}{2}$$ hollowed out of it as shown in the figure given below.

The moment of inertia of the resulting body about Y-axis is

A

$$\frac{{\pi \rho {R^5}}}{4}$$

B

$$\frac{{5\pi \rho {R^5}}}{{12}}$$

C

$$\frac{{7\pi \rho {R^5}}}{{12}}$$

D

$$\frac{{3\pi \rho {R^5}}}{4}$$

According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through center of gravity (i.e. $${I_{\text{P}}}$$) is given by (where, A = Area of the section, $${I_{\text{G}}}$$ = Moment of inertia of the section about an axis passing through its C.G. and h = Distance between C.G. and the parallel axis.)

A

$${I_{\text{P}}} = {I_{\text{G}}} + {\text{A}}{{\text{h}}^2}$$

B

$${I_{\text{P}}} = {I_{\text{G}}} - {\text{A}}{{\text{h}}^2}$$

C

$${I_{\text{P}}} = \frac{{{I_{\text{G}}}}}{{{\text{A}}{{\text{h}}^2}}}$$

D

$${I_{\text{P}}} = \frac{{{\text{A}}{{\text{h}}^2}}}{{{I_{\text{G}}}}}$$

A disc is rotating with ω =10rad/s about a fixed central axis which is perpendicular to its plane. The disc has a mass =2kg & radius =10cm. A small particle of mass 100gm is put slowly on the disc’s outer circumference. There is sufficient friction between the disc and the particle. What will be the new angular velocity of the system?

A

10 rad/s

B

9.09 rad/s

C

9.51 rad/s

D

Can’t conserve angular momentum because of friction

In order to have a connecting rod equally strong in buckling about X-axis and Y-axis, the moment of inertia about X-axis should be __________ the moment of inertia about Y-axis.

A

Equal to

B

Twice

C

Three times

D

Four times

The moment of inertia of a planar disc about a diameter is 8kgm2. What is the moment of inertia about an axis passing through its centre and perpendicular to the plane of disc?

A

8kgm2

B

16kgm2

C

4kgm2

D

2√2kgm2

The parallel axis theorem can add any angle varied moment of inertias to give the perpendicular moment of inertia and it can be used in the determination of the moment of inertia about inclined axis.

A

True

B

False