The moment of inertia of a thin rod of mass 'm' and length '$$l$$', about an axis through its centre of gravity and perpendicular to its length is

A hollow rod has frictionless inner walls. Inside the rod are two small spheres that are kept on either side of the centre of rod. The rod is rotated about its central axis which is perpendicular to the plane of rotation. If the rod had been rotated by an impulsive torque which gave it an instantaneous angular velocity of 5rad/s, what will be the angular velocity after some finite but long time ‘t’? Assume no external forces act on rod after the impulse. Also state what will happen to the spheres in the centre of the rod. The rod has a mass =2kg & length =10cm. The two spheres have a mass =1kg each.

A

2.5 rad/s, spheres will stay where they are as there is no friction to move them

B

0 rad/s, spheres will move outwards and decrease velocity of rod to zero

C

1.25 rad/s, spheres will move to opposite ends of rod

D

2.5 rad/s, spheres will move to opposite ends of rod

A small connecting rod of mass 1.5 kg is suspended in a horizontal plane by two wires 1.25 m long. The wires are attached to the rod at points 120 mm on either side of the centre of gravity. If the rod makes 20 oscillations in 40 seconds, find the mass moment of inertia of the rod about a vertical axis through the centre of gravity.

A

0.014 kg-m2

B

0.015 kg-m2

C

0.016 kg-m2

D

0.017 kg-m2

Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m Find the equivalent length L of a simple pendulum swung about an axis.

A

1.35 m

B

1.42 m

C

1.48 m

D

1.50 m

For a steam engine, the following data is given: Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m calculate inertia force at θ=30 degrees from IDC.

A

19000 N

B

19064 N

C

19032 N

D

20064 N

Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m Find angular acceleration of connecting rod in rad/s2.

A

16.782

B

17.824

C

15.142

D

17.161

A particle of mass m is attached to a thin uniform rod of length a and mass 4m. The distance of the particle from the centre of mass of the rod is $$\frac{a}{2}.$$
Classical Mechanics mcq question image

The moment of inertia of the combination about an axis passing through a normal to the rod is

A

$$\frac{{64}}{{48}}m{a^2}$$

B

$$\frac{{91}}{{48}}m{a^2}$$

C

$$\frac{{27}}{{48}}m{a^2}$$

D

$$\frac{{51}}{{48}}m{a^2}$$

From the data given: Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m Find the correcting couple in N-m?

A

52.7

B

49.5

C

59.5

D

56.5

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. $$I$$P) is given by (where, A = Area of the section, $$I$$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

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

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

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

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.
Classical Mechanics mcq question image

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

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

The moment of inertia of a thin rod of mass 'm' and length '$$l$$', about an axis through its centre of gravity and perpendicular to its length is

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