A rod PQ of mass M and length L is hinged at end P.The rod is kept horizontal by a massless string tied to point Q as shown in figure. When string is cut, the initial angular acceleration of the rod is

g/L
2g/L
^2g⁄_3L
^3g⁄_2L

Solution

Three masses are placed on the x-axis : 300 g at origin, 500g at x= 40 cm and 400 g at x = 70 cm. The distance of the centre of mass from the origin is

40 cm
45 cm
50 cm
30 cm

Solution

The moment of inertia of a uniform circular disc is maximum about an axis perpendicular to the disc and passing through

Solution

According to parallel axis theorem of the moment of Inertia

I = I_cm + md²

d is maximum for point B so I_max about B.

The instantaneous angular position of a point on a rotating wheel is given by the equation θ(t) = 2t³– 6t². The torque on the wheel becomes zero at

t = 1s
t= 0.5s
t= 0.25 s
t = 2s

Solution

The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is I₀. Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is

I₀+ ML²/2
I₀+ ML²/4
I₀+ 2ML²
I₀+ ML²

Solution

By theorem of parallel axes,

I = I_cm + Md²

I = I₀ + M (L/2)²= I₀ + ML²/4

A circular disk of moment of inertia I_t is rotating in a horizontal plane, its symmetry axis, with a constant angular speed ω_i. Another disk of moment of inertia I_b is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed ω_f. The energy lost by the initially rotating disk to friction is

\(\frac{1}{2}\frac{I_{b}^{2}}{(I_{t}+I_{b})}\omega _{i}^{2}\)
\(\frac{I_{t}^{2}}{(I_{t}+I_{b})}\omega _{i}^{2}\)
\(\frac{I_{t}-I_{b}}{(I_{t}+I_{b})}\omega _{i}^{2}\)
\(\frac{1}{2}\frac{I_{b}I_{t}}{(I_{t}+I_{b})}\omega _{i}^{2}\)

Solution

A tube of length L is filled completely with an incompressible liquid of mass M and closed at both ends.The tube is then rotated in a horizontal plane about one of its ends with uniform angular speed ω. What is the force exerted by the liquid at the other end?

\(\frac{ML\omega ^{2}}{2}\)
MLω²
\(\frac{ML\omega ^{2}}{4}\)
\(\frac{ML\omega ^{2}}{8}\)

Solution

Tube may be treated as a particle of mass M at distance L/2 from one end.

Centripetal force = Mrω² = \(\frac{ML\omega ^{2}}{2}\)

A drum of radius R and mass M, rolls down without slipping along an inclined plane of angle θ. The frictional force

dissipates energy as heat.
decreases the rotational motion.
decreases the rotational and translational motion.
converts translational energy to rotational energy

Solution

Net work done by frictional force when drum rolls down without slipping is zero.

W_net = 0, W_trans. + W_rot.= 0

ΔK_trans.+ ΔK_rot. = 0

ΔK_trans= –ΔK_rot.

i.e., converts translation energy to rotational energy.

Two bodies have their moments of inertia I and 2I respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momenta will be in the ratio

2 : 1
1 : 2
√2 : 1
1 : √2

Solution

The moment of inertia of a uniform circular disc of radius ‘R’and mass ‘M’ about an axis passing from the edge of the disc and normal to the disc is

MR²
¹⁄₂ MR²
³⁄₂ MR²
⁷⁄₂ MR²

Solution

M.I. of a uniform circular disc of radius ‘R’ and mass ‘M’ about an axis passing through C.M.and normal to the disc is

I_CM.=¹⁄₂ MR²

From parallel axis theorem

I_T = I_CM. + MR² = ¹⁄₂ MR² + MR² = ³⁄₂ MR²

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