Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle ABC of side/cm (as shown in the figure). The moment of inertia of the system about a line AX perpendicular to AB and in the plane of ABC, in gram-cm² units will be

³⁄₂ ml²
³⁄₄ ml²
2 ml²
⁵⁄₄ ml²

Solution

I_AX = m(AB)²+ m(OC)² = ml² + m (l cos 60º)2= ml² + ml²/4 = 5/4 ml²

A round disc of moment of inertia I₂ about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia I₁ rotating with an angular velocity ω about the same axis. The final angular velocity of the combination of discs is

\(\frac{(I_{1}+I_{2})\omega }{I_{1}}\)
\(\frac{I_{2}\omega }{I_{1}+I_{2}}\)
ω
\(\frac{I_{1}\omega }{I_{1}+I_{2}}\)

Solution

Angular momentum will be conserved

I₁ω= I₁ω' + I₂ω' ⇒ \(\frac{I_{1}\omega }{I_{1}+I_{2}}\)

The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is

1 :√2
1 : 3
2 : 1
√5 : √6

Solution

Consider a system of two particles having masses m₁ and m₂ . If the particle of mass m₁ is pushed towards the centre of mass particles through a distance d, by what distance would the particle of mass m₂ move so as to keep the mass centre of particles at the original position?

\(\frac{m_{2}}{m_{1}}d\)
\(\frac{m_{1}}{m_{1}+m_{2}}d\)
\(\frac{m_{1}}{m_{2}}d\)
d

Solution

m₁d = m₂ d₂ ⇒ d₂ = \(\frac{m_{1}}{m_{2}}d\)

A wheel having moment of inertia 2 kg-m² about its vertical axis, rotates at the rate of 60 rpm about this axis, The torque which can stop the wheel’s rotation in one minute would be

^π⁄₁₈ Nm
^2π⁄₁₅ Nm
^π⁄₁₂ Nm
^π⁄₁₅ Nm

Solution

An annular ring with inner and outer radii R₁ and R₂ is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring , ^F₁⁄_F₂ is

\(\left ( \frac{R_{1}}{R_{2}} \right )^{2}\)
\(\frac{R_{2}}{R_{1}}\)
\(\frac{R\, 1}{R\, 2}\)
1

Solution

In a bicycle, the radius of rear wheel is twice the radius of front wheel. If r_F and r_F are the radii, v_r and v_r are the speed of top most points of wheel. Then

v_r = 2 v_F
v_F = 2 v_r
v_r = v_F
v_F > v_r

Solution

The velocity of the top point of the wheel is twice that of centre of mass. And the speed of centre of mass is same for both the wheels.

A sphere rolls down on an inclined plane of inclination θ.What is the acceleration as the sphere reaches bottom?

⁵⁄₇g sin θ
³⁄₅g sin θ
²⁄₇g sin θ
²⁄₅g sin θ

Solution

A toy car rolls down the inclined plane as shown in the fig.It loops at the bottom. What is the relation between H and h?

^H⁄_h=2
^H⁄_h=3
^H⁄_h=4
^H⁄_h=5

Solution

Velocity at the bottom and top of the circle is \(\sqrt{5gr}\) and \(\sqrt{gr}\). Therefore (1/2)M(5gr) = MgH and (1/2) M (gr) = Mgh.

Fig. shows a disc rolling on a horizontal plane with linear velocity v. Its linear velocity is v and angular velocity is ω.Which of the following gives the velocity of the particle P on the rim of the disc

v (1 + cos θ)
v (1 - cos θ)
v (1 + sin θ)
v (1 - sin θ)

Solution

Velocity of P = (NP)ω = (NM + MP)ω

= r(r + sin θ)ω = v(1 + sinθ)

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