## ABC is a plane lamina of the shape of an equilateral triangle . D, E are the mid points AB , AC and G is the centroid of the lamina …

Q: ABC is a plane lamina of the shape of an equilateral triangle . D, E are the mid points AB , AC and G is the centroid of the lamina . Moment of inertia of the lamina about an axis passing through G and perpendicular to the plane ABC is Io . If part ADE is removed the moment of inertia of the remaining part about the same axis is $\frac{N I_o}{16}$ where N is an integer . The value of N is …

Click to See Solution :
Ans: (15)
Sol:

## A uniform rod of length l is pivoted at one of its ends on a vertical shaft of negligible radius . When the shaft rotates ….

Q: A uniform rod of length l is pivoted at one of its ends on a vertical shaft of negligible radius . When the shaft rotates at angular speed ω the rod makes an angle θ with it . To find θ equate the rate of change of angular momentum (direction going into paper) $\frac{m l^2}{12} \omega^2 sin\theta cos\theta$ about the center of mass (CM) to the torque provided by the horizontal and vertical forces FH and FV about the CM . The value of θ is then such that

(a) $\displaystyle cos\theta = \frac{2 g}{3 l \omega^2}$

(b) $\displaystyle cos\theta = \frac{g}{2 l \omega^2}$

(c) $\displaystyle cos\theta = \frac{g}{l \omega^2}$

(d) $\displaystyle cos\theta = \frac{3g}{2 l \omega^2}$

Click to See Solution :
Ans: (d)

Sol: By taking an element of length dx at a distance x from O .

Torque , dτ = dm × (x sinθ) ω2 x cosθ

$\displaystyle \int d\tau = \int_{0}^{l} \frac{m}{l} \times \omega^2 sin\theta \times cos\theta x^2 dx$

$\displaystyle \tau = \frac{m}{l} \times \omega^2 sin\theta \times cos\theta [\frac{x^3}{3}]_{0}^{l}$

$\displaystyle m g \frac{l}{2}sin\theta = \frac{m l^2 \omega^2 sin\theta cos\theta }{3}$

$\displaystyle cos\theta = \frac{3 g}{2 l \omega^2}$

## Two uniform circular discs are rotating independently in the same direction around their common axis passing through there centers ….

Q: Two uniform circular discs are rotating independently in the same direction around their common axis passing through there centers . The moment of Inertia and angular velocity of the first disc are 0.1 kg m2 and 10 rad/s respectively while those for the second one are 0.2 kg m-2 and 5 rad/s respectively . At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed . he kinetic energy of the combined system is

(a) 2/3 J

(b) 20/3 J

(c) 10/3 J

(d) 5/3 J

Click to See Solution :
Ans: (b)

Sol: Applying Conservation of Angular momentum ,

$\displaystyle I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2)\omega$

$\displaystyle \omega = \frac{I_1 \omega_1 + I_2 \omega_2 }{I_1 + I_2}$

Kinetic Energy of Combined System ,

$\displaystyle K = \frac{1}{2}(I_1 + I_2)\omega^2$

$\displaystyle K = \frac{1}{2}(I_1 + I_2)(\frac{I_1 \omega_1 + I_2 \omega_2 }{I_1 + I_2} )^2$

$\displaystyle K = \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{2(I_1 + I_2)}$

$\displaystyle K = \frac{(0.1 \times 10 + 0.2 \times 5)^2}{2(0.1 + 0.2)} = \frac{20}{3}J$

## A uniformly thick wheel with moment on Inertia I and radius R is free to rotate about its center of mass …

Q: A uniformly thick wheel with moment on Inertia I and radius R is free to rotate about its center of mass . A massless string is wrapped over its rim and two blocks of masses m1 and m2 (m1 > m2) are attached to the ends of the string . The system is released from rest . The angular speed of the wheel when m1 decends by a distance h is

(a) $\displaystyle [\frac{2(m_1 – m_2)g h}{(m_1 + m_2)R^2 + I}]^{1/2}$

(b) $\displaystyle [\frac{(m_1 + m_2)}{(m_1 + m_2)R^2 + I}]^{1/2}g h$

(c) $\displaystyle [\frac{2(m_1 + m_2)g h}{(m_1 + m_2)R^2 + I}]^{1/2}$

(d) $\displaystyle [\frac{2(m_1 – m_2)}{(m_1 + m_2)R^2 + I}]^{1/2}g h$

Click to See Solution :
Ans: (a)

Sol:

For block m1 ,

$\displaystyle m_1 g – T_1 = m_1 a$ …(i)

For block m2 ,

$\displaystyle T_2 – m_2 g = m_2 a$ …(ii)

$\displaystyle (T_1 – T_2 )R = I \alpha$ …(iii)

$\displaystyle (m_1 – m_2)g + (T_2 – T_1) = (m_1 + m_2 )a$

$\displaystyle a = \frac{(m_1 – m_2)g}{(m_1 + m_2)} + \frac{T_2 -T_1}{(m_1 + m_2)}$

$\displaystyle a = \alpha R$

$\displaystyle \alpha = \frac{(m_1 – m_2)g}{(m_1 + m_2)R} + \frac{T_2 -T_1}{(m_1 + m_2)R}$

$\displaystyle \alpha = \frac{(m_1 – m_2)g R}{I + (m_1 + m_2)R^2 }$

The time taken by m1 to decends height h1

$\displaystyle t = \sqrt{\frac{2h}{a}} = \sqrt{\frac{2h}{R \alpha}}$

Angular speed $\displaystyle \omega = \alpha t$

$\displaystyle \omega = \sqrt{\frac{2h \alpha}{R}}$

$\displaystyle \omega = [\frac{2(m_1 – m_2)g h}{(m_1 + m_2)R^2 + I}]^{1/2}$

## Three solid spheres each of mass m and diameter d are suck together such that the lines connecting the center form an equilateral triangle ….

Q: Three solid spheres each of mass m and diameter d are suck together such that the lines connecting the center form an equilateral triangle of side of length d . The ratio Io/IA of moment of Inertia Io of the system about an axis passing the centroid and about center of any of the spheres IA and perpendicular to the plane of the triangle is

(a) $\displaystyle \frac{15}{13}$

(b) $\displaystyle \frac{13}{15}$

(c) $\displaystyle \frac{13}{23}$

(d) $\displaystyle \frac{23}{13}$

Click to See Solution :
Ans: (c)

Sol: Let d is the diameter of each sphere and m be the mass of each sphere .

$\displaystyle CD = \sqrt{d^2 -\frac{d^2}{4}} = \frac{\sqrt{3}d}{2}$

$\displaystyle CO = \frac{2}{3} \times \frac{\sqrt{3}d}{2} = \frac{d}{\sqrt{3}}$

Moment of Inertia about O (using || axis theorem )

$\displaystyle I_o = 3 \times [ \frac{2}{5}m(\frac{d}{2})^2 + m (\frac{d}{\sqrt{3}})^2 ]$

$\displaystyle I_o = \frac{13}{10} m d^2$

$\displaystyle I_A = 2 \times [ \frac{2}{5}m(\frac{d}{2})^2 + m d^2 ] + \frac{2}{5} m (\frac{d}{2})^2$

$\displaystyle I_A = \frac{23}{10} m d^2$

$\displaystyle \frac{I_o}{I_A} = \frac{13}{23}$