Net force = $mg \sin \theta$

As a cylinder rolls down an incline, the static friction force acts up the plane. This torque causes the object to rotate, thereby increasing its rotational kinetic energy at the expense of potential energy.

The condition for slipping on a turntable is that the maximum static friction equals the required centripetal force, which means omega^2 r = constant. If the angular velocity omega is doubled, its square increases by a factor of 4. Therefore, the radius r must decrease by a factor of 4 to maintain the same limiting acceleration, giving 40 cm / 4 = 10 cm.

Acceleration a = g*sin(theta) / (1 + I/MR^2). The object with the smallest I/MR^2 has the largest acceleration and reaches the bottom first. I/MR^2 values: SS (2/5), SC (1/2), HC (1). Order: SS, SC, HC.

Toppling occurs when the torque due to pseudo-force (ma) about the edge exceeds the torque due to gravity. The condition is m*a*h/2 = m*g*r. With a = d^2x/dt^2 = 2bt, we get m*(2bt)*h/2 = m*g*r, so t = rg/bh.

The velocity of a point on the rim is v_p = v_cm + omega x r. For vertical velocity, the horizontal components must cancel. The height y = R - v/omega is a standard result for the point where horizontal velocity is zero.

The center of mass of the rod moves in a circle of radius L/2 with angular velocity omega. The centripetal force required to keep the center of mass moving in this circle is provided by the horizontal force from the axle. Using the formula F = m omega^2 r_cm, where r_cm = L/2, we get F = (1/2) m omega^2 L directed inward toward the axis of rotation.

Acceleration a = g*sin(theta) / (1 + I/MR^2). The sphere has I = (2/5)MR^2, so a = g*sin(theta) / 1.4. The disc has I = (1/2)MR^2, so a = g*sin(theta) / 1.5. The sphere has higher acceleration and arrives first.

By conservation of angular momentum, I1*omega1 = I2*omega2. Given I2 = (2/5)I1, then omega2 = (5/2)omega1 = 2.5 * 40 = 100 rev/min. KE2 = (1/2)I2*omega2^2 = (1/2)(2/5)I1(2.5*omega1)^2 = 2.5 * KE1.

The whole system does not continue to rotate about P because the moment is balanced. Thus option B is wrong. And the center of mass of the system lies at pivot point P. Thus option C and D are wrong. As the force applied at the two points of suspension is different $\tau$ is different.

$\begin{array}{l} You\, \, have\, \, to\, \, use\, \, the\, \, equation, \ \; { { { \omega  } }^{ { 2 } } }\; ={ { { \omega  } } _{ { 0 } } }^{ { 2 } }\; +{ { 2\alpha \theta  } }\; \, \, for\, \, finding\, \, the\, \, angular\, \, acceleration\; \, \alpha \, \, and \ hence\, \, the\, \, number\, \, of\, \, further\, \, rotations. \ Note\, \, that\, \, this\, \, equation\, \, is\, \, the\, \, rotational\, \, analogue\, \, of\, \, the\, equation \ { v^{ 2 } }\; ={ v _{ 0 } }^{ 2 }+2as{ {  } }(or,\; { v^{ 2 } }\; ={ u^{ 2 } }\; +2as)\, \, in\, \, linear\, \, motion. \ Since\, \, the\, \, angular\, \, velocity\, \, has\, \, reduce\, \, to\, \, half\, \, of\, \, the\, \, initial\, \, value\, \, { \omega _{ 0 } }\, \, after\, \, 36\, \, rotations,\, \, we\, \, have \ { \left( { { \omega _{ 0\;  } }/2 } \right) _{ \;  } }^{ 2 }={ \omega _{ 0 } }^{ 2 }+2\alpha \times 36\, \, from\, \, which\, \; \alpha =--\; { \omega _{ 0 } }^{ 2 }/96 \ \left[ { We\, \, have\, \, expressed\, \, the\, \, angular\, \, displacement\, \, \theta \, \, in\, \, rotations\, \, itself\, \, for\, \, convenience } \right]  \ If\, \, the\, \, additional\, \, number\, \, of\, \, rotations\, \, is\, \, x,\, \, we\, \, have \ 0={ \left( { { \omega _{ 0\;  } }/2 } \right) _{ \;  } }^{ 2 }\; +\; 2\alpha x\; =\; { \left( { { \omega _{ 0\;  } }/2 } \right) _{ \;  } }^{ 2 }\; +\; 2\times (--\; { \omega _{ 0 } }^{ 2 }/96)x \ This\, \, gives, \ x\; =12 \end{array}$