Total kinetic energy for a rolling body is the sum of translational and rotational kinetic energy, given by K.E.(total) = (1/2)mv^2 + (1/2)Iomega^2. For a solid sphere, the moment of inertia I = (2/5)mr^2 and v = r omega. Substituting this in gives K.E.(rotational) = (1/5)mv^2 and K.E.(total) = (7/10)mv^2, leading to a ratio of 2/7.

Analogue of mass in rotational motion is moment of inertia. It plays the same role as mass plays in translational motion.

Newton's second law of motion in kinematics is F =  ma. To express the same in rotation, replace mass by moment of inertia I AND linear acceleration a by angular acceleration $\alpha$. Thus, Newton's law of motion becomes, Torque $\tau = I \alpha$

Similarly work done in kinematics is given by W = F.S. To express the same in rotation, replace Force by torque $\tau$ AND linear displacement by angular displacement $\theta$. Thus, Work done becomes, Torque $W= \tau \theta$

Newton's second law for rotation states that the net torque acting on an object is equal to the rate of change of its angular momentum, expressed as tau = dL/dt.

In rotation, m is replaced by I and v by  $\omega$. Thus the expression for kinetic energy becomes $I \omega^2/2$

The correct option is thus option (b)

Net force = $mg \sin \theta$

For half a revolution, the center moves forward by piR. The point initially at the bottom moves to the top, so its vertical displacement is 2R. The total displacement is sqrt((piR)^2 + (2R)^2) = Rsqrt(pi^2 + 4). The angle is tan^-1(vertical/horizontal) = tan^-1(2R/piR) = tan^-1(2/pi).

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.

Using conservation of energy: mgh = (1/2)mv^2 + (1/2)I(omega)^2. Since v = omega*r, omega = v/r. So mgh = (1/2)mv^2 + (1/2)I(v/r)^2. Solving for v gives v = sqrt(2mgh / (m + I/r^2)).

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 mah/2 = mgr. With a = d^2x/dt^2 = 2bt, we get m*(2bt)h/2 = mg*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.

Using work-energy theorem: Initial KE = (1/2)Mv^2 + (1/2)I(omega)^2. Since v = omegaa, KE = (1/2)(M + I/a^2)v^2. The distance d = v^2 / (2gsin(theta)). Substituting v = omegaa gives the result.

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 = gsin(theta) / (1 + I/MR^2). The sphere has I = (2/5)MR^2, so a = gsin(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.

since the body is given initial translational velocity so it will move forward while coming in pure rolling condition.

By conservation of angular momentum, I1omega1 = I2omega2. Given I2 = (2/5)I1, then omega2 = (5/2)omega1 = 2.5 * 40 = 100 rev/min. KE2 = (1/2)I2omega2^2 = (1/2)(2/5)I1*(2.5*omega1)^2 = 2.5 * KE1.