Q.A rigid object is rolling down an inclined plane. Derive expressions for the acceleration along the track and the speed after falling through a certain vertical distance.

Energy conservation ($Mgh=\frac{1}{2}Mv^2(1+K^2/R^2)$) gives $v=\sqrt{2gh/(1+K^2/R^2)}$; combining with $v^2=2as$ and $s=h/\sin\theta$ gives $a=g\sin\theta/(1+K^2/R^2)$. …

Consider a rigid object of mass M, radius R and radius of gyration K, released from rest and rolling WITHOUT slipping down an incline of angle $\theta$. As it descends through vertical height h, its gravitational PE converts entirely into rolling KE (no energy lost to friction, since static friction at a non-slipping contact point does no net work): $Mgh=\frac{1}{2}Mv^2\left(1+\frac{K^2}{R^2}\right)$ Solving for v: $v^2=\frac{2gh}{1+K^2/R^2} \quad\Rightarrow\quad v=\sqrt{\frac{2gh}{1+K^2/R^2}} \qquad \text{--- (1.19)}$ The distance actually travelled ALONG the incline while falling through height h is $s=\dfrac{h}{\sin\theta}$. Since the object starts from rest and reaches this speed v over distance s under a constant linear acceleration a, the kinematic relation $v^2=u^2+2as$ (with $u=0$) gives $a=\frac{v^2}{2s}=\frac{1}{2}\cdot\frac{2gh}{1+K^2/R^2}\cdot\frac{1}{s}=\frac{gh}{s(1+K^2/R^2)}=\frac{g\sin\theta}{1+K^2/R^2} \qquad \text{--- (1.20)}$ usi …

Apply energy conservation (loss in gravitational PE = gain in total rolling KE) to find v in terms of the vertical drop h, then combine with the kinematic relation $v^2=2 …

Using $v=\sqrt{2gh}$ or $a=g\sin\theta$ (the pure-sliding values) for a rolling object, forgetting that part of the released PE goes into ROTATIONAL kinetic ene …