From an inclined plane,a sphere,a disc,a ring | vedclass

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(C) For an object rolling down an inclined plane without slipping,the velocity $v$ at the bottom is given by $v = \sqrt{\frac{2gh}{1 + k^2/r^2}}$,wher

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Sphere,disc,shell,ring

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(C) For an object rolling down an inclined plane without slipping,the velocity $v$ at the bottom is given by $v = \sqrt{\frac{2gh}{1 + k^2/r^2}}$,where $k$ is the radius of gyration.To reach the base first,the object must have the highest velocity,which corresponds to the smallest value of the ratio $k^2/r^2$.For a sphere: $I = \frac{2}{5}mr^2 \Rightarrow k^2/r^2 = 0.4$For a disc: $I = \frac{1}{2}mr^2 \Rightarrow k^2/r^2 = 0.5$For a shell: $I = \frac{2}{3}mr^2 \Rightarrow k^2/r^2 = 0.67$For a ring: $I = mr^2 \Rightarrow k^2/r^2 = 1.0$Comparing the ratios: $0.4 Since the velocity is inversely proportional to the factor $\sqrt{1 + k^2/r^2}$,the object with the smallest $k^2/r^2$ reaches the bottom first.Therefore,the order of reaching the base is: Sphere,disc,shell,ring.

From an inclined plane,a sphere,a disc,a ring,and a shell are rolled without slipping. The order of their reaching the base will be:

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