Two disks A and B of equal mass m and radius R roll without slipping from rest down identical inclined planes. A has all mass concentrated at its rim, while B is uniform. At the bottom, the ratio of centre speeds of A to B is
Energy conservation gives $mgh=mv^2/2+I(v/R)^2/2$, hence $v^2=2gh/[1+I/(mR^2)]$. A has $I_A=mR^2$ and B has $I_B=mR^2/2$. Thus $v_A/v_B=\sqrt{(1+1/2)/(1+1)}=\sqrt{3/4}$, A. The rim-loaded disk uses a larger fraction of its energy for rotation.
More Impulse and momentum, energy formulations questions
GATE 2024 ME Q24– A ram in the form of a rectangular body of size l = 9 m and b = 2 m is suspended by two…
GATE 2023 ME Q37– A spherical ball weighing 2 kg is dropped from a height of 4.9 m onto an immovable rigid…
GATE 2022 ME (ME1) Q31– The plane of the figure represents a horizontal plane. A thin rigid rod at rest is pivoted…
GATE 2022 ME (ME2) Q37– A rigid homogeneous uniform block of mass 1 kg, height h=0.4 m and width b=0.3 m is pinned…
GATE 2020 ME (ME1) Q39– A 2 kg block rests on a smooth horizontal surface (top view) and is attached to a massless…
GATE 2020 ME (ME2) Q50– A rigid 10 kg block at 2 m/s travels 1 m on a frictionless horizontal surface and strikes…