Questions
A uniform solid disk of radius R and mass M is free to rotate on a frictionless pivot through a point on its rim. If the disk is released from rest in the position shown in figure. The speed of the lowest point on the disk in the dashed position is
detailed solution
Correct option is A
To identify the change in gravitational energy, think of the height through which the centre of mass falls. From the parallel-axis theorem, the moment of inertia of the disk about the pivot point on the circumference isI = ICM+MD2 = 12MR2+MR2 = 32MR2The pivot point is fixed, so the kinetic energy is entirely rotational around the pivot. The equation for the isolated system (energy) model(K+Uf) = (K+U)ffor the disk-Earth system becomes0+MgR = 12(32MR2)ω2+0Solving for ω, ω = 4g3RAt the lowest point on the rim, v = 2Rω =4 Rg3Talk to our academic expert!
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A uniform thin rod of length l and mass m is hinged at a distance l/4 from one of the end and released from horizontal position as shown in Fig. The angular velocity of the rod as it passes the vertical position is
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