A uniform cylindrical hoop is released from rest at a height of \(0.25 \mathrm{~m}\) near the top of an inclined plane ( Figure 8.23). If the hoop rolls down the plane without slipping and no energy is lost due to friction, what is the linear speed of the cylinder's center of mass at the bottom of the incline?

THINKING IT THROUGH. Here, gravitational potential energy is converted into kinetic energy - both rotational and translational. The conservation of (mechanical) energy applies, since no energy is lost to friction. The initial kinetic energy is zero since the hoop is released from rest.

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R @ VCM h = 0.25 m