I. Express the kinetic energy of the body rotating about an...

i. Express the kinetic energy of the body rotating about an axis through its center of mass. ii. Express the relationship between linear velocity and angular velocity. As shown in the figure, a spherical shell of mass M = 7 kg and radius R = 0.3 m rotates without friction on a vertical bearing. A light string wound on the equator of the spherical crust, after passing through a pulley with a moment of inertia I = 0.16 kg.m² and a radius of r = 0.2 m, was attached to an object of mass m = 3 kg that can fall freely under the effect of gravity. The object was initially at rest. It was then released to let the object fall. Using the principle conservation of energy, find its linear velocity after it falls through a height h = 1 m. (Assume the moment of inertia of the sphere to be 2/3 MR²).

Transcript

00:01 This question is about rotational kinetic energy.

00:04 Say we have a solid sphere moving with translational velocity of three meters per second towards an inclined plane.

00:11 We want to know how high the sphere will go up the plane in two cases.

00:15 Case a, where the plane is frictionless, so the sphere continues to rotate until it reaches its maximum height, or case b, where friction is involved, so the rotational motion stops at the maximum height.

00:28 In most problems with planes, energies you could place to start.

00:32 So let's start with that.

00:35 Initially, the sphere is at ground level.

00:41 So it has no gravitational potential energy, just kinetic energy.

00:47 But since it's a rotating object, it has a translational part, let's denote t, and a rotational part, which i'll denote r.

00:55 Translational plus rotational.

00:57 The translational part is 1⁄2mv squared, and the rotational part is one -half i which is the moment of inertia omega squared or omega is the angular velocity for a sphere know that i sphere equals two -fifths m r squared now we're not given an angular velocity we're given a translational velocity so it would be nice if we could eliminate this angular velocity in favor of translational velocity.

01:33 We can do that.

01:34 Translational velocity, v equals r times omega, or equivalently, equivalently, omega equals v over r.

01:42 So we can substitute these two things in to get a better expression.

01:46 We have one -half mv squared plus one -half times two -fifths m -r -squared for the i, and omega is v squared over r -squared.

02:01 Being careful not to forget this square here.

02:05 Simplifying a bit, one -half mb squared plus one -fifth m -r's cancel out b squared.

02:18 So this is, again, the translational part and the rotational part initially.

02:25 We can, of course, combine them, so they're just both mb squared, 5 tenths plus 2 -tenths is 7 -tenths.

02:34 Mv squared.

02:37 So this is the initial energy.

02:40 So now let's visit case a...

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