Two spheres are each rotating at an angular speed of 229...

Two spheres are each rotating at an angular speed of 22.9 rad/s about axes that pass through their centers. Each has a radius of 0.480 m and a mass of 1.69 kg. However, as the figure shows, one is solid and the other is a thin-walled spherical shell. Suddenly, a net external torque due to friction (magnitude = 0.110 N · m) begins to act on each sphere and slows the motion down. How long does it take (a) the solid sphere and (b) the thin-walled sphere to come to a halt?

Transcript

00:01 Hello, the question is taken from physics and the question is two spheres are each rotating at an angular speed of 22 .9 radiant per second.

00:11 Okay.

00:13 And about axis that passes through their center.

00:20 Each has a radius of so let me first write the angular speed that is 22 .9 radian per second.

00:30 And the radius of them are each has a radius of 0 .4 .8.

00:34 Meter so r is equal to 0 .48 meter and the mass is 1 .69 kg then however as the figure shows one is solid and other is thin walled spherical shell suddenly a net torque due to friction friction torque will be given as 0 .110 newton meter this is the value of the torque that is applied on it begins to act on easy sphere and slows the motion down how does it take the solid spare and the thin vault sphere to come to a hole so for solid sphere we know that we know that first we need to evaluate the moment of inertia and then we can use the formula for angular value angular speed and by evaluating the angular acceleration in order to evaluate the time okay so from here for solid spare moment of inertia is equal to 2 by 5 m r square so that will be 2 by 5 mass is equal to 1 .69 the radius is given to us that is 0 .48 square and also it is equal to torque is equal i into alpha this implies alpha is equal to torque over i so let me evaluate and torque is 0 .11 newton meter so let me first evaluate the value of moment of inertia and then evaluate the value of alpha so that is 0 .4 into 1 .69 into 0 .48 square so that is 0 .156 kilogram meter square and so the value of the angular acceleration will be multiply 0 .11 so that is 0 .706 radian per second square that is the required value of the angular acceleration and also given that omega is equal to omega 0 plus alpha t so this implies t will be equal to omega over alpha so that will be angular speed is second 22 .9 divided by 0 .706 so the time will be 32 .42 second which is the required value of the time it will take to stop and in second part for the thin spherical shell the moment of inertia i is equal to 2 by 3 m r square so that will be 2 by 3 mass is equal to 1 .69 into 0 .48 square so alpha is equal to torque over moment of inertia torque is 0 .6.

04:12 1.

04:14 Divided by moment of inertia, let me solve this 2 by 3, multiply 1 .69 into 0 .48 whole square.

04:24 So that is 0 .2596...

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