A uniform sphere with mass 28.0 kg and radius 0.380 m is rotating...

Question

A uniform sphere with mass 28.0 kg and radius 0.380 m is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere. If the kinetic energy of the sphere is 236 J, what is the tangential velocity of a point on the rim of the sphere?

Accepted Answer

Hello everyone. So in this problem homogeneous metal sphere rotates about aboriginal access at its center of mass at a constant velocity. The rotational kinetic energy of the sphere is 150 jewels. And it's asking what is the tangential velocity of a point on the rim of the sphere? If the diameter of the sphere is 20 cm and its mass is one. So from the problem statement, We know that the kinetic energy is equal to 150 tools. We know that the diameter of the sphere is 20 cm, which is also equal to .2 m. That the mass is equal to one kg. We want to find a tangential velocity B. Call. Now we can solve for me by the relation R times omega where r. Is the radius of the sphere and omega is the angular philosophy. So if we're able to solve our omega were able to solve for the tangential velocity, we can relate the kinetic energy with omega. As we recall. The equation for kinetic energy is one to have I omega squared. We can rearrange this equation to get an equation for omega, which is the square root of two times the kinetic energy over I we know generally that the equation for the moment of inertia of a metal sphere about its central axis Is to 5th MR two. So we can make the substitution in this omega equation on the square root of two tons of kinetic energy to theft, M. R square. So the tubes cancel out and we're left with the square root of five times the kinetic energy over our sweet. Now we have the kinetic energy we have em and we know that our is simply equal to D over two Which is .1 years. So we can make this substitution in salt. Omega now Square of five times the kinetic energy, which is 150 jewels over. Um which is one kg Times are which is .1 square. Let me get that. Omega is equal to 273.9 waiting this person. And now we can use this tangential velocity equation to solve directly for where V is equal to R times another. We have R. And omega Where R is equal to .1 m And Omega is equal to 273.9/s. After reading this, we get that B is approximately equal to 27. m per seven. His answer choice, I hope this will have a great day.

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