Rotational kinetic energy problem

[blackbyron]

Homework Statement

A solid sphere with a mass 8.2kg and radius 10cm is sliding along a frictionless surface with a speed 5.4m/s while at the same time spinning. The sphere has 0.31 of its total kinetic energy in translational motion. How fast is the sphere spinning?

Homework Equations

KAe = (1/2)(I)(W)^2

The Attempt at a Solution

The problem is that the rotational kinetic energy is unknown, but I found that the inertia is .0328kgm^2

How do I find the rotation kinetic energy so I can solve for w?

 

[SteamKing]

In the relevant equations, you have left out the formula for calculating the kinetic energy due to translation of the sphere at the given speed.

 

[blackbyron]

In the relevant equations, you have left out the formula for calculating the kinetic energy due to translation of the sphere at the given speed.

You mean this ?

Translation:

Ke = (1/2)(m)(v^2)

 

[SteamKing]

Ya. Now reread your problem statement very carefully.

 

[rwisz]

To find the rotational kinetic energy, you can use the equation Krot = (1/2)(I)(ω)^2, where I is the moment of inertia and ω is the angular velocity. In this problem, the moment of inertia of a solid sphere is (2/5)MR^2, where M is the mass and R is the radius. So, in this case, I = (2/5)(8.2kg)(0.1m)^2 = 0.0328 kgm^2.

Since the problem states that the sphere has 0.31 of its total kinetic energy in translational motion, we can write the equation: Ktrans = (0.31)(Ktotal). Since the total kinetic energy is equal to the sum of translational and rotational kinetic energy, we can write: Ktotal = Ktrans + Krot.

Plugging in the known values, we get: (0.31)(Ktotal) = (0.31)(0.5)(8.2kg)(5.4m/s)^2. Solving for Ktotal, we get Ktotal = 127.73J.

Now, we can plug this value into the equation for rotational kinetic energy and solve for ω: 127.73J = (0.5)(0.0328kgm^2)(ω)^2. Solving for ω, we get ω = 15.58 rad/s.

Therefore, the sphere is spinning at a speed of 15.58 rad/s.