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Find the total kinetic energy Ktot of a dumbbell of mass m when it is rotating with angular speed Ï and its center of mass is moving translationally with speed v. Denote the dumbbell's moment of inertia about its center of mass by Icm. Note: If you approximate the spheres as point masses of mass m/2 each located a distance r from the center and ignore the moment of inertia of the connecting rod, then the moment of inertia of the dumbbell is given by Icm=mr^2, but this fact will not be necessary for this problem.

Required:
Find the total kinetic energy Ktot of the dumbell.

Answer :

Answer:

Explanation:

Total kinetic energy of dumbbell = translational kinetic energy + Rotational kinetic energy

Translational kinetic energy = 1/2 m v²

Rotational kinetic energy = 1/2 x moment of inertia about CM  x angular speed

= 1/2  x Icm x Ï ²

Total kinetic energy

Ktot = 1/2 m v² + 1/2  x Icm x Ï ²

The total kinetic energy K(tot) of the object is :

K(tot) = 1/2 m v² + 1/2  Icm Ï ²

Finding the total kinetic energy:

The total kinetic energy of the object is the sum of the translational kinetic energy and rotational kinetic energy.

let the total kinetic energy be K(tot)

The translational kinetic energy be KE and

the rotational kinetic energy be K(rot)

So,

K(tot) = KE + K(rot)

The translational kinetic energy is given by:

KE = 1/2 mv²

where m is the mass and v is the velocity

The rotational kinetic energy is given by:

K(rot) = 1/2  Icm Ï ²

where Icm is the moment of inertia and Ï is the angular speed

So the total kinetic energy is:

K(tot) = 1/2 m v² + 1/2  Icm Ï ²

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