A light rod has two m = 5 kg masses attached the distance d = .2 cm apart, each an equal distance from the center point P. The assembly is rotated around point P at a rate 75 radians per second....

A light rod has two m = 5 kg masses attached the distance d = .2 cm apart, each an equal distance from the center point P. The assembly is rotated around point P at a rate 75 radians per second.

A mechanism on the assembly releases the masses so that they slide to the end and are the distance L = 1 cm apart.

1. What is the assembly's kinetic energy multiplied by as a result of this process?

Expert Answers
ishpiro eNotes educator| Certified Educator

The kinetic energy of the assembly comes from the rotation of the masses around the central point P. The rotational energy of each mass can be found as

`(Iomega^2)/2` , where I is the moment of inertia the mass around P and `omega` is the angular velocity of rotation.

The moment of inertia for a point mass is `I = mr^2` , where r is the distance between the mass and the point around which it rotates. Since the masses are distance d apart, and are equal distance from the center point P, the distance between each mass and P is d/2. The moment of inertia of each mass is then

`I = m(d/2)^2 = (md^2)/4`

The angular velocity of rotation relates to the frequency of rotation as

`omega = 2pif` . For now, let's express the rotational energy of each mass in terms of `omega` . It will be

`(Iomega^2)/2 = (md^2omega^2)/8` .

The kinetic energy of the assembly is the sum of the rotational energy of each mass. Since they are identical, the kinetic energy equals twice the above expression:

`K = (md^2omega^2)/4` .

When the masses are released, the moment of inertia of each mass will become

`I' = m(L/2)^2 = (mL^2)/4` .

The angular velocity will also change, in a way so that the angular momentum of the system is conserved. The angular momentum before the masses are released has to equal the angular momentum after they slide to the end:

`2Iomega = 2I'omega'` .

From here, the angular velocity after the masses are at the end is

`omega' = (Iomega)/(I')`

or `omega' = d^2/L^2 omega` .

The new rotational energy of each mass therefore will be

`"(I'(omega')^2)/2`

Plugging in the expressions for the new moment of inertia and angular velocity results in

`1/2 (mL^2)/4*(d^2/L^2 omega)^2 = (md^4)/(8L^2) omega^2`

The final kinetic energy of the assembly will be twice the above expression:

`K' =(4L^2) omega^2 /(md^4)`

Now the ratio of the final and original kinetic energies can be found:

`(K')/K = (md^4omega^2)/(4L^2) / (md^2omega^2)/4`

`(K')/K = d^2/L^2`

`(K')/K = (0.2 cm)^2/(1 cm)^2 = 1/25`

 

The assembly's kinetic energy is multiplied by 1/25 as the result of this process.

 

 

 

ishpiro eNotes educator| Certified Educator

Correction:

The ratio of the final and original kinetic energies is

`(K')/K = (md^4omega^2)/(4L^2) / (md^2omega^2)/4`

`(K')/K = d^2/L^2`

`(K')/K = (0.2 cm)^2/(1 cm)^2 = 1/25`

The assembly's kinetic energy is multiplied by 1/25 as the result of this process.