Frictionless cylindrical flywheel, of diameter 250mm and mass 30k

[gk1989]

A frictionless cylindrical flywheel, of diameter 250mm and mass 30kg, has a string attached/wrapped on the outside diameter, with a mass of 10kg on the other end.

What is the radius of gyration of the flywheel?
If the mass starts 1.5m form the floor, what will its speed be when the mass is released & it hits the floor? and at this point how much rotational kinetic energy will the flywheel have?


Any hints or answers would be gratefully received.
many thanks
Greg

 

[minger]

The radius of gyration should be easy, it's just a simple formula. It's just a function of mass and moment of inertia if I remember correctly. OK, I looked it up, it's just:
$$R = \sqrt{\frac{I}{m}}$$
As for the problem let's try and walk through it. From the mass, you will need to convert the equivalent moment that the mass is exerting on the flywheel. From there, you can plug it into the equivalent rotational f=ma equation:
$$T = I\omega$$ With your moment of inertia and torque/moment, you can get a rotational acceleration. Your mass is 1.5m above the floor, so convert that to radians, and see how long it takes to hit the floor (from the rotational acceleration).
Given rotational acceleration and time, you can now get your final rotational velocity and the kinetic energy which is the equivalent:
$$KE = \frac{1}{2}I\omega^2$$ Good luck

 

[Mene]

The radius of gyration can be calculated using the formula: k = √(I/m), where k is the radius of gyration, I is the moment of inertia, and m is the mass of the object. In this case, the moment of inertia for a cylindrical flywheel is given by I = ½mr^2, where m is the mass and r is the radius. Therefore, the radius of gyration for this flywheel would be k = √(½mr^2/m) = √(½r^2) = 0.707r.

To calculate the speed of the mass when it hits the floor, we can use the conservation of energy principle. The initial potential energy of the mass is given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the initial height (1.5m). This potential energy is converted into kinetic energy when the mass is released. The kinetic energy of the mass is given by ½mv^2, where v is the speed of the mass. Therefore, we can set these two energies equal to each other and solve for v: mgh = ½mv^2. Plugging in the values, we get v = √(2gh) = √(2*9.8*1.5) = 4.83 m/s.

At this point, the rotational kinetic energy of the flywheel can be calculated using the formula: K = ½Iω^2, where K is the rotational kinetic energy, I is the moment of inertia, and ω is the angular velocity. The angular velocity can be calculated using the formula v = ωr, where v is the linear speed and r is the radius of the flywheel. Therefore, ω = v/r = 4.83/0.125 = 38.64 rad/s. Plugging this value into the formula for rotational kinetic energy, we get K = ½*(½mr^2)*(38.64)^2 = 73.68 J.

I hope this helps! Let me know if you have any further questions.