Use conservation of energy to determine the angular speed of the spool shown in Figure P8.36 after the 3.00 kg?

The law of conservation of energy tells us that the potential energy the bucket 'lost' during its fall will have been 'given' to the spool in the form of kinetic rotational energy. So we will look at both sides of that equation:

First the fall of the bucket;

Potential energy in this case, will equal: E = m*h*g, with m mass, h the difference in height before the fall and after, and g = 9.81 m/s²

This gives us E = 119.1915 J (Joules)

Now we look at the rotation of the spool;

Kinetic rotational energy equals: E = (1/2)*I*w, with I the moment of inertia, which is in the case of a cilinder rotating around its symmetry axis, equals (1/2)*m*r², and with w the angular velocity.

So, both have to be equal: 119.1915 J = (1/4)*(5.00kg)*(0.600m)² * (w)

Solving for w:

w = 264.87 rad/s

Conversion of GPE = mgh = 3*9.80 one*3.fifty 5 = ? N.m into KE = a million/2 Iw^2; sparkling up for w = sqrt(2GPE/I) I = kMr^2 the position ok = your disk coefficient (i imagine it really is advantageous = a million/2, yet seem it up), M is the mass of the spool. You got here upon GPE, ok, M, and r are given you'll locate w = ? radian/sec the angular %.

.