Just a little mechanical energy?

Question

Given a system of two masses m1 and m2 which are attached by a very light string over a solid pulley of mass M and radius R. The smaller mass m1 is held at the ground level. If the m2 is suspended h meters above the ground what will the velocity be of mass m2 as it hits the ground (after the mass m1 is released).

Please show work.

Answer 1

Sounds familiar...

The energy of the system before the release is just the potential energy of m2:

E_initial = PE_m2 = m2*g*h

Now when the system is set in motion the total energy is the potential energy of the smaller mass, the kinetic energy of the masses together (which are moving at the same speed), and the kinetic energy of the pulley. So in order of mention:

E_final = PE_m1 + KE_m1_m2 + KE_pulley

= m1*g*h + ½*(m1+m2)*v² + ½*I*ω²

= m1*g*h + ½*(m1+m2)*v² + ½*(½*M*R²)*(v/R)²

= m1*g*h + [ 2*(m1+m2) + M]*v²/4

Now set E_initial = E_final and you get:

√ { [ 4*(m2 - m1)*g*h ] / [ 2*(m1+m2) + M ] } = v

Not too ugly!

Hope that helps.

Answer 2

"doug_donaghue", you did not take into account the mass of the pulley....apparently, it is not negligible, otherwise it would not have been specified that the pulley has "mass M and radius R".

The rotational inertia of a solid cylinder is give as,
I = 1/2 MR^2
Where I is the rotational inertia, M is the mass of the cylinder, and R is the radius of the cylinder (also radius of rotation).

When mass 2 is allowed to drop it gives up its gravitational potential energy. This potential energy from mass 2 goes into kinetic energy of the masses, potential energy of mass 1, as well as rotational kinetic energy of the pulley.

KE = 1/2 mv^2
PE = mgh
KE_rotational = 1/2 I*w^2
where m is the mass, g is the gravitational acceleration, v is the velocity, I is the rotational inertia, and w (Greek letter, lower case, omega) is the angular velocity.

The potential energy given up by mass 2 goes into several other forms of energy and since we are working with purely symbolic values, this gets turned into a rather ugly algebra problem rather quickly. I think that with only the values we have defined (m1, m2, M, R, and h, [and the assumed g]), the final answer is going to be ugly.

But to simplify it a bit,
we know that the translational velocity of the pulley must equal the translational velocity of both masses (since they are tied together with a string)..this simplifies things quite a bit by eliminating unknown individual velocities (both angular and translational) for each mass.
After this, we can solve for a velocity v in terms of only things which we defined in the question.


EDIT:
The revised response given by "doug_donaghue" goes to show you that there is usually more than 1 correct method to go about solving these problems. Whereas I used the law of conservation of energy, "doug_donaghue" went another route and used the angular acceleration which results from the net torque exerted on the pulley. Both answers should be equally correct numerically and unit-wise, although symbolically they may look different.

Answer 3

The net force acting on the system will be m1g - m2g or
F = g(m2-m1) where g = 9.8 m/s²
The angular acceleration of the pulley (in Rad/s²) is given by
α=RF/I where I is the moment of inertia and (assuming a pulley of constant density) is
I = RM/2.
The distance traveled by m2 is then
given by Rw where w is the angular displacement of the pulley and is equal to αt²/2 where t is the time in seconds. Then calculating the angular velocity as Θ = αt the velocity of m2 is seen to be RΘ.

The system you describe is also known in classical mechanics as an 'Atwood Machine'


Doug

Ed. Yes, mrjeffy, I caught that right *after* I posted the answer (when else do you see such things ☺) The correct form is shown above and it's really not all that bad since the angular acceleration of the pulley is what controls the overall acceleration of the two masses. The masses only provide a constant force acting on the rotational inertia of the pulley.

Answer 4

Acceleration times the sum of the masses is equal to the difference of the two masses times gravitational force ... less friction force of the pulley
(see ATWOOD's machine)

Answer 5

"autos convert electric powered power to mechanical power. turbines, the opposite, convert electric powered enegy to mechanical power." You propose, "turbines, the opposite, convert mechanical power to electric powered power." it fairly is via fact maximum actual procedures are (theoretically) reversible.

Answer 6

if the pully is friction less than velocity is

V=(2*g*h)^1/2