Generally, frictional losses are more predominant for the machines being not 100% efficient. This friction leads to the loss of energy in the form of heat, into the surroundings. Some of the supplied energy may be utilised to change the entropy (measure of randomness of the particles) of the system.
Answer:
The process which has friction
Explanation:
The entropy is simply the change in the state of the things or the molecules in the system. It is simply the change in the energy of the system with a focus on the atoms in the system. This is also known as the internal energy of the system and is given the symbol, G. The friction contributes to the change in the energy of the system. This is because friction generates another form of energy - that is heat energy. This energy causes the internal temperature id the system to increase. Hence the greater change in the temperature.
Answer:
![\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}](https://tex.z-dn.net/?f=%5Comega%20%3D%5Cfrac%7B24%7D%7B1.14375%7D%3D20.983%5Cfrac%7Brad%7D%7Bs%7D)
Explanation:
Previous concepts
Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:
![H_o =r x mv=rxL](https://tex.z-dn.net/?f=H_o%20%3Dr%20x%20mv%3DrxL)
Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =
MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:
MO = H˙ O
Principle of Angular Impulse and Momentum
The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:
![\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1](https://tex.z-dn.net/?f=%5Cint_%7Bt_1%7D%5E%7Bt_2%7DM_O%20dt%20%3D%20%5Cint_%7Bt_1%7D%5E%7Bt_2%7DH_O%20dt%3DH_0t2%20-H_0t1)
Solution to the problem
For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is
".
If we analyze the staritning point we see that the initial velocity can be founded like this:
![v_o =\omega r_{OIC}=\omega (0.15m)](https://tex.z-dn.net/?f=v_o%20%3D%5Comega%20r_%7BOIC%7D%3D%5Comega%20%280.15m%29)
And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:
![H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af](https://tex.z-dn.net/?f=H_Ai%20%2B%5Csum%20%5Cint_%7Bt_i%7D%5E%7Bt_f%7D%20M_A%20dt%20%3DH_Af)
![0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)](https://tex.z-dn.net/?f=0%2B%5Csum%20%5Cint_%7B0%7D%5E%7B4%7D%2020t%20%280.15m%29%20dt%20%3D0.46875%20%5Comega%20%2B%2030kg%5B%5Comega%280.15m%29%5D%280.15m%29)
And if we integrate the left part and we simplify the right part we have
![1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega](https://tex.z-dn.net/?f=1.5%284%5E2%29-1.5%280%5E2%29%20%3D%200.46875%5Comega%20%2B0.675%5Comega%3D1.14375%5Comega)
And if we solve for
we got:
![\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}](https://tex.z-dn.net/?f=%5Comega%20%3D%5Cfrac%7B24%7D%7B1.14375%7D%3D20.983%5Cfrac%7Brad%7D%7Bs%7D)