Answer :
Answer:
[tex]\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}[/tex]
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:
[tex]H_o =r x mv=rxL[/tex]
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:
[tex]\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1[/tex]
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 [tex]I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2[/tex]".
If we analyze the staritning point we see that the initial velocity can be founded like this:
[tex]v_o =\omega r_{OIC}=\omega (0.15m)[/tex]
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:
[tex]H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af[/tex]
[tex]0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)[/tex]
And if we integrate the left part and we simplify the right part we have
[tex]1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega[/tex]
And if we solve for [tex]\omega[/tex] we got:
[tex]\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}[/tex]
