Question

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s starting from rest. Gear rack B is fixed to the horizontal plane and the gear's radius of gyration about its mass center O is Ko=125mm

Answer 1

`\omega =(24)/(1.14375)=20.983(rad)/(s)`

Step-by-step 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`

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`

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
`I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2`".

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)`

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`

`0+\sum \int_(0)^(4) 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)`

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`

And if we solve for
`\omega` we got:

`\omega =(24)/(1.14375)=20.983(rad)/(s)`