The image attached that is supposed to be attached to the question is shown in the first file below.
Answer:
t = 2.19 seconds
Explanation:
The free body diagram showing the center of mass A and B is attached in the second diagram below.
NOTE : that from the second diagram; Mass A and B do not have any acceleration
Taking the moment about wheel A:
[tex]\sum M_A = I_A \alpha _A[/tex]
[tex]-f(r_A) = I_A \alpha _A ----- (1)[/tex]
The equilibrium forces in the y-direction is 0
i.e
[tex]F_y = 0[/tex]
So;
[tex]N +T sin 30^0 -W_A = 0 ----- (2)[/tex]
The equilibrium forces in the x-direction is as follows:
[tex]\sum F_x = 0[/tex]
[tex]Tcos 30^0 + f= 0 -----(3)[/tex]
The kinetic friction f can be expressed as :
[tex]f = \mu _k N[/tex]
From above equation (2) and equation (3);
[tex]N + [\dfrac{-f}{cos 30^0}]sin 30^0 -150 =0[/tex]
[tex]N - \mu _k N \ tan 30^0 -150 =0[/tex]
[tex]N = \dfrac{150}{1-0.3 \ tan 30^0}[/tex]
N = 181.423 lb
Similarly; from equation(1)
[tex]\alpha_A = - \dfrac{f(r_A)}{I_A}[/tex]
[tex]\alpha _A = \dfrac{-\mu_k N(r_A)}{I_A}[/tex]
[tex]\alpha _A = \dfrac{-0.3*181.423*1.25}{\frac{150}{32.2}*I^2}[/tex]
[tex]\alpha _A =-14.6045 \ rad/s^2[/tex]
However; from the kinematics ; as moments are constant ; so is the angular acceleration is constant )
Thus;
[tex]\omega _A - \omega_o^A = \alpha_A t[/tex]
[tex]\omega _A = \omega_o^A + \alpha_A t[/tex]
[tex]\omega _A = 100 -14.6045 \ t ---- (4)[/tex]
Let's take a look at wheel B now;
Taking the moment about wheel B from the equation of motion:
[tex]\sum M_B = I_B \alpha _B[/tex]
[tex]f(r_B) = I_B \alpha _B[/tex]
[tex]\mu_k N (r_B) = I_B \alpha_B[/tex]
[tex]\mu_k N (r_B) = \dfrac{W_B}{g}* k^2_B \alpha_B[/tex]
[tex]\alpha_B = \dfrac{0.3*181.423*1}{\frac{100}{32.2}*0.75^2}[/tex]
[tex]\alpha = 31.1563 \ rad/s^2[/tex]
Again; from the kinematics; as the moments are constant which lead to the angular accleration;
[tex]\omega _B = \omega _o^B + \alpha _B \ t[/tex]
[tex]\omega _B =0 + 31.156 \ t-----(5)[/tex]
From equation 4 and 5 which attain the same angular velocity; we have;
[tex]\omega^A = \omega^B[/tex]
100 - 14.6045 t = 31.1563 t
100 = 31.1563 t + 14.6045 t
100 = 45.761 t
t = 100/45.761
t = 2.19 seconds
