Question

Wheels A and B have weights of 150 lb and 100 lb , respectively. Initially, wheel A rotates clockwise with a constant angular velocity of 100 / A   rad s and wheel B is at rest. If A is brought into contact with B, determine the time required for both wheels to attain the same angular velocity. The coefficient of kinetic friction between the two wheels is 0.3 k  and the radii of gyration of A and B about their respective centers of mass are 1 A k ft  and 0.75 B k ft . Neglect the weight of link AC.

Answer 1

The image attached that is supposed to be attached to the question is shown in the first file below.

Answer:

t = 2.19 seconds

Step-by-step 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:

`\sum M_A = I_A \alpha _A`

`-f(r_A) = I_A \alpha _A ----- (1)`

The equilibrium forces in the y-direction is 0

i.e

`F_y = 0`

So;

`N +T sin 30^0 -W_A = 0 ----- (2)`

The equilibrium forces in the x-direction is as follows:

`\sum F_x = 0`

`Tcos 30^0 + f= 0 -----(3)`

The kinetic friction f can be expressed as :

`f = \mu _k N`

From above equation (2) and equation (3);

`N + [(-f)/(cos 30^0)]sin 30^0 -150 =0`

`N - \mu _k N \ tan 30^0 -150 =0`

`N = (150)/(1-0.3 \ tan 30^0)`

N = 181.423 lb

Similarly; from equation(1)

`\alpha_A = - (f(r_A))/(I_A)`

`\alpha _A = (-\mu_k N(r_A))/(I_A)`

`\alpha _A = (-0.3*181.423*1.25)/((150)/(32.2)*I^2)`

`\alpha _A =-14.6045 \ rad/s^2`

However; from the kinematics ; as moments are constant ; so is the angular acceleration is constant )

Thus;

`\omega _A - \omega_o^A = \alpha_A t`

`\omega _A = \omega_o^A + \alpha_A t`

`\omega _A = 100 -14.6045 \ t ---- (4)`

Let's take a look at wheel B now;

Taking the moment about wheel B from the equation of motion:

`\sum M_B = I_B \alpha _B`

`f(r_B) = I_B \alpha _B`

`\mu_k N (r_B) = I_B \alpha_B`

`\mu_k N (r_B) = (W_B)/(g)* k^2_B \alpha_B`

`\alpha_B = (0.3*181.423*1)/((100)/(32.2)*0.75^2)`

`\alpha = 31.1563 \ rad/s^2`

Again; from the kinematics; as the moments are constant which lead to the angular accleration;

`\omega _B = \omega _o^B + \alpha _B \ t`

`\omega _B =0 + 31.156 \ t-----(5)`

From equation 4 and 5 which attain the same angular velocity; we have;

`\omega^A = \omega^B`

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