How do you do this problem, I set it up with the intial and final kinetic energi

Question

How do you do this problem, I set it up with the intial and final kinetic energies and the spring final, getting a squareroot of a negative.

The mass center of the 7-kg unbalanced wheel is located G. The mass miment of incritia of the wheel about its center O is l0 = 0.3 kg middot m2. In thc position shown. the angular velocity of the wheel is 3 rad/s clockwisc and the horizonal spring is undeformed. Find the angular velocity of the wheel aftcr it has rolled clockwise through 180degree from this position. Assume that the whecl does not slip

Explanation / Answer

F = Friction force

After theta deg rotation, torque = -mg*(0.1*sin theta) + F*r

and distance travelled = theta*r

Torque = I*alpha

and alpha = a/r

-mg*(0.1*sin theta) + F*r = 0.3*a/r

-7*9.81*0.1*sin theta + F*0.2 = 0.3*a/0.2

-6.867*siin theta + F*0.2 = 1.5*a.............(1)

Spring deflection x = distance travelled = theta*r = 0.2*theta

Spring force = kx = 75*0.2*theta = 15*theta

Also, Newton's 2nd law gives, F - kx = ma

F - 15*theta = 7*a...............(2)

From 1 and 2,

-6.867*sin theta + (15*theta + 7a)*0.2 = 1.5*a

-6.867*sin theta + 15*theta = 0.1*a

Putting a = r*alpha = r*(dw / dt) = r*(dw / d theta)*(d theta / dt) = r*(dw / d theta)*w

or a = r*w*(dw / d theta) = 0.2*w*(dw / d theta)

-6.867*sin theta + 15*theta = 0.1*0.2*w*(dw / d theta)

w dw = (-343.35*sin theta + 750*theta) d theta

Integrating, w^2 /2 = (343.35*cos theta + 750*theta^2 /2) + C

At tehta = 0 we have w = 3 rad/s.

Thus, C = -338.85

Thus, w^2 /2 = (343.35*cos theta + 750*theta^2 /2) -338.85

When theta = 180 deg or pi rad = 3.14 rad we get

w^2 /2 = (343.35*cos pi + 750*3.14^2 /2) -338.85

w = 77.65 rad/s

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