Problem 3 The 60kg flywheel has a radius of gyration about its center of mass of

Question

Problem 3 The 60kg flywheel has a radius of gyration about its center of mass of 260mm It rotates with a constant angular velocity of 1300rev/min before the brake is applied Figure 1 of 1 0.5 m m 0.2 m 0.3 m Part A If the coefficient of kinetic friction between the brake pad B and the wheel's rim is Auk 0.5, determine the constant force at must be applied to the braking mechanism's handle in order to stop the wheel in 150 revolutions. Figure 1 Express your answer with the appropriate units Value Units Submit My Answers Give U Provide Feedback Continue

Explanation / Answer

Here ,

theta = 150 revolution

theta = 942.3 rad

wi = 1300 rev/min

wi = 136.11 rad/s

Here , Using third equation of motion ,

wi^2 = 2*theta*a

a = 136.11^2/(2*942.3)

a = 9.83 rad/s^2

Now ,

as I = mk^2

I = 60*.26^2

I = 4.056 Kg.m^2

Now ,

Torque = F*r = I*a

N*0.5 * 0.3 = 4.056 * 9.83

N =265.8 N

Now , balacing Moment about A

P*1.5 - 0.2*0.5*265.8 -1*265.8 = 0

P = 194.9 N

the force P is 194.9 N

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