Gear A has a mass of 1 kg and a radius of gyration of 30 mm; gear B has a mass of 4 kg and a radius of gyration of 75 mm; gear C has a mass of 9 kg and a radius of gyration of 100 mm. The system is at rest when a couple M0 of constant magnitude 4 N, m is applied to gear C. Assuming that no slipping occurs between the gears, determine the number of revolutions required for disk A to reach an angular velocity of 288 rpm. (Round the final answer to three decimal places.)

The number of revolutions required for disk A to reach an angular velocity of 288 rpm is rev.

(4.5125 * 10^-3 kg.m^2)ω_A^2

Explanation:

solution:

Moments of inertia:

I = mk^2

Gear A: I_A = (1)(0.030 m)^2 = 0.9*10^-3 kg.m^2

Gear B: I_B = (4)(0.075 m)^2 = 22.5*10^-3 kg.m^2

Gear C: I_C = (9)(0.100 m)^2 = 90*10^-3 kg.m^2

Let r_A be the radius of gear A, r_1 the outer radius of gear B, r_2 the inner radius of gear B, and r_C the radius of gear C.

r_A=50 mm

r_1 =100 mm

r_2 =50 mm

r_C=150 mm

At the contact point between gears A and B,

r_1*ω_b = r_A*ω_A

ω_b = r_A/r_1*ω_A

       = 0.5ω_A

At the contact point between gear B and C.

At the contact point between gears A and B,

r_C*ω_C = r_2*ω_B

ω_C = r_2/r_C*ω_B

       = 0.1667ω_A

kinetic energy T = 1/2*I_A*ω_A^2+1/2*I_B*ω_B^2+1/2*I_C*ω_C^2

                           =(4.5125 * 10^-3 kg.m^2)ω_A^2

answer:

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explanation: