A uniform disk with mass m = 8.78 kg and radius R = 1.42 m lies in the x-y plane

Question

A uniform disk with mass m = 8.78 kg and radius R = 1.42 m lies in the x-y plane and centered at the origin. Three forces act in the +y-direction on the disk: 1) a force 319 N at the edge of the disk on the +x-axis, 2) a force 319 N at the edge of the disk on the A uniform disk with mass m = 8.78 kg and radius R = 1.42 m lies in the x-y plane and centered at the origin. Three forces act in the +y-direction on the disk: 1) a force 319 N at the edge of the disk on the +x-axis, 2) a force 319 N at the edge of the disk on the

Explanation / Answer

1) TF1 = torque of force F1   = R * 319 *sin(90) = 1.42*319*1 = 452.98    az^2 Nm

      magnitude =452.98 Nm

2) TF2 = torques due to F2 = R1*F2*sin(0) = 0

3) TF3 = toruqe due to F3 = - R*F3*cos(39) az^ = -352 az

            magnitude = 352

T = net torque = TF1 + TF2 +TF3 = 452.98 az^ + 0 -352 aZ^   = 100.9 aZ^

4)   X component of net toruqe = 0

5)   y component of net torque =0

6 ) z component net torque =   100.9

7)   I = momnet of inertia = mR^2/2 =8.78 *(1.41)^2/2 =8.6 kg m^2

      a =agular accceleration = T/I = 100.9/8.6 =11.74 rad/s^2

8) at t= 1.7 s

    W =angular velocity = a*t = 11.74*1.7 =19.95 rad/s

ratational kinetic energy = I*w^2/2 = 8.86* (19.95)^2/2 =1711.41

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.