A uniform disk with mass m-9.39 kg and radius R 1.3 m lies in the x-y plane and

Question

A uniform disk with mass m-9.39 kg and radius R 1.3 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 311 N at the edge of the disk on the +x-axis, 2) a force 311 N at the edge of the disk on the y axis, and 3) a force 311 N acts at the edge of the disk at an angle 0-34 above the -x-axis. 1) What is the magnitude of the torque on the disk about the z axis due to F1? N-m Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question 2) What is the magnitude of the torque on the disk about the z axis due to F2? 0 You currently have 1 submissions for this question, Only 10 submission are allowed. N-m Submit

Explanation / Answer

1)Torque due to f1 will be:

tau = F1 r1 sin(theta)

tau = 311 x 1.3 x sin90

tau = 404.3 N-m

2)tau = 0

Since, theta = 180 deg ; sin(180) = 0

tau = 0

3)tau = F3 r3 cos(theta)

tau = 311 x 1.3 x cos34 = 335.2 N-m

Hence, tau = 335.2 Nm

4)0

5)0

6)tau = 404.3 - 335.2 = 69.1 Nm

Hence, tau = 69.1 Nm

7)we know that

tau = I alpha => alpha = tau/I

I = 1/2 m r^2 = 0.5 x 9.39 x 1.3^2 = 7.93 kg-m^2

tau = 69.1/7.93 = 8.71 rad/s^2

hence, tau = 8.71 rad/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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