A horizontal rod with a mass of 6.0 kg and a length of 2.3 m is free to rotate a

Question

A horizontal rod with a mass of 6.0 kg and a length of 2.3 m is free to rotate about a horizontal axle that is perpendicular to the rod and passes through its center. Someone is pushing straight down on the left side of the rod with a force of 180 N at a point which is 75 cm to the left of the center of the rod. Someone else is pushing down on the right side of the rod with a force of 120 N at a point which is 44 cm to the right of the center of the rod. What is the resulting angular acceleration of the rod? a) 21 rad/s^2 b) 31 rad/s^2 c) 41 rad/s^2 d) 51 rad/s^2

Explanation / Answer

net torque on rod = 180 (0.75) - `120 (0.44) = 135- 52.8 = 82.2 N -m

Torque= I ( alpha0

I of rod= 1/12 ( 6) (2.3)^2= 2.645

82.2 = 2.645 ( alpha)

alpha= 31 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.