A 1m long stiff massless rod is free to rotate around a pivot indicated by the d

Question

A 1m long stiff massless rod is free to rotate around a pivot indicated by the dot. Assume Fa=8.0N, FB=32.0N, and Fc=24.0N. There is a 30degree between FB and Fc. The rod has a moment of inertia of 5.0kg m2.

A. What is the torque around the pivot due to force A?

B. What is the torque around the pivot due to force B?

C. What is the torque around the pivot due to force C?

D. What is the net angular acceleration of the rod?

E. If the bars starts from rest, what will the angular velocity of the bar be after 7 seconds?

Note: the radius between FA to the pivot is 25cm. From the pivot to FB and Fc is 75cm.

Explanation / Answer

A)   torque around the pivot due to force A = 8 * 0.25

                                                                    = 2 N.m

B)   torque around the pivot due to force B = 32 * 0.75

                                                                     = 24 N.m

C)   torque around the pivot due to force C = 24 * 0.75 * sin120

                                                                     = - 15.588 N .m

D)     net angular acceleration of the rod   =   ( 24 + 2 - 15.588)/5

                                                                 =   2.082 rad/sec2

E)   angular velocity of the bar be   =   2.082 * 7

                                                        = 14.574 rad/sec

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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