Question

14. A thin rod (mass lkg and length Im) with its rotation axis perpendicular to its length has a moment of incrtia of 1/12 ML2. Two forces are applied to the rod. The points of application and the directions of the forces are indicated in the diagram below. The orientations of the forces remain constant relative to the rod as the rod rotates and produce a constant torque on the rod. through its centre and Rotation axis (a) If the force magnitudes are Fi 10 N and F2 10 N and torque on the rod? 30°. What is the net (4 marks) (b) If the rod starts off from rest, what is the angular velocity after 5 seconds? (4 marks) TURN OVER PHYS101 - 12- (c) Ifthe rotation is stopped and the rotation axis is changed to be at the end of the rod at which the force Fi is applicd, what will be the angular velocity of the rod after 5 econds assuming that everything else rer ns the me?

Answer 1

Albout point 'o', -he inerbia of Yod is ML F. is unter clock.wise toraue Rotahon Let's say the axis and kclock - wise torque be Posibve negahve A F, 10 N F 10 N &=30. Given Fi I m. M Ikg L to unter clock-wise tovque forces produce o both Ca) About (F* 늘)• (F,* 늘~ Ga4) Net toraue od To is on (r. lo x 9.33 N-m 금지지 The inertia of Tod is 1K8.m2 (b) 9.33 angula To accelerabion of 4he Yod is Sp the (12) 11.96 Yade Since starts trom rest, initial angular speed angular velocity t 5 Sec, the be After is Cill 559.8 radsec Fabout the torque due when Yotahon changed to point to axis is A bill be Zero The torque due will be 8.66 N-m angulay accelerahion will be A So where 'A' iner hal about Sc 8-66 So a= 25.98 rad/sec, So andulav Velocity alter S sec isill be w, 0 + (2 5.9 8) (5) = 129.9 rad/sec