Class Management | Help Chapter 11 Rotational Motion Dynamics Extension Begin Da
ID: 1793572 • Letter: C
Question
Class Management | Help Chapter 11 Rotational Motion Dynamics Extension Begin Date: 11:23/2017 10:00:00 PM 11/27/2017 11:59:00 PM Due Date: 1127/2017 11:59:00 PM End Date: (10%) Problem 5: To stud torque experimentally. vou apply a force to a beam. One end of the beam is attached to a pivot that allows the beam to rotate freely. The pivot is taken as the oriin or your coordinate system. You apply a force ofF- FFyJ+ F kat a pointrryJ+rkon the beam. 33% Part a Enter a vector expression for the resilting torque. In terms of the unit vectors i k and the components of F and r Grade Summary 0% Potential 100 Atternpls reTanng: 6 (1% per attempt) detailed vlew F. Submit Hint I give up Hints: 100 deduction per hint. Hints Feedback: 1 % deduction per feedback. 33% Part 1) Calculate the magnitude of the torque in newton meters, when the components of the position and force vectors have the values rx = 4.7 m. ry = 0.065 m, r,-0.025 m. F,--2.6 . Fy = 8.8 N, F,-1.2 N. 33% Part (c) If the moment of nertia of the bcam with respcct to the pivot is 1-79 kgm. calculate the magnitude of the angular acceleration of the beam about the pivot, in radiaus per second squared.Explanation / Answer
a)The torque is given by cross product of vectors r and F
Torque,T = (rxi+ryj+rzk)x(Fxi+Fyj+Fzk) = (ryFz-rzFy)i + (rzFx-rxFz)j+(rxFy-ryFx)k
b)For the values given,the cross product = -0.142i-5.705j+41.529k Nm
Magnitude of torque is = sqrt(0.1422+5.7052+41.5292)=41.92 Nm
c)Angular acceleration = Torque/moment of inertia = 41.92/79=0.53 rad/s2