(890) Problem 8: A uniform thin rod of mass m = 1.3 kg and length L = 1.4 m can

Question

(890) Problem 8: A uniform thin rod of mass m = 1.3 kg and length L = 1.4 m can rotate about an axle through its center. Four forces are acting on it as shown in the figure. Their magnitudes are F1-7.5 N, F2-1.5 N, F3 13.5 N and F4 16.5 N F2 acts a distance d-0.11 m from the center of mass 45° 4 ©theexpertta.com 20% Part (a) Calculate the magnitude 1 of the torque due to force F1, in newton meters. Grade Summary 090 100% Potential cos0 cotanasin acos(0 atanacotansinhO cosh tanh)cotanh) Submissions Attempts remaining: 10 (0% per attempt) detailed view sin 0 END Degrees Radians Submit Hint I give up! Hints: 1% deduction per hint. Hints remaining: 2 Feedback: 3% deduction per feedback. 2006 Part (b) Calculate the magnitude 2 of the torque due to force F2 in newton meters 2096 Part (c) Calculate the magnitude 3 of the torque due to force F3 in newton meters 2006 Part (d) Calculate the magnitude 4 of the torque due to force F4 in newton meters 20% Part (e) Calculate the angular acceleration of the thin rod about its center of mass in radians per square second. Let the counter clockwise direction be positivie

Explanation / Answer

torque = force*perpendicular distance of the application point of force from the axis about which torque needs to be measured

part a:

for F1, perpendicular distance=half of the length of the rod

=1.4/2=0.7 m

then torque=T1=force*perpendicular distance

=7.5*0.7=5.25 N.m

part b:

perpendicular distance of force from the axis=d=0.11 m

component of F2 perpendicular to the rod (as the component of the force along the rod will be passing through the center and torque will be zero)

=F2*sin(45)=1.0607 N

then torque due to F2=1.0607*0.11=0.11667 N.m

part c:

as F3 passes through the center, perpendicular distance=0

hence torque due to F3=0

part d:

as F4 passes through the center, perpendicular distance=0

hence torque due to F4=0

part e:

torque due to F1 is in clockwise direction while torque due to F2 is in anticlockwise direction

total torque, taking anti clockwise direction to be positive=0.11667-5.25

=-5.1333 N.m

moment of inertia of the rod about its center=(1/12)*mass*length^2

=(1/12)*1.3*1.4^2

=0.21233 kg.m^2

then angular acceleration=total torque/moment of inertia

=-5.1333/0.21233

=-24.176 rad/s^2

magnitude of angular acceleration is 24.176 rad/s^2 and it is in clockwise direction.

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