A uniform horizontal rod of mass2.6kg and length 0.47 m is free to pivot about o

Question

A uniform horizontal rod of mass2.6kg and length 0.47 m is free to pivot about oneend as shown. The moment of inertia of the rod about an axisperpendicular to the rod and through the center of mass is givenby I =ml^2 / 12

If a 6.4 N force at an angle of 58degrees to the horizontal acts on the rod as shown, what is the magnitude ofthe resulting angular acceleration about the pivot point? Theacceleration of gravity is 9.8 m/s^2 . Answer in units ofrad/s^2

A uniform horizontal rod of mass2.6kg and length 0.47 m is free to pivot about oneend as shown. The moment of inertia of the rod about an axisperpendicular to the rod and through the center of mass is givenby I =ml^2 / 12

If a 6.4 N force at an angle of 58degrees to the horizontal acts on the rod as shown, what is the magnitude ofthe resulting angular acceleration about the pivot point? Theacceleration of gravity is 9.8 m/s^2 . Answer in units ofrad/s^2

Explanation / Answer

    Mass of the rod, M = 2.6 kg     Length of the rod, l = 0.47 m     Force applied, F = 6.4 N     Angle made by the force, =58o     Acceleration due to gravity, g = 9.8 m/s^2     Moment of inertia about an axis through pivot, I= M L2 / 3                                                                             = 2.6 * 0.472 / 3                                                                             = 0.1914 kg m^2     Torque about pivot, = F L Sin                                      = 6.4 * 0.47 * sin 58                                      = 2.55 N.m     But, Torque, = I     Angular acceleration, = / I                                        = 2.55 / 0.1914                                        = 13.33 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.