Part 2. Torque and angular motion. 3. A uniform rod of length and mass M is atta

Question

Part 2. Torque and angular motion. 3. A uniform rod of length and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane, as shown in the figure to the right. The rod is released from rest in the horizontal position. Calculate: (a) the initial torque acting on the rod; (b) the initial angular acceleration of the rod (c) the initial linear (tangential) acceleration of its right end; (d) the torque acting on the rod after it swings down by 45° (Hint: is the torque constant?); (e) the angular velocity of the rod after it swings down by 45° (Hint:is the torque constant?). CM 7.00cm F 1-4.5kg

Explanation / Answer

3a] Torque = mg*L/2 = 0.5 mgL = 0.5*4.5*9.8*0.07 = 1.5435 Nm

b] initial angular acceleration alpha = torque/moment of inertia

=  0.5 mgL/(1/3 mL^2) = 3g/2L = 3*9.8/(2*0.07) = 210 rad/s^2

c] a = alpha *L = 3g/2L *L = 1.5g = 1.5*9.8 = 14.7 m/s^2

d] Torque = mg*L/2 cos 45 degree = 0.5 mgL cos 45 degree= 0.5*4.5*9.8*0.07 cos 45 degree

= 1.0914 Nm

e] angular acceleration alpha = torque/moment of inertia

=  0.5 mgL cos 45 degree /(1/3 mL^2) = 3g cos 45 degree/2L

= 3*9.8*cos 45 degree/(2*0.07)

= 148.49 rad/s^2

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