A rod of length L = 2.00 m is pivoted about a point that is L/3 from the left end, and it can rotate in the horizontal plane. The rod has a nonuniform mass density 7.00x⁴+3.00, where x=0 at the pivot. A force of magnitude 20.4 N is applied to the rod a distance of 0.500 m from the end farthest from the pivot. The force makes an angle of 65 with the rod. This is the only torque acting on the rod.

a)Discuss with your peers and write down a plan for how to calculate the tangential acceleration of a point that is 0.250 m from the pivot. Then execute your plan to find the solution, showing your work and annotating each step.

b)What is the value of the tangential acceleration you found in part (a), in m/s²?

c)Assume that this torque acts for a total of 0.320 seconds. Assume the bar starts from rest. Use an annotated step-by-step solution for the final tangential speed of a point 0.250 m from the axis to show the answer.

d)Enter the tangential speed you found in part (c) in m/s.

e)Use an annotated step-by-step solution for the total mass of the rod to show the answer.

f)Enter the mass that you found in part (e) in kg.