Rotational Dynamics Quiz V1 PSI AP Physics I Name_______________________________ 1. A force of 32 N is applied to a doorknob at an angle of 820 to the line connecting the doorknob to the door hinges (axis of rotation). The doorknob is 0.87 m from the hinges. What is the torque applied to the hinges? 2. A thin rod of length 0.62 m rests on a fulcrum at its midpoint. A mass of 0.16 kg is placed at one end of the rod. Where must a mass of 0.23 kg be placed to balance the rod (α = 0 rad/s2)? 3. Two students are on opposite sides of a merry-go-round of radius 2.3 m and mass 98 kg. One pushes tangent to the outer edge in the clockwise direction with a force of 120 N. The other pushes in the counter-clockwise direction with a force of 96 N. Model the merry-go-round as a disc (I=1/2MR2). What is the angular acceleration of the merry-go-round (magnitude and direction)? 4. Two masses of 2.5 kg and 4.8 kg are attached to either end of a massless rod of length 0.92 m. Compute the moment of inertia for: a) The rod is rotated about its midpoint. b) The rod is rotated at a point 0.11 m from the 2.5 kg mass. c) The rod is rotated about the position of the 4.8 kg mass. Rotational Dynamics Quiz V1 PSI AP Physics I ANSWER KEY 22 points 1. τ=rFsinθ τ=(0.87m)(32N)sin820 τ = 27.6 N-m +1 for equation +1 for substitution +1 for correct answer with units 2. Στ=0 Στ=r1F1-r2F2=0 r2=(r1m1g)/m2g=(r1m1)/m2 r2 = ((0.31m)(.16kg))/(.23kg) r2 = 0.22 m +1 for equation 3. Στ=Iα Στ=r1F1-r2F2=Iα α=(r(F1-F2))/I +1 for τ equation +1 for algebraic manipulation +1 for substitution +1 for correct answer with units +1 for algebraic manipulation I=1/2MR2 I=1/2(98kg)(2.3m)2 I=260kg-m2 +1 for I equation +1 for substitution +1 for correct answer with units α=(2.3m(96N-120N)/(260kg-m2) α = -0.21 rad/s2 – negative sign indicates clockwise rotation. +1 for substitution +1 for correct magnitude with units +1 for correct direction 4. I=Σmiri2 +1 for equation a. Imp=(2.5kg)(.46m)2+(4.8kg)(.46m)2 Imp = 1.5 kg-m2 +1 for substitution +1 for correct answer with units b. I.11=(2.5kg)(.11m)2+(4.8kg)(.81m)2 I.11 = 3.2 kg-m2 +1 for substitution +1 for correct answer with units c. I4.8=(2.5kg)(.92m)2+(4.8kg)(0m)2 I4.8 = 2.1 kg-m2 +1 for substitution +1 for correct answer with units