quiz rotational dynamics

Transcription

quiz rotational dynamics
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