Rotational Inertia AP Physics 1 FRQ – Exam Style Questions etc.
Rotational Inertia AP Physics 1 FRQ
Unit 5: Torque and Rotational Dynamics
Weightage : 10-15%
Exam Style Practice Questions, Rotational Inertia AP Physics 1 FRQ
Question
The horizontal uniform rod shown above has length 0.60 m and mass 2.0 kg. The left end of the rod is attached to a vertical support by a frictionless hinge that allows the rod to swing up or down. The right end of the rod is supported by a cord that makes an angle of 30° with the rod. A spring scale of negligible mass measures the tension in the cord. A 0.50 kg block is also attached to the right end of the rod.
a. On the diagram below, draw and label vectors to represent all the forces acting on the rod. Show each force vector originating at its point of application.
b. Calculate the reading on the spring scale.
The rotational inertia of a rod about its center is \(\frac{1}{12}ML^{2}\), where M is the mass of the rod and L is its length.
c. Calculate the rotational inertia of the rod-block system about the hinge.
d. If the cord that supports the rod is cut near the end of the rod, calculate the initial angular acceleration of the rod-block system about the hinge
Answer/Explanation
Ans:
a. 
b. \(\sum \) τ= 0
About the hinge: TL sin 30º – mgL – Mg(L/2) = 0 gives T = 29 N
c. Itotal = Irod + Iblock where Irod,end = Icm + MD2 = ML2/12 + M(L/2)2 = ML2/3
Itotal = ML2/3 + mL2 = 0.42 kg-m2
d. \(\sum \) τ = Iα
mgL + MgL/2 = Iα gives α = 21 rad/s2
Question
A cloth tape is wound around the outside of a uniform solid cylinder (mass M, radius R) and fastened to the ceiling as shown in the diagram above. The cylinder is held with the tape vertical and then released from rest. As the cylinder descends, it unwinds from the tape without slipping. The moment of inertia of a uniform solid cylinder about its center is ½MR2
.
a. On the circle above draw vectors showing all the forces acting on the cylinder after it is released. Label each force clearly.
b. In terms of g, find the downward acceleration of the center of the cylinder as it unrolls from the tape.
c. While descending, does the center of the cylinder move toward the left, toward the right, or straight down? Explain.
Answer/Explanation
Ans:
a. 
b. \(\sum \)τ= Iα (about center of mass) (one could also choose about the point at which the tape comes off the cylinder)
TR = ½ MR2×(a/R)
T = ½ Ma
\(\sum \)F = ma
Mg – T = Ma
Mg = 3Ma/2
a = 2g/3
c. As there are no horizontal forces, the cylinder moves straight down.