Systems and Center of Mass AP Physics C Mechanics FRQ – Exam Style Questions etc.
Systems and Center of Mass AP Physics C Mechanics FRQ
Unit 2: Force and Translational Dynamics
Weightage : 20-15%
Question
Object A is a long, thin, uniform rod of mass M and length 2L that is free to rotate about a pivot of negligible friction at its left end, as shown above.
(a) Using integral calculus, derive an expression to show that the rotational inertia IA of object A about the pivot is given by \(\frac{4}{3}ML^{2}\) .
Object B of total mass M is formed by attaching two thin, uniform, identical rods of length L at a right angle to each other. Object B is held in place, as shown above. Express your answers in part (b) in terms of L .
(b) Determine the following for the given coordinate system shown in the figure.
i. The x-coordinate of the center of mass of object B
ii. The y-coordinate of the center of mass of object B
Object B has a rotational inertia of IB about its pivot.
(c) Is the value of IB greater than, less than, or equal to IA ?
_____ Greater than _____ Less than _____ Equal to
Justify your answer.
Object B is released from rest and begins to rotate about its pivot.
(d) On the axes below, sketch graphs of the magnitude of the angular acceleration α and the angular speed ω of object B as functions of time t from the time it is released to the time its center of mass reaches its lowest point.
(e) While object B rotates from the horizontal position down through the angle θ shown above, is the magnitude of its angular acceleration increasing, decreasing, or not changing?
_____ Increasing _____ Decreasing _____ Not changing
Justify your answer.
Object B rotates through the position shown above.
(f) Derive an expression for the angular speed of object B when it is in the position shown above. Express your answer in terms of M , L , IB , and physical constants, as appropriate.
Answer/Explanation
Ans:
(a)
I = MR2 \(dm = \frac{m}{2L}dx\)
dI = dmr2 \(I = \int_{0}^{2L}\frac{m}{2L}r^{2}dr\)
\(I = \frac{m}{2L}\int_{0}^{2L}r^{2}dr\)
\(I = \frac{m}{2L}_{4}^{0}\textrm{}\left ( \frac{1}{5}r ^{3}|_{0}^{2L}\right )\)
\(I = \frac{m8L^{2}}{6L}=\frac{4}{3}ML^{2}\)
(b)
i. Each bunch mass m/2
\(X_{am}=\frac{\sum x_{i}m_{i}}{m_{total}}=\frac{\left ( \frac{L}{2} \right )\left ( \frac{M}{2} \right )+L\left ( \frac{M}{2} \right )}{m}=\frac{\frac{3ml}{4}}{m}=\frac{3}{4}L\)
ii.
\(Y_{am}=\frac{\sum x_{i}m_{i}}{m_{total}}=\frac{\left ( \frac{L}{2} \right )\left ( \frac{M}{2} \right )+L\left ( \frac{M}{2} \right )}{m}=\frac{\frac{3ml}{4}}{m}=\frac{3}{4}L\)
(c)
√ Less than
Object as must is distributed further anny thus the phot up to distance 2l1 while object B mast is distributed closer to the point.
(d)
(e)
√ Decreasing
As it rotates towards eunhilibrium, the ampronent of gravity tangent to the motion of its center of mass devengds, so its aneleatun also devenges.
(f)
PE los!
\(mgh =\left ( \frac{1}{2}L \right )(10Lm = 5ML)\)
averted to statements kBs \(w = \sqrt{\frac{10ML}{I_{B}}}\)
1/2 Iw2 = 5ML
Iw2 = 10 ML