AP Physics C Mechanics Systems and Center of Mass FRQ

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%

AP Physics C Mechanics Exam Style Questions – All Topics

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

Scroll to Top