Tutorial Sheet 5: 3D Kinematics Answers

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Question 1

The aeroplane’s angular velocity vector relative to an earth-fixed reference frame, expressed in terms of the body-fixed coordinate system shown, is $\omega$ = 0.62i+0.45j−0.23k rad/s. The coordinates of point A of the airplane are (3.6, 0.8, −1.2) m. What is the velocity of point A relative to the velocity of the aeroplane’s center of mass?


Answer

\[v_A=v_O+\omega \times r_{A/O} \\ = 0 + \begin{vmatrix} i & j & k\\ 0.62 & 0.45 & -0.23 \\ 3.6 & 0.8 & -1.2 \end{vmatrix} \\ = -0.356i-0.084j-1.12k \text{ m/s}\]

Question 2

The angular velocity of the cube relative to the primary reference frame, expressed in terms of the body-fixed coordinate system shown is $\omega$=−6.4i+8.2j+12k rad/s.The velocity of the center of mass G of the cube relative to the primary reference frame at the instant shown is $v_G$=26i+14j+32k m/s. What is the velocity of point A of the cube relative to the primary reference frame at the instant shown?


Answer

The vector A to G is

\[i+j+k\]

Then

\[v_A=v_G+\omega \times r_{A/G} \\ v_A= 26i+14j+32k + \begin{vmatrix} i & j & k\\ -6.4 & 8.2 & 12 \\ 1 & 1 & 1 \end{vmatrix} \\ v_A = 22.2i+32.4j+17.4k \text{ m/s}\]

Question 3

Using the cube in Q2, the coordinate system shown is fixed with respect to the cube. The angular velocity of the cube relative to the primary reference frame, $\omega$=−6.4i+8.2j+12k rad/s, is constant.The acceleration of the center of mass G of the cube relative to the primary reference frame at the instant shown is $a_G$=136i+76j−48k m/s $^2$. What is the acceleration of point A of the cube relative to the primary reference frame at the instant shown?

Answer

There is constant angular velocity so alpha is 0. The acceleration can be found

\[a_A=a_G+\alpha \times r_{A/G}+ \omega \times(\omega\times r_{A/G}) \\ a_A=(136i+76j-48k) +0 + (-6.4i + 8.2j + 12k ) \times \begin{vmatrix} i & j & k\\ -6.4 & 8.2 & 12 \\ 1 & 1 & 1 \end{vmatrix} \\ a_A= 136i+76j-48k + \begin{vmatrix} i & j & k\\ -6.4 & 8.2 & 12 \\ -3.8 & 18.4 & -14.6 \end{vmatrix} \\ v_A = -204.5i-63.04j-135k \text{ m/s}\]

Question 4

The origin of the secondary coordinate system shown is fixed to the center of mass G of the cube. The velocity of the center of mass G of the cube relative to the primary reference frame at the instant shown is $v_G$=26i+14j+32 m/s. The cube is rotating relative to the secondary coordinate system with angular velocity $\omega_{rel}$ =6.2i−5j+8.8k rad/s. The secondary coordinate system is rotating relative to the primary reference frame with angular velocity $\omega$ = 2.2i+4j−3.6k rad/s.

(a) What is the velocity of point A of the cube relative to the primary reference frame at the instant shown?
(b) If the components of the vectors $\omega_{rel}$ and are constant, what is the cube’s angular acceleration relative to the primary reference frame?


Answer

There is a lot going on here, so annotate


(a)

\[\omega = \Omega+\omega_{rel} \\ \omega = (22i+4j-3.6k)+(6.2i-5j+8.8k) \\ \omega = 8.4i-j+5.2k\] \[v_A=v_G+\omega \times r_{A/G} \\ v_A= 26i+14j+32k + \begin{vmatrix} i & j & k\\ 8.4 & -1 & 5.2 \\ 1 & 1 & 1 \end{vmatrix} \\ v_A = 19.8i+10.8j+41.4k \text{ m/s}\]

(b)

\[\alpha = \frac{d\omega}{dt}+\Omega\times \omega_{rel} \\ \alpha = 0 + \begin{vmatrix} i & j & k\\ 2.2 & 4 & -3.6 \\ 6.2 & -5 & 8.8 \end{vmatrix} \\ \alpha = 17.2i-41.7j-35.8k \text{ rad/s}^2\]

Question 5

Relative to an earth-fixed reference frame, points A and B of the rigid parallelepiped are fixed and it rotates about the axis AB with an angular velocity of 30 rad/s. Determine the velocities of points C and D relative to the earth-fixed reference frame.


Answer

We need to know the vector $\omega$ as we only know the overall magnitude. This can be found using the unit vector of the axis of rotation.

\[(30)\frac{0.4i + 0.2j - 0.4k}{\sqrt{0.4^2+0.2^2+0.4^2}} \\ = 20i+10j-20k\]

Then solving for velocity of C and D

\[v_C = v_A+\omega\times r_{C/A} \\ v_C = 0 + \begin{vmatrix} i & j & k\\ 20 & 10 & -20 \\ 0 & 0.2 & 0 \end{vmatrix} \\ v_C = 4i+4k\] \[v_D = v_A+\omega\times r_{C/A} \\ v_D = 0 + \begin{vmatrix} i & j & k\\ 20 & 10 & -20 \\ 0.4 & 0.2 & 0 \end{vmatrix} \\ v_D = 4i-8j\]

Question 6

Using the parallelepiped in Q5, relative to the xyz coordinate system shown, points A and B of the rigid parallelepiped are fixed and the parallelepiped rotates about the axis AB with an angular velocity of 30 rad/s. Relative to an earth fixed reference frame, point A is fixed and the xyz coordinate system rotates with angular velocity −5i+8j+6k rad/s. Determine the velocities of points C and D relative to the earth-fixed reference frame.

Answer

Again, we need to find $\omega$, but this time it becomes $\omega_{rel}$ as the coordinate system is rotating as well.

\[(30)\frac{0.4i + 0.2j - 0.4k}{\sqrt{0.4^2+0.2^2+0.4^2}} \\ \omega_{rel}= 20i+10j-20k\]

$\omega$ is then

\[\omega = \Omega+\omega_{rel} \\ \omega = (-5i+8j+6k)+(20i+10j-20k) \\ \omega = 15i+18j-14k\]

Then solving for velocity of C and D

\[v_C = v_A+\omega\times r_{C/A} \\ v_C = 0 + \begin{vmatrix} i & j & k\\ 15 & 18 & -14 \\ 0 & 0.2 & 0 \end{vmatrix} \\ v_C = 2.8i+3k\] \[v_D = v_A+\omega\times r_{C/A} \\ v_D = 0 + \begin{vmatrix} i & j & k\\ 15 & 18 & -14 \\ 0.4 & 0.2 & 0 \end{vmatrix} \\ v_D = 2.8i-5.6j-4.2k\]

Question 7

Relative to an earth-fixed reference frame, the vertical shaft rotates about its axis with angular velocity $\omega_O$=4 rad/s. The secondary xyz coordinate system is fixed with respect to the shaft and its origin is stationary. Relative to the secondary coordinate system, the disk (radius 8 cm) rotates with constant angular velocity $\omega_d$=6 rad/s. At the instant shown, determine the velocity of point A

(a) Relative to the secondary reference frame.
(b) Relative to the earth-fixed reference frame.


Answer


(a)

\[v_A=v_O+\omega_{rel} \times r_{A/O} \\ v_A=0+ \begin{vmatrix} i & j & k\\ 6 & 0 & 0 \\ 0 & 8\sin(45) & 8\cos(45) \end{vmatrix} \\ v_A = -33.9j+33.9k \text{ cm/s}\]

(b)

\[\omega = \Omega+\omega_{rel} \\ \omega = 4j+6i \\ v_A=v_O+\omega \times r_{A/G} \\ v_A=0+ \begin{vmatrix} i & j & k\\ 6 & 4 & 0 \\ 0 & 8\sin(45) & 8\cos(45) \end{vmatrix} \\ v_A = 22.6i-33.9j+33.9k \text{ cm/s}\]

Question 8

The object in figure (a) is supported by bearings at A and B in figure (b). The horizontal circular disk is supported by a vertical shaft that rotates with angular velocity $\omega_O$=6 rad/s. The horizontal bar rotates with angular velocity $\omega$=10 rad/s. At the instant shown,

(a) What is the velocity relative to an earth-fixed reference frame of the end C of the vertical bar?
(b) What is the angular acceleration vector of the object relative to an earth-fixed reference frame?
(c) What is the acceleration relative to an earth-fixed reference frame of the end C of the vertical bar?


Answer


(a)

\[\omega = \Omega+\omega_{rel} \\ \omega = 6j+10i\] \[v_C = v_O+\omega \times r_{C/O} \\ v_C = 0+\begin{vmatrix} i & j & k\\ 10 & 6 & 0 \\ 0.1 & 0.1 & 0 \end{vmatrix} \\ v_C = 0.4k \text{ m/s}\]

(b)

\[\alpha = \frac{dw}{dt} + \Omega\times\omega \\ \alpha = 0+\begin{vmatrix} i & j & k\\ 0 & 6 & 0 \\ 10 & 6 & 0 \end{vmatrix} \\ \alpha = -60k \text{ rad/s}^2\]

(c) We’ve worked out the last sum in part (a) so you can shortcut it

\[a_C = a_O+\alpha\times r_{C/O}+\omega\times(\omega\times r_{C/O}) \\ a_C = a_O+ \begin{vmatrix} i & j & k\\ 0 & 0 & -60 \\ 0.1 & 0.1 & 0 \end{vmatrix} + \begin{vmatrix} i & j & k\\ 10 & 6 & 0 \\ 0 & 0 & 0.4 \end{vmatrix} \\ a_C = 8.4i-10j \text{ m/s}^2\]

Question 9

The point of the spinning top remains at a fixed point on the floor, which is the origin O of the secondary reference frame shown. The top’s angular velocity vector relative to the secondary reference frame, $\omega_{rel}$=50k rad/s, is constant. The angular velocity vector of the secondary reference frame relative to an earth-fixed primary reference frame is $\omega$=2j+5.6k rad/s. The components of this vector are constant. (Notice that it is expressed in terms of the secondary reference frame.)

(a) Determine the velocity relative to the earth-fixed reference frame of the point of the top with coordinates (0, 20, 30) mm.
(b) What is the top’s angular acceleration vector relative to the earth-fixed reference frame
(c) Determine the acceleration relative to the earth fixed reference frame of the point of the top with coordinates (0, 20, 30) mm


Answer

(a)

\[\omega = \Omega + \omega_{rel} \\ \omega = 50k+2j+5.6k \\ \omega = 2j+55.6k\] \[v_A = v_O+\omega\times r_{A/O} \\ v_A = 0+\begin{vmatrix} i & j & k\\ 0 & 2 & 55.6 \\ 0 & 0.02 & 0.03 \end{vmatrix} \\ v_A = -1.05i \text{ m/s}\]

(b)

\[\alpha = \frac{dw}{dt} + \Omega\times\omega \\ \alpha = 0+\begin{vmatrix} i & j & k\\ 0 & 2 & 5.6 \\ 0 & 2 & 55.6 \end{vmatrix} \\ \alpha = 100i \text{ rad/s}^2\]

(c) Shortcut like in the previous question

\[a_A = a_C+\alpha\times r_{A/C}+\omega\times(\omega\times r_{A/C}) \\ a_A = 0 + \begin{vmatrix} i & j & k\\ 100 & 0 & 0 \\ 0 & 0.02 & 0.03 \end{vmatrix} + \begin{vmatrix} i & j & k\\ 0 & 2 & 55.6 \\ -1.05 & 0 & 0 \end{vmatrix} \\ a_C = -61.4j+4.1k \text{ m/s}^2\]

Question 10

The radius of the circular disk is R=0.2 m, and b=0.3 m. The disk rotates with angular velocity $\omega_d$=6 rad/s relative to the horizontal bar. The horizontal bar rotates with angular velocity $\omega_b$=4 rad/s relative to the vertical shaft, and the vertical shaft rotates with angular velocity $\omega_O$=2 rad/s relative to an earth-fixed reference frame. Assume that the secondary reference frame shown is fixed with respect to the horizontal bar.

(a) What is the angular velocity vector $\omega_{rel}$ of the disk relative to the secondary reference frame?
(b) Determine the velocity relative to the earth-fixed reference frame of point P, which is the uppermost point of the disk.


Answer

There’s a lot of moving parts, so I’m going to label it so algebra symbols don’t get mixed up.


(a) This is simply

\[\omega_{rel} = 6i \text{ rad/s}\]

(b) The angular velocity of the disk to the primary reference frame is

\[\omega = \omega_{rel}+\omega_{rel}+\Omega \\ \omega = 6i+2j+4k\]

The velocity of M (middle of the disk) is

\[v_M = v_O+\Omega\times r_{C/O} \\ v_M = 0 + \begin{vmatrix} i & j & k\\ 0 & 2 & 4 \\ 0.3 & 0 & 0 \end{vmatrix} \\ v_M=1.2j-0.6k\]

The velocity of P is

\[v_P = v_M+\omega\times r_{P/M} \\ v_P = 1.2j-0.6k + \begin{vmatrix} i & j & k\\ 6 & 2 & 4 \\ 0 & 0.2 & 0 \end{vmatrix} \\ v_P = -0.8i+1.2j+0.6k \text{ m/s}\]