Tutorial Sheet 2: Planar Kinematics with Acceleration Answers

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Question 1

The rigid body rotates about the z axis with counterclockwise angular velocity ω = 4 rad/s and counterclockwise angular acceleration α = 2 rad/s $^2$. The distance $r_{A/B}$ = 0.6m.

(a) What are the rigid body’s angular velocity and angular acceleration vectors?
(b) Determine the acceleration of point A relative to point B.


Answer

(a) Using the thumb rule, both are positive hence

\[\omega = 4k \\ \alpha = 2k\]

(b)

\[a_{A/B}=a_B+\alpha\times r_{A/B}+\omega\times(\omega\times r_{A/B}) \\ a_{A/B}=0+2k\times 0.6i+4k\times(4k\times 0.6i) \\ a_{A/B} = -9.6i+1.2j \text{ m/s}^2\]

or

\[a_{A/B}=a_B+\alpha\times r_{A/B}-\omega^2 r_{A/B} \\ a_{A/B}=0+2k\times 0.6i - 4^2. 0.6i \\ a_{A/B} = -9.6i+1.2j \text{ m/s}^2\]

Question 2

The helicopter is in planar motion in the xy plane. At the instant shown, the position of its center of mass G is x=2 m, y=2.5 m, its velocity is $v_G$ = 12i + 4j m/s, and its acceleration is $a_G$ = 2i + 3j m/s $^2$. The position of point T where the tail rotor is mounted is x=-3.5 m, y=4.5 m. The helicopter’s angular velocity is 0.2 rad/s clockwise, and its angular acceleration is 0.1 rad/s $^2$ counter-clockwise.

What is the acceleration of point T?


Answer

Draw it out


It then follows

\[a_{T}=a_G+\alpha\times r_{T/G}-\omega^2 r_{T/G} \\ a_{T}=2i+3j+\begin{vmatrix} i & j & k\\ 0 & 0 & 0.1 \\ -5.5 & 2 & 0 \end{vmatrix} -0.2^2(-5.5i+2j) \\ a_T= 2.02i+2.37j \text{ m/s}^2\]

Question 3

The bar rotates with a counterclockwise angular velocity of 5 rad/s and a counterclockwise angular acceleration of 30 rad/s $^2$. Determine the acceleration of A using $ a_{A}=a_B+\alpha\times r_{A/B}+\omega\times(\omega\times r_{A/B}) $.


Answer

For this question, you use the full expansion version of the equation meaning a lot of cross products!

Assuming B as the axis of rotation

\[a_{A}=a_B+\alpha\times r_{A/B}+\omega\times(\omega\times r_{A/B}) \\ a_A = \begin{vmatrix} i & j & k\\ 0 & 0 & 30 \\ 2\cos(30) & 2\sin(30) & 0 \end{vmatrix} + 5k\times \begin{vmatrix} i & j & k\\ 0 & 0 & 5 \\ 2\cos(30) & 2\sin(30) & 0 \end{vmatrix} \\ = -30i+52j+ \begin{vmatrix} i & j & k\\ 0 & 0 & 5 \\ -5 & 8.66 & 0 \end{vmatrix} \\ = -73.3i+27j \text{ m/s}^2\]

Question 4

If $\omega_{AB}$=2rad/s, $\alpha_{AB}$=2rad/s $^2$, $\omega_{BC}$=−1rad/s, and $\alpha_{BC}$=−4rad/s $^2$, what is the acceleration of point C where the scoop of the excavator is attached?


Answer

The acceleration of B is

\[a_{B}=a_A+\alpha_{AB}\times r_{B/A}-\omega_{AB}^2 r_{B/A} \\ 0+\begin{vmatrix} i & j & k\\ 0 & 0 & 2 \\ 3 & 3.9 & 0 \end{vmatrix} - 2^(3i+3.9j) \\\ = -19.8i-9.6j \text{ m/s} ^2\] \[a_{C}=a_B+\alpha_{BC}\times r_{C/B}-\omega_{BC}^2 r_{C/B} \\ -19.8i-9.6j +\begin{vmatrix} i & j & k\\ 0 & 0 & -4 \\ 2.3 & -0.5 & 0 \end{vmatrix} - 1^2(2.3i-0.5j) \\\ = -24.1i-18.3j \text{ m/s} ^2\]

Question 5

The length of the bar is L = 4 m and the angle $\theta$ = 30°. The bar’s angular velocity is $\omega$ = 1.8 rad/s and its angular acceleration is $\alpha$ = 6 rad/s $^2$. The endpoints of the bar slide on the plane surfaces. Determine the acceleration of the midpoint G.


Answer

Drawing and labelling the situation calling the top point A and bottom B


Solve, taking into account the constraints of the movements imparted by the walls and floor

\[a_{A}=a_B+\alpha\times r_{A/B}-\omega^2 r_{A/B} \\ = a_B + \begin{vmatrix} i & j & k\\ 0 & 0 & 6 \\ -4\cos(60) & 4\sin(60) & 0 \end{vmatrix} - 1.8^2(-4\cos(60)i+ 4\sin(60)j) \\ a_Aj = a_Bi - 14.28i -23.21j\]

Equating components

\[(i) 0 = a_B -14.28 \rightarrow a_B=14.28 \text{ m/s} ^2\\ (j) a_A = -23.21 \rightarrow a_A=-23.21 \text{ m/s} ^2\]

Now we can use either points to work out G

\[a_{G}=a_A+\alpha\times r_{G/A}-\omega^2 r_{G/A} \\ a_G= -23.21j + \begin{vmatrix} i & j & k\\ 0 & 0 & 6 \\ 2\cos(60) & -2\sin(60) & 0 \end{vmatrix} - 1.8^2(2\cos(60)i-2\sin(60)j) \\ a_G = 7.15i-11.6 \text{ m/s} ^2\]

Question 6

The angular velocity’s magnitude $\omega_{AB}$ = 6 rad/s. If the acceleration of the slider C is zero at the instant shown, what is the angular acceleration $\alpha_{AB}$?


Answer

To find the acceleration, we need to know all angular velocities

\[v_B = v_A + \omega_{AB} \times r_{B/A} \\ = 0+ -6k\times(0.04i+0.04j)\\=0.24i-0.24j\] \[v_C = v_B + \omega_{BC} \times r_{C/B} \\ = 0.24i-0.24j + \omega_{BC} k\times(0.1i-0.07j)\\= (0.24+0.07\omega_{BC})i+(0.24+0.1\omega_{BC})j\]

Component analysis (remember C cannot move in i)

\[(j) 0 = 0.24 + 0.1\omega_{BC} \rightarrow \omega_{BC}=-2.4 \text{ rad/s}\]

From here we can find the acceleration by finding two different expressions for the movement of B; note because of the diagram we have put $\alpha_{AB}$ as negative

\[a_{B}=a_A+\alpha_{AB}\times r_{B/A}-\omega_{AB}^2 r_{B/A} \\ = 0- \alpha_{AB}k\times(0.04i+0.04j)-6^2(0.04i+0.04j) \\ a_{B}= (0.04 \alpha_{AB}-1.44)i+(-0.04\alpha_{AB}-1.44)j\] \[a_{B}=a_C+\alpha_{BC}\times r_{B/C}-\omega_{BC}^2 r_{B/C} \\ = 0- \alpha_{BC}k\times(-0.1i+0.07j)-(-2.4)^2(-0.1+0.07j) \\ a_{B}= (-0.07\alpha_{BC}+0.576)i+(-0.1\alpha_{BC}-0.4)j\]

Set components equal

\[(i) 0.04 \alpha_{AB}-1.44 = -0.07\alpha_{BC}+0.576 \\ 0.04\alpha_{AB}+0.07\alpha_{BC}=2.016 \\ (j) -0.04\alpha_{AB}-1.44 = -0.1\alpha_{BC}-0.4 \\ -0.04\alpha_{AB}+0.1\alpha_{BC}=1.037\]

Via simultaneous equations

\[\alpha_{AB}=19 \text{ m/s} ^2 \text{ clockwise}\]

Question 7

At the instant shown, the piston’s velocity and acceleration are $v_C$ =−14i m/s and $a_C$ = −2200i m/s $^2$. What is the angular acceleration of the crank AB?


Answer

Similar to the last question, we can use the fact C can only move in i and that we need to solve for $\omega $ first

\[v_B=v_A +\omega_{AB} \times r_{B/A} \\ 0+\omega_{AB}k \times (0.05i+0.05j) \\ = -0.05 \omega_{AB}i + 0.05 \omega_{AB}j\] \[v_B=v_C +\omega_{BC} \times r_{B/C} \\ -14i+\omega_{BC}k \times (-0.175i+0.05j) \\ = -14i -0.05 \omega_{BC}i -0.175 \omega_{BC}j\]

Then analyse looking at each component of velocity

\[(i) -0.05 \omega_{AB}+0.05 \omega_{BC} =-14 \\ (j) 0.05 \omega_{AB}=-0.175 \omega_{BC}\]

And using simultaneous equations

\[\omega_{AB}=218 \text{ rad/s, } \omega_{BC}=-62.2 \text{ rad/s}\]

This time I will solve it in a slightly different way, instead of finding two expressions for $a_B$, finding $a_B$ and using it as the reference point to find $a_C$. Both ways work so use whichever is most intuitive to you!

\[a_{B}=a_A+\alpha_{AB}\times r_{B/A}-\omega_{AB}^2 r_{B/A} \\ = 0+\alpha_{AB}k\times(0.05i+0.05j)-218^2(0.05i+0.05j) \\ = (-0.05\alpha_{AB}-2376)i+(0.05\alpha_{AB}-2376)j\]

Then using this as the reference point to find C

\[a_{C}=a_B+\alpha_{BC}\times r_{C/B}-\omega_{BC}^2 r_{C/B} \\ -2200i = (-0.05\alpha_{AB}-2376)i+(0.05\alpha_{AB}-2376)j+ \alpha_{BC}k\times(0.175i-0.05j)-(-62.2)^2(0.175i-0.05j) \\ -2200i= (-0.05\alpha_{AB}-2376 + 0.05\alpha_{BC}-678)i+(0.05\alpha_{AB}-2376+0.175\alpha_{BC}+193)j\]

Equating components

\[(i) -2200=-0.05\alpha_{AB}+ 0.05\alpha_{BC}-3054\\ (j) 0= 0.05\alpha_{AB}+0.175\alpha_{BC}-2183\]

Solving we find

\[\alpha_{AB}=-3580 \text{ rad/s}^2\]

Hence AB rotates 3580 rad/s $^2 $ clockwise.

Question 8

If arm AB has a constant clockwise angular velocity of 0.8 rad/s, arm BC has a constant angular velocity of 0.2 rad/s, and arm CD remains vertical, what is the acceleration of part D?


Answer

Because both AB and BC have constant angular velocity, $\alpha = 0$ for both.

Acceleration of B can be expressed as

\[a_{B}=a_A+\alpha_{AB}\times r_{B/A}-\omega_{AB}^2 r_{B/A} \\ = 0+0\times (0.3\cos(50)i+0.3\sin(50)) - 0.8^2(0.3\cos(50)i+0.3\sin(50)j) \\ = 0.123i-0.147j\]

Acceleration of C can be expressed as

\[a_{C}=a_B+\alpha_{BC}\times r_{C/B}-\omega_{BC}^2 r_{C/B} \\ = 0.123i-0.147j + 0\times (0.3\cos(15)i+0.3\sin(15)) - 0.8^2(0.3\cos(15)i+0.3\sin(15)j) \\ = -0.135i-0.144j\]

Given CD remains vertical, it is just translating the same as point D

\[\alpha_C=\alpha_D= -0.135i-0.144j \text{ m/s}^2\]

Question 9

Point A of the rolling disk is moving toward the right and accelerating toward the right. The magnitude of the velocity of point C is 2 m/s, and the magnitude of the acceleration of point C is 14 m/s $^2$. Determine the angular acceleration of the disk.


Answer


First find the velocity of the centre of the disk. because it is rolling, it only has i component, hence

\[v_a=r\omega i\]

Next find velocity of C. Intuitively this must only have j velocity component meaning the velocity must be

\[v_C = r\omega i + r\omega j\]

But we can prove it mathematically with

\[v_C = v_A + r\times\omega \\ = r\omega i + \begin{vmatrix} i & j & k\\ 0 & 0 & -\omega \\ -0.3 & 0 & 0 \end{vmatrix} = r\omega i + r\omega j\]

From here, we know the magnitude of velocity of C is 2 so we can calculate the angular velocity using basic absolute value vector calculations

\[2= \sqrt{(r\omega)^2+(r\omega)^2} \\ 4= r^2 \omega^2+r^2 \omega^2 \\ 4 = 2r^2\omega^2 \\ 4 = 0.18\omega^2 \\ \omega = 4.71 \text{ rad/s}\]

Then use a similar method to find angular acceleration

\[a_{C}=a_A+\alpha\times r_{C/A}-\omega^2 r_{C/A} \\ a_{C}=0.3\alpha i+\begin{vmatrix} i & j & k\\ 0 & 0 & -\alpha \\ -0.3 & 0 & 0 \end{vmatrix}-|4.71|^2 (-0.3) \\ a_C = 0.3\alpha i + 6.66 i + 0.3\alpha j\]

Now for the magnitude

\[14 = \sqrt{(0.3\alpha+6.6)^2+0.3\alpha^2}\]

Expanding and solving the equation

\[\alpha = 20 \text{ m/s}^2 \text { in the negative direction}\]

Question 10

The disk rolls on the circular surface with a constant clockwise angular velocity of 1 rad/s. What are the accelerations of points A and B?


Answer

First find the acceleration of the center of the disk (call it O). There is an instantaneous center where the circle meets the ground so

\[v_B=0 \\ v_0=v_B+\omega k \times r_{O/B} = -1k \times 0.4j \\ v_O = 0.4i\]

The speed of the disk along the circular path is not changing. Remember O is moving on a circular path with radius 1.6 m total. We can use the normal acceleration formula from the centre of the system.

\[a_O = \frac{v_O^2}{r} = \frac{-0.16}{0.4+1.2} \\ = -0.1\text{ m/s}^2 \text{ in the j direction}\]

We don’t use $a_n=r\omega$ in this situation as this is used for calculating normal acceleration at the rim due to a disk’s angular velocity (essentially projecting a different frame and center) whereas we are finding simply the linear velocity along a circular path.

Then A can be found with

\[a_A = a_O+\alpha\times r_{A/O}-\omega^2 r_{A/O} \\ a_A = -0.1j+0-1^2(0.4j) \\ a_A=-0.5j\text{ m/s}^2\]

And B

\[a_B = a_O+\alpha\times r_{B/O}-\omega^2 r_{B/O} \\ a_B = -0.1j+0-1^2(-0.4j) \\ a_B=0.3j\text{ m/s}^2\]