Chapter 7, Problem 7.1P

A point at a 6.5-in radius is on a body that is in pure rotation with $\omega =100$ rad/sec and a constant $a=-500$ ${\text{rad/sec}}^{\text{2}}$ at point $A$. The rotation center is at the origin of a coordinate system. When the point is at position $A$, its position vector makes a $45°$ angle with the $X$ axis. It takes 0.01 sec to reach point $B$. Draw this system to some convenient scale, calculate the $\text{θ}$ and $\text{ω}$ of position $B$, and:

1. Write an expression for the particle's acceleration vector in position $A$ using complex number notation, in both polar and cartesian forms.
2. Write an expression for the particle's acceleration vector in position $B$ using complex number notation, in both polar and cartesian forms.
3. Write a vector equation for the acceleration difference between points $B$ and $A$. Substitute the complex number notation for the vectors in this equation and solve for the acceleration difference numerically.
4. Check the result of part c with a graphical method.

To determine

An expression for the particle’s acceleration vector in position A.

Answer to Problem 7.1P

Acceleration in polar form is $A_A=-32500j{e}^{j\left(\frac{\pi }{4}\right)}-65000e^{j\left(\frac{\pi }{4}\right)}$ .

Acceleration in cartesian form is $A_A=\left(-43664-48260j\right)$ .

Explanation of Solution

Calculation:

Compute the angular velocity at point B.

  ${\omega }_B={\omega }_A+\alpha t$

Here, ${\omega }_B$ is the angular velocity at point B, ${\omega }_A$ is the initial rotation speed, is the acceleration of rotation, and t is the time to reach point A.

Substitute $-500rad/{\sec }^2$ for $\alpha$ , 0.01 sec for t and 100 rad/sec for ${\omega }_A$ .

  $\begin{array}{c}{\omega }_B=\left(-500\times 0.01\right)+100\\ =95rad/\sec \end{array}$

Calculate the angular position at point B.

  $\left({\theta }_B-{\theta }_A\right)=\frac{1}{2}\left(\alpha t^2\right)+{\omega }_{A}t$

Here, ${\theta }_B$ is the angular position at point B and ${\theta }_A$ is the angular position at point A.

Substitute $-500rad/{\sec }^2$ for $\alpha$ , 0.01 sec for t and 100 rad/sec for ${\omega }_A$ and 45^o for ${\theta }_A$ .

  $\begin{array}{c}\left({\theta }_B-{\theta }_A\right)=\left[\frac{1}{2}\left(\left(-500\right)\times {\left(0.01\right)}^2\right)+\left(100\times 0.01\right)\right]\\ =0.975rad\left(\frac{180°}{\pi \text{rad}}\right)\\ =55.863°\\ {\theta }_B=55.863°+45°\\ =100.863°\end{array}$

Compute the magnitude of the normal component of acceleration at point A.

  $a_A^n=R{\omega }_A^2$

Here, $a_{An}$ is the magnitude of the normal component of acceleration at point A. Substitute the value of R and ${\omega }_A$ .

  $\begin{array}{c}{a}_A^n=6.5{\left(100\right)}^2\\ =65000\text{in}/{\sec }^2\end{array}$

Compute the direction of the normal component of acceleration at point A.

  $\begin{array}{c}{\theta }_A^n={\theta }_A+180°\\ =45°+180°\\ =225°\end{array}$

Compute the magnitude of the tangential component of acceleration at point A.

  $a_A^t=R\alpha$

Here, $a_A^t$ is the magnitude of the tangential component of acceleration at point A. Substitute the value of R and $\alpha$ .

  $\begin{array}{c}{a}_A^t=6.5{\left(-500\right)}^2\\ =3250\text{in}/{\sec }^2\end{array}$

Compute the direction of the tangential component of acceleration at point A.

  $\begin{array}{c}{\theta }_A^t={\theta }_A-90°\\ =45°-90°\\ =-45°\end{array}$

Compute the magnitude of the normal component of acceleration at point B.

  $a_B^n=R{\omega }_B^2$

Here, $a_{Bn}$ is the magnitude of the normal component of acceleration at point B. Substitute the value of R and ${\omega }_B$ .

  $\begin{array}{c}{a}_B^n=6.5{\left(95\right)}^2\\ =58663\text{in}/{\sec }^2\end{array}$

Compute the direction of the normal component of acceleration at point B.

  $\begin{array}{c}{\theta }_B^n={\theta }_B+180°\\ =100.863°+180°\\ =280.863°\end{array}$

Compute the magnitude of the tangential component of acceleration at point B.

  $a_B^t=R\alpha$

Here, $a_B^t$ is the magnitude of the tangential component of acceleration at point B. Substitute the value of R and $\alpha$ .

  $\begin{array}{c}{a}_A^t=6.5{\left(-500\right)}^2\\ =3250\text{in}/{\sec }^2\end{array}$

Compute the direction of the tangential component of acceleration at point A.

  $\begin{array}{c}{\theta }_A^t={\theta }_A-90°\\ =100.863°-90°\\ =10.863°\end{array}$

Draw the position vector diagram.

  

Computer polar form by using the formula.

  $\begin{array}{c}{R}_A=R·e^{j{\theta }_A}\\ =6.5\times e^{j\frac{\pi }{4}}\end{array}$

Calculate the velocity at point A.

  $\begin{array}{c}{V}_A=Rj{\omega }_A{e}^{j{\theta }_A}\\ =6.5\times j\times 100\times e^{j\frac{\pi }{4}}\\ =650j{e}^{j\frac{\pi }{4}}\end{array}$

Compute the acceleration vector in position A by the following formula.

  $\begin{array}{l}{A}_A=Rj\alpha e^{j{\theta }_A}-Rj{\omega }_A^2{e}^{j{\theta }_A}\\ =6.5\times j\times \left(-500\right)\times e^{j\left(\frac{\pi }{4}\right)}-6.5\times \left({100}^2\right)\times e^{j\left(\frac{\pi }{4}\right)}\\ A_A=-32500j{e}^{j\left(\frac{\pi }{4}\right)}-65000e^{j\left(\frac{\pi }{4}\right)}\end{array}$

Compute the acceleration vector of position A in cartesian form.

  $\begin{array}{c}{A}_A=R\alpha \left(-\sin \left({\theta }_A\right)+j\cos \left({\theta }_A\right)\right)-R{\omega }_A^2\left(\cos \left({\theta }_A\right)+j\sin \left({\theta }_A\right)\right)\\ A_A=\left[\left[6.5\times -500\left(-\sin \left(45°\right)+j\cos \left({45}^{°}\right)\right)\right]-\left[6.5\times {100}^2\left(\left(-\cos \left(45°\right)+j\sin \left({45}^{°}\right)\right)\right)\right]\right]\\ A_A=\left(-43664-48260j\right)\\ =\sqrt{{\left(-43664\right)}^2+{\left(-48260\right)}^2}\\ =65081.27\text{in}/{\sec }^2\end{array}$

Therefore, the acceleration of point A is $65081.27\text{in}/{\sec }^2$ .

To determine

An expression for the particle’s acceleration vector in position B.

Answer to Problem 7.1P

Acceleration in polar form is $A_B=-32500j{e}^{j\left(0.56035\pi \right)}-58662.5e^{j\left(0.56035\pi \right)}$ .

Acceleration in cartesian form is $A_B=\left(-14248-56999j\right)$ .

Explanation of Solution

Calculation:

Compute the angular velocity at point B.

  ${\omega }_B={\omega }_A+\alpha t$

Here, ${\omega }_B$ is the angular velocity at point B, ${\omega }_A$ is the initial rotation speed, is the acceleration of rotation, and t is the time to reach point A.

Substitute $-500rad/{\sec }^2$ for $\alpha$ , 0.01 sec for t and 100 rad/sec for ${\omega }_A$ .

  $\begin{array}{c}{\omega }_B=\left(-500\times 0.01\right)+100\\ =95rad/\sec \end{array}$

Calculate the angular position at point B.

  $\left({\theta }_B-{\theta }_A\right)=\frac{1}{2}\left(\alpha t^2\right)+{\omega }_{A}t$

Here, ${\theta }_B$ is the angular position at point B and ${\theta }_A$ is the angular position at point A.

Substitute $-500rad/{\sec }^2$ for $\alpha$ , 0.01 sec for t and 100 rad/sec for ${\omega }_A$ and 45^o for ${\theta }_A$ .

  $\begin{array}{c}\left({\theta }_B-{\theta }_A\right)=\left[\frac{1}{2}\left(\left(-500\right)\times {\left(0.01\right)}^2\right)+\left(100\times 0.01\right)\right]\\ =0.975rad\left(\frac{180°}{\pi \text{rad}}\right)\\ =55.863°\\ {\theta }_B=55.863°+45°\\ =100.863°\end{array}$

Compute the magnitude of the normal component of acceleration at point A.

  $a_A^n=R{\omega }_A^2$

Here, $a_{An}$ is the magnitude of the normal component of acceleration at point A. Substitute the value of R and ${\omega }_A$ .

  $\begin{array}{c}{a}_A^n=6.5{\left(100\right)}^2\\ =65000\text{in}/{\sec }^2\end{array}$

Compute the direction of the normal component of acceleration at point A.

  $\begin{array}{c}{\theta }_A^n={\theta }_A+180°\\ =45°+180°\\ =225°\end{array}$

Compute the magnitude of the tangential component of acceleration at point A.

  $a_A^t=R\alpha$

Here, $a_A^t$ is the magnitude of the tangential component of acceleration at point A. Substitute the value of R and $\alpha$ .

  $\begin{array}{c}{a}_A^t=6.5{\left(-500\right)}^2\\ =3250\text{in}/{\sec }^2\end{array}$

Compute the direction of the tangential component of acceleration at point A.

  $\begin{array}{c}{\theta }_A^t={\theta }_A-90°\\ =45°-90°\\ =-45°\end{array}$

Compute the magnitude of the normal component of acceleration at point B.

  $a_B^n=R{\omega }_B^2$

Here, $a_{Bn}$ is the magnitude of the normal component of acceleration at point B. Substitute the value of R and ${\omega }_B$ .

  $\begin{array}{c}{a}_B^n=6.5{\left(95\right)}^2\\ =58663\text{in}/{\sec }^2\end{array}$

Compute the direction of the normal component of acceleration at point B.

  $\begin{array}{c}{\theta }_B^n={\theta }_B+180°\\ =100.863°+180°\\ =280.863°\end{array}$

Compute the magnitude of the tangential component of acceleration at point B.

  $a_B^t=R\alpha$

Here, $a_B^t$ is the magnitude of the tangential component of acceleration at point B. Substitute the value of R and $\alpha$ .

  $\begin{array}{c}{a}_A^t=6.5{\left(-500\right)}^2\\ =3250\text{in}/{\sec }^2\end{array}$

Compute the direction of the tangential component of acceleration at point A.

  $\begin{array}{c}{\theta }_A^t={\theta }_A-90°\\ =100.863°-90°\\ =10.863°\end{array}$

Draw the position vector diagram.

  

Compute the position vector of point B.

  $\begin{array}{l}{R}_B={\mathrm{Re}}^{j{\theta }_B}\\ =6.5e^{j\left(100.863\times \frac{\pi }{180°}\right)}\end{array}$

Compute the polar form.

  $\begin{array}{c}{V}_B=6.5\times j\times 95\times e^{j\left(100.863\times \frac{\pi }{180°}\right)}\\ =617.5\times j\times e^{j\left(100.863\times \frac{\pi }{180°}\right)}\end{array}$

Compute the acceleration vector in position B by the following formula.

  $\begin{array}{l}{A}_B=Rj\alpha e^{j{\theta }_B}-Rj{\omega }_B^2{e}^{j{\theta }_B}\\ =6.5\times j\times \left(-500\right)\times e^{j\left(100.863\times \frac{\pi }{180}\right)}-6.5\times \left({95}^2\right)\times e^{j\left(100.863\times \frac{\pi }{180}\right)}\\ A_B=-32500j{e}^{j\left(0.56035\pi \right)}-58662.5e^{j\left(0.56035\pi \right)}\end{array}$

Compute the acceleration vector of position B in cartesian form.

  $\begin{array}{c}{A}_B=R\alpha \left(-\sin \left({\theta }_B\right)+j\cos \left({\theta }_B\right)\right)-R{\omega }_A^2\left(\cos \left({\theta }_B\right)+j\sin \left({\theta }_B\right)\right)\\ A_B=\left[\left[6.5\times -500\left(-\sin \left(100.863°\right)+j\cos \left({100.863}^{°}\right)\right)\right]-\left[6.5\times {95}^2\left(\left(-\cos \left(100.863°\right)+j\sin \left({100.863}^{°}\right)\right)\right)\right]\right]\\ A_B=\left(-14248-56999j\right)\\ =\sqrt{{\left(-14248\right)}^2+{\left(-56999\right)}^2}\\ =58752.79\text{in}/{\sec }^2\end{array}$

To determine

A vector equation for the acceleration difference between points B and A.

Explanation of Solution

Calculation

Calculate the difference in acceleration between the given points.

  $\begin{array}{c}{A}_{BA}=A_B-A_A\\ =\left(14248-56999j\right)-\left(-43664-48260j\right)\\ =\left(57912-8739j\right)\end{array}$

Therefore, the position difference is $\left(57912-8739j\right)$ .

To determine

The answer from the graphical method.

Explanation of Solution

Calculation Solve the equation $A_B^t+A_B^n=A_A^t+A_A^n+A_{BA}$ , use velocity scale of $10000\text{in}/{\sec }^2$ .

Construct the acceleration polygon following the below steps.

1. Locate point O_A.
2. Draw 0.325 in line from the point O_A. This represents the velocity of the point $A_A^t$ in a 45-degree direction.
3. From the tip of the above line, draw a line of 6.508 in. This represents the acceleration of the point $A_A^n$ in the indicated direction.
4. From the point O_A, draw a 0.325 line which represents the velocity of the point $A_B^t$ in 100.863^o direction.
5. From the tip of the point $A_B^t$ draw a perpendicular line of 5.875 which represents the acceleration of point $A_B^n$ in the indicated direction.
6. From the tip of point $A_A^n$ draw a perpendicular line of 5.875 which represents the acceleration of point $A_{BA}$ in the indicated direction.