Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 2PP from Chapter 16 from Hibbeler's Engineering Mechanics.
We are given the systems with various dimensions.
We are asked to determine the location of the instantaneous center of zero velocity for finding the velocity of point B.
(a)
The diagram of the system is shown as:

We have the radius of disk is $R = 2\;{\rm{m}}$.
We have the angular velocity of disk is $\omega = 8\;{\rm{rad/s}}$.
We have the angle of point B from horizontal is $\theta = 45^\circ $
The formula to calculate the velocity at point B is,
\[{v_B} = \omega \times r\]The formula to calculate the distance r is,
\[{r^2} = {\left( {R\cos \theta } \right)^2} + {\left( {R + R\sin \theta } \right)^2}\]Substitute the values in the above expression.
\[\begin{array}{l} {r^2} = {\left( {2\cos 45^\circ \;{\rm{m}}} \right)^2} + {\left( {2\;{\rm{m}} + 2{\rm{sin}}45^\circ \;{\rm{m}}} \right)^2}\\ r = \sqrt {13.6\;{{\rm{m}}^2}} \\ r = 3.7\;{\rm{m}} \end{array}\](b)
The diagram of the system is shown as:

We have the position vector of point A is ${r_A} = \left( {2i} \right)\;{\rm{m}}$.
We have the angular velocity is $\omega = - 4k\;{\rm{rad/s}}$.
We have the vertical distance of point B from instantaneous center is $r = 0.3\;{\rm{m}}$.
The formula to calculate the velocity at point A is,
\[{v_A} = \omega \times {r_A}\]Substitute the values in the above expression.
\[\begin{array}{l} {v_A} = \left( { - 4k} \right)\;{\rm{rad/s}} \times \left( {2i} \right)\;{\rm{m}}\\ {v_A} = \left( { - 8j} \right)\;{\rm{m/s}} \end{array}\]Here, negative sign shows that the direction of velocity is downward.
The formula to calculate the velocity at point B is,
\[{v_B} = r\omega \]Substitute the values in the above expression.
\[{v_B} = \left( {0.3\;{\rm{m}}} \right)\omega \](c)
The diagram of the system is shown as:

We have the position vector of point A is ${r_A} = \left( {0.5j} \right)\;{\rm{m}}$.
We have the position vector of point A with respect to B is ${r_{BA}} = \left( {1.5\cos 30^\circ i + 1.5\sin 30^\circ j} \right)\;{\rm{m}} = \left( {1.3i + 0.75j} \right)\;{\rm{m}}$ We have the angular velocity is $\omega = - 4k\;{\rm{rad/s}}$.
The formula to calculate the velocity at point A is,
\[{v_A} = \omega \times {r_A}\]Substitute the values in the above expression.
\[\begin{array}{l} {v_A} = \left( { - 4k} \right)\;{\rm{rad/s}} \times \left( {0.5j} \right)\;{\rm{m}}\\ {v_A} = \left( {2i} \right)\;{\rm{m/s}} \end{array}\]The formula to calculate the velocity at point B is,
\[{v_B} = {v_A} + {\omega _{AB}}\left( {{r_{BA}}} \right)\]Substitute the values in the above expression.
\[\begin{array}{l} {v_B}i = \left( {2i} \right)\;{\rm{m/s}} + \left( {{\omega _{AB}}} \right)\left( {1.3i + 0.75j} \right)\;{\rm{m}}\\ {v_B}i = \left( {2 - 1.3{\omega _{AB}}} \right)i + 0.75{\omega _{AB}}j \end{array}\]On comparing j coefficients, we get:
\[{\omega _{AB}} = 0\]Substitute the value of ${\omega _{AB}}$ in expression of ${v_B}i$.
\[{v_B} = {v_B} = \left( {2i} \right)\;{\rm{m/s}}\](d)
The diagram of the system is shown as:

We have the velocity of point C is ${v_C} = 4\;{\rm{m/s}}$.
We have the distance between point B and C is $y = 1\;{\rm{m}}$.
The formula to calculate the velocity at point B by using similar triangle theorem is,
\[\frac{{{v_B}}}{y} = \frac{{{v_C}}}{d}\]Substitute the values in the above expression.
\[\frac{{{v_B}}}{{1\;{\rm{m}}}} = \frac{{4\;{\rm{m/s}}}}{d}\]Thus, the required instantaneous velocity is drawn and obtained for expression velocity at point B.
(e)
The diagram of the system is shown as:

We have the position vector of point A is ${r_A} = \left( {0.5j} \right)\;{\rm{m}}$.
We have the angular velocity is $\omega = - 3k\;{\rm{rad/s}}$.
The formula to calculate the velocity at point A is,
\[{v_A} = \omega \times {r_A}\]Substitute the values in the above expression.
\[\begin{array}{l} {v_A} = \left( { - 3k} \right)\;{\rm{rad/s}} \times \left( {0.5j} \right)\;{\rm{m}}\\ {v_A} = \left( { - 1.5i} \right)\;{\rm{m/s}} \end{array}\]Here, negative sign shows that the direction of velocity is towards left.
The diagram of instantaneous velocity diagram of the disk is shown as,

(f)
The diagram of the system is shown as:

We have the position vector of point A is ${r_A} = \left( { - 0.5\cos 30^\circ i + 0.5\sin 30^\circ j} \right)\;{\rm{m}} = \left( { - 0.433i + 0.25j} \right)\;{\rm{m}}$.
We have the angular velocity is $\omega = - 6k\;{\rm{rad/s}}$.
The formula to calculate the velocity at point A is,
\[{v_A} = \omega \times {r_A}\]Substitute the values in the above expression.
\[{v_A} = \left( { - 6k} \right)\;{\rm{rad/s}} \times \left( { - 0.433i + 0.25j} \right)\;{\rm{m}}\]