Step 1

We are given the constant angular velocity of the slotted link is $\dot \theta = 3\;{\rm{rad/s}}$, and the relation between the short distance and spiral guide is $r = \left( {0.4\theta } \right)\;{\rm{m}}$.

We are asked to determine the radial and transverse components of the velocity and acceleration of P at the instant $\theta = \frac{\pi }{3}\;{\rm{rad}}$.

 
Step 2

The diagram of the system is shown as:

We have the length of the slotted link is $L = 0.5\;{\rm{m}}$.

 
Step 3

The expression to calculate the first derivative of r is,

\[\begin{array}{c} \dot r = \frac{d}{{dt}}\left( r \right)\\ \dot r = \frac{d}{{dt}}\left( {0.4\theta } \right)\\ \dot r = 0.4\dot \theta \end{array}\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{c} \dot r = 0.4\left( 3 \right)\\ \dot r = 1.2\,{\rm{m/s}} \end{array}\]
 
Step 5

The expression to calculate the second derivative of r is,

\[\begin{array}{c} \ddot r = \frac{d}{{dt}}\left( {\dot r} \right)\\ \ddot r = \frac{d}{{dt}}\left( {1.2} \right)\\ \ddot r = 0 \end{array}\]
 
Step 6

The formula to calculate the radial component of velocity is,

\[{v_r} = \dot r\]
 
Step 7

Substitute the values in the above expression.

\[{v_r} = 1.2\,{\rm{m/s}}\]
 
Step 8

The formula to calculate the transverse component of velocity is,

\[\begin{array}{l} {v_\theta } = r\dot \theta \\ {v_\theta } = 0.4\theta \dot \theta \end{array}\]
 
Step 9

Substitute the values in the above expression.

\[\begin{array}{c} {v_\theta } = 0.4\left( {\frac{\pi }{3}\;{\rm{rad}}} \right)\left( {3\;{\rm{rad/s}}} \right)\\ {v_\theta } = 1.26\;{\rm{m/s}} \end{array}\]
 
Step 10

The formula to calculate the radial component of acceleration is,

\[\begin{array}{c} {a_r} = \ddot r - r{\left( {\dot \theta } \right)^2}\\ {a_r} = \ddot r - \left( {0.4\theta } \right){\left( {\dot \theta } \right)^2} \end{array}\]
 
Step 11

Substitute the values in the above expression.

\[\begin{array}{c} {a_r} = 0 - 0.4\left( {\frac{\pi }{3}\;{\rm{rad}}} \right){\left( {3\;{\rm{rad/s}}} \right)^2}\\ {a_r} = - 3.77\;{\rm{m/}}{{\rm{s}}^2} \end{array}\]
 
Step 12

The expression to calculate the transverse component of acceleration is,

\[\begin{array}{c} {a_\theta } = r\ddot \theta + 2\dot r\dot \theta \\ {a_\theta } = \left( {0.4\theta } \right)\left[ {\frac{d}{{dt}}\dot \theta } \right] + 2\dot r\dot \theta \end{array}\]
 
Step 13

Substitute the values in the above expression.

\[\begin{array}{c} {a_\theta } = \left( {0.4 \times \frac{\pi }{3}\;{\rm{rad}}} \right)\left[ {\frac{d}{{dt}}\left( {3\;{\rm{rad/s}}} \right)} \right] + 2\left( {1.2\,{\rm{m/s}}} \right)\left( {3\;{\rm{rad/s}}} \right)\\ {a_\theta } = 0 + 7.2\;{\rm{m/}}{{\rm{s}}^2}\\ {a_\theta } = 7.2\;{\rm{m/}}{{\rm{s}}^2} \end{array}\]