Step 1

We are given the expression for the cardioid path is $r = 25\left( {1 - \cos \theta } \right)\;{\rm{ft}}$, the angular velocity of the bucket is $\dot \theta = 2\;{\rm{rad/s}}$, and the angular accelerationof the bucket is $\ddot \theta = 0.2\;{\rm{rad/}}{{\rm{s}}^2}$.

We are asked to determine the magnitude of the velocity and acceleration of the bucket when $\theta = 120^\circ $.

 
Step 2

The diagram of the system is shown as:

 
Step 3

The expression to calculate the first derivative of r is,

\[\begin{array}{c} \dot r = \frac{d}{{dt}}\left( r \right)\\ \dot r = \frac{d}{{dt}}\left[ {25\left( {1 - \cos \theta } \right)} \right]\\ \dot r = 25\sin \theta \dot \theta \end{array}\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{c} \dot r = 25\sin 120^\circ \left( {2\;{\rm{rad/s}}} \right)\\ \dot r = 43.30\,{\rm{ft/s}} \end{array}\]
 
Step 5

The expression to calculate the second derivative of r is,

\[\begin{array}{c} \ddot r = \frac{d}{{dt}}\left( {\dot r} \right)\\ \ddot r = \frac{d}{{dt}}\left( {25\sin \theta \dot \theta } \right)\\ \ddot r = 25\left[ {\cos \theta {{\left( {\dot \theta } \right)}^2} + \sin \theta \ddot \theta } \right] \end{array}\]
 
Step 6

Substitute the values in the above expression.

\[\begin{array}{c} \ddot r = 25\left[ {\cos 120^\circ {{\left( {2\;{\rm{rad/s}}} \right)}^2} + \sin 120^\circ \left( {0.2\;{\rm{rad/}}{{\rm{s}}^2}} \right)} \right]\\ \ddot r = - 45.67\;{\rm{ft/}}{{\rm{s}}^2} \end{array}\]
 
Step 7

The formula to calculate the radial component of velocity is,

\[{v_r} = \dot r\]
 
Step 8

Substitute the values in the above expression.

\[{v_r} = 43.30\,{\rm{ft/s}}\]
 
Step 9

The formula to calculate the transverse component of velocity is,

\[\begin{array}{l} {v_\theta } = r\dot \theta \\ {v_\theta } = \left[ {25\left( {1 - \cos \theta } \right)} \right]\dot \theta \end{array}\]
 
Step 10

Substitute the values in the above expression.

\[\begin{array}{c} {v_\theta } = \left[ {25\left( {1 - \cos 120^\circ } \right)} \right]\left( {2\;{\rm{rad/s}}} \right)\\ {v_\theta } = 75\;{\rm{ft/s}} \end{array}\]
 
Step 11

The formula to calculate the velocityof the bucket is,

\[v = \sqrt {{{\left( {{v_r}} \right)}^2} + {{\left( {{v_\theta }} \right)}^2}} \]
 
Step 12

Substitute the values in the above expression.

\[\begin{array}{c} v = \sqrt {{{\left( {43.30\,{\rm{ft/s}}} \right)}^2} + {{\left( {75\;{\rm{ft/s}}} \right)}^2}} \\ v = 86.60\;{\rm{ft/s}} \end{array}\]
 
Step 13

The formula to calculate the radial component of acceleration is,

\[\begin{array}{l} {a_r} = \ddot r - r{\left( {\dot \theta } \right)^2}\\ {a_r} = \ddot r - \left[ {25\left( {1 - \cos \theta } \right)} \right]{\left( {\dot \theta } \right)^2} \end{array}\]
 
Step 14

Substitute the values in the above expression.

\[\begin{array}{c} {a_r} = - 45.67\;{\rm{ft/}}{{\rm{s}}^2} - \left[ {25\left( {1 - \cos 120^\circ } \right)} \right]{\left( {2\;{\rm{rad/s}}} \right)^2}\\ {a_r} = - 45.67\;{\rm{ft/}}{{\rm{s}}^2} - 150\;{\rm{ft/}}{{\rm{s}}^2}\\ {a_r} = 195.67\;{\rm{ft/}}{{\rm{s}}^2} \end{array}\]
 
Step 15

The expression to calculate the transverse component of acceleration is,

\[\begin{array}{l} {a_\theta } = r\ddot \theta + 2\dot r\dot \theta \\ {a_\theta } = \left[ {25\left( {1 - \cos \theta } \right)} \right]\ddot \theta + 2\dot r\dot \theta \end{array}\]
 
Step 16

Substitute the values in the above expression.

\[\begin{array}{c} {a_\theta } = \left[ {25\left( {1 - \cos 120^\circ } \right)} \right]\left( {0.2\;{\rm{rad/}}{{\rm{s}}^2}} \right) + 2\left( {43.30\,{\rm{ft/s}}} \right)\left( {2\;{\rm{rad/s}}} \right)\\ {a_\theta } = 7.5\;{\rm{ft/}}{{\rm{s}}^2} + 173.2\;{\rm{ft/}}{{\rm{s}}^2}\\ {a_\theta } = 180.7\;{\rm{ft/}}{{\rm{s}}^2} \end{array}\]
 
Step 17

The expression to calculate the acceleration of the bucket is,

\[a = \sqrt {{{\left( {{a_r}} \right)}^2} + {{\left( {{a_\theta }} \right)}^2}} \]
 
Step 18

Substitute the values in the above expression.

\[\begin{array}{c} a = \sqrt {{{\left( {195.67\;{\rm{ft/}}{{\rm{s}}^2}} \right)}^2} + {{\left( {180.7\;{\rm{ft/}}{{\rm{s}}^2}} \right)}^2}} \\ a = 266.34\;{\rm{ft/}}{{\rm{s}}^2} \end{array}\]