Question 11.11: A bar of length 2l and mass M is fixed at point A, so that i...

In order to determine F_{r} , one calculates the torque D_{A} with respect to the center of gravity of the bar, caused by F_{r} .

The torque with respect to the fixed point A is

D_{A} = −Fl =\Theta_{A}\dot{ω} ,                                 (11.43)

since the constraints do not contribute to D_{A}. The angular acceleration of the bar \dot{ω} then follows from (11.43):

\dot{ω}= \frac{D_{A}}{\Theta_{A}}=− \frac{Fl}{\Theta_{A}},                                 (11.44)

where \Theta_{A} is the moment of inertia of the bar with respect to A. Since the moment of inertia \Theta_{S} with respect to the center of gravity S is easily calculated as

\Theta_{S}=\int\limits_{−l}^{l}{\varrho  r^{2}  dV} = \frac{1}{3}Ml^{2},                                         (11.45)

one immediately gets for \Theta_{A} by means of Steiner’s theorem

\Theta_{A}= \Theta_{S}+Ml^{2} = \frac{1}{3} Ml^{2} + Ml^{2} = \frac{4}{3} Ml^{2}.                                 (11.46)

Equation (11.46) inserted into (11.44) leads to

\dot{ω}=− \frac{Fl}{\Theta_{A}} =−\frac{3}{4} \frac{F}{Ml}.                                       (11.47)

Since (11.47) must be correct, independent of the point from which the torque is being calculated, from the knowledge of the torque with respect to the center of gravity S,

D_{s} =−F_{r}  l,                                       (11.48)

and hence of the angular acceleration

\dot{ω} = \frac{D_{s}}{\Theta_{s}}=−\frac{3F_{r} l}{Ml^{2}}=−\frac{3F_{r}}{Ml},                                 (11.49)

one can calculate the reaction force F_{r} , by comparing (11.47) and (11.49):