Does a pivot exert an additional force, normal to the centripetal force, on an object rotating about it?

Question

A thin plank of mass $M$ and length $l$ is pivoted at one end. The plank is released at 60$^{\circ}$ from the vertical. What is the magnitude and direction of the force on the pivot when the plank is horizontal?

I have a doubt: does the pivot exert an upwards force on the plank when the plank is horizontal? I think the answer to this is yes, because:

The torque on the plank about the pivot, when the plank is horizontal, is

$$\tau=\frac{Mgl}{2}$$

And the plank's angular acceleration at that point will be

$$\alpha=\frac{\tau}{I}=\frac{3g}{2l}$$

So the vertical acceleration of the plank's CM will be

$$a=\alpha R=\frac{3g}{4} $$

Therefore, the upwards force exerted by the pivot on the plank is

$$F_{up}=\frac{mg}{4}$$

Is this reasoning ok for this case? (I'm doubtful of this part because there was a post on PF where this force was not mentioned.)

Answer 1

Yes, spot on! Another way of proving that this is right is through d'Alembert's principle. This involves the drastic transformation of Newton's 2nd Law of:

$$F_{net} = ma_G$$

(where $a_G$ is the acceleration of the centre of mass)

into:

$$F_{net} - ma_G = 0$$

Indeed this may seem very little has changed by this algebraic manipulation. However, it is worth noting that the equation is now in the following form:

$$F' = 0$$

In other words, it is possible to treat this dynamics problem as a statics problem by defining an equivalent but stationary system with a net force of $F' = F_{net} - ma_G$. Therefore, we can solve this new system by enforcing equilibrium equations in order to determine the dynamics of the original system.

This can be extended to the rotational version of 2nd law, for the equilibrium equation for torque:

$$\tau' = \tau_{net} - J\alpha = 0$$

For the problem in the question, the usual dynamic system is given by:

The equivalent stationary system is:

Note that forces and moments are shown in red and accelerations are shown in blue.

$I_G$ is the moment of inertia of the rod taken about its centre of mass.

Answer 2

A free body diagram is your friend ($c= \frac{\ell}{2}$)

The general case above yields the following equations of motion

$$\begin{align} A_x & = m \ddot{x}_C \\ A_y - m g & = m \ddot{y}_C \\ A_x c \sin \theta - A_y c \cos \theta &= I_C \ddot{\theta} \end{align} $$

Since the pivot does not move, the acceleration of the center of mass $(\ddot{x}_C,\,\ddot{y}_C)$ is

$$ \begin{align} \ddot{x}_C & = - c\dot{\theta} \cos\theta - c \ddot{\theta} \sin\theta \\ \ddot{y}_C & = - c \dot{\theta}^2 \sin \theta + c \ddot{\theta} \cos\theta\end{align} $$

Using the motion of the center of mass in the equations of motion, and setting $\theta=0$ yields

$$ \left. \begin{align} A_x & = - m c \dot{\theta}^2 \\ A_y &= m c \ddot{\theta} + m g \\ -A_y c &= I_C \ddot{\theta} \end{align} \right\} \begin{aligned} \ddot{\theta} &= -\frac{m c g}{I_C + m c^2} \\ A_x &= -m c \dot{\theta}^2 \\ A_y &= \frac{I_C}{I_C+m c^2} m g \end{aligned} $$

using $c=\frac{\ell}{2}$ and $I_C =\frac{m}{12} \ell^2$ yields a familiar value $$ A_y = \frac{m g}{4}$$

To find the rotational speed $\dot{\theta}$ you use conservation of energy to relate initial potential energy to final kinetic energy.

Answer 3

Yes! Your reasoning is absolutely correct! There will be two forces exerted by the pivot: one in the horizontal direction and the other in the vertical direction!