# Acceleration of pivoted rod

As is evident from the figure, we have a rod of length 'L' and mass 'm' pivoted at one of its ends. The rod is free to rotate about the pivot. The rod is released from a horizontal position as shown in the figure. We are interested in finding accelerations of two points, the center of mass and the extreme end(towards right) right at the moment when the rod is released.

For Center of Mass:

let acceleration of COM be a(cm), then using the Torque-Angular acceleration relationship,

$$\tau=I\alpha$$ $$mg\frac{L}{2}=(\frac{mL^2}{12} + \frac{mL^2}{4})\alpha$$ $$\alpha=\frac{3g}{2L}$$ $$a(cm)=r\alpha=\frac{L}{2}\alpha=\frac{3g}{4}$$

For end Point:

There will be two accelerations for the point, centripetal and tangential. But at the moment the rod is just released, angular velocity($$\omega$$) will be zero hence tangential acceleration will be the only acceleration.

$$a=r\alpha$$ $$a=L\frac{3g}{2L}$$ $$a=1.5g$$

Now the Question: In the downward direction, only gravitational force is acting. The acceleration of COM is less than $$g$$ and that is understandable because the pivot will also exert some force which accounts for the external force acting on the rod. But why is the acceleration of end point greater than the acceleration due to gravity, $$g$$ ?

Interesting question, once you look beyond the fact that the end point is further away from the pivot and thus it has higher tangential acceleration. The interesting thing is to look at where the tangential acceleration is exactly $$g$$. This special point is at a distance $$x$$ from the pivot

This point is found by solving $$g = x \, \alpha$$ and the solution is

$$x = (\tfrac{L}{2}) + \frac{I}{m (\tfrac{L}{2}) } = \tfrac{L}{2} + \tfrac{L}{6} = \tfrac{2}{3} L$$

This point is called the axis of percussion of the rod.

Any point beyond $$x$$ will accelerate more than $$g$$ and any point inbound of $$x$$ less than $$g$$ due to the kinematics of the problem. So what is so special about $$x$$?

A force through $$x$$ would rotate the rod without any reaction from the pivot. But there is another definition of the axis of percussion. It is the exact point where an impulse applied that is equal but opposite to momentum the body would stop translating and rotating at the same time.

There is a connection between the point where acceleration is $$g$$ and the axis of percussion (IAP). The connection is that point $$x$$ is where you would place the effective mass $$m_x$$ to transform this problem into a lumped mass problem. Replace the rod with distributed mass with a point particle with mass $$m_x = \tfrac{3}{4} m$$ at the IAP and the device should be dynamically equivalent. The dynamic response of a lumped mass is that of free fall with $$g$$.

How do we find the effective mass $$m_x$$. The easiest way is do a kinetic energy equivalence between a rotating rod from one end and a translating moving mass

$$\frac{1}{2} m_x v^2 = \frac{1}{2} I_{\rm pivot} \omega^2$$

where $$v_x = \omega\, x$$, $$I_{\rm pivot} = \tfrac{m}{3} L^2$$ and $$x$$ as defined above. Plug the above and solve for $$m_x = \tfrac{3}{4} m$$.

This mean that if the moving rod hit you in the head at point $$x$$, you would feel like a mass $$m_x$$ hit you. The further away $$x$$ is, the less the effective mass. But the effective mass of a rotating object is another discussion to be had somewhere else.

• So how does the acceleration of different points on the rod becomes greater than 'g' the further one moves away from IAP? Mar 9 at 12:19
• @VanhiV - all the points rotate together around the pivot. They all share the same $\omega$ and thus the same $\alpha$. That is the kinematics of the problem. The speed of each point is given by $v = x\,\omega$ and taking the time derivative at any instant $\dot{v} = x\, \alpha$. The pivot has zero acceleration and the further away you are from the pivot the higher the acceleration. That is the nature of rotation. Mar 9 at 13:41
• But rotation is occurring due to the torque produced, the force for which is the force of gravity. When the rod is in freefall, it will accelerate with 'g' but here the rod is under constraints, so how can it accelerate faster than 'g' which is the only external force providing the torque? Mar 9 at 18:22
• @VanhiV you are trying to use $F=m a$ and Newton's laws for points not the center of gravity. Remember the not force on a body only describes how the center of mass moves, and nothing more. To describe the motion of points not the center of mass, you need to add the rotational motion $\tau = I \alpha$ and in this case the pivot support provides the torque necessary to rotate the object. Mar 9 at 19:48
• @VanhiV - also the concept of kinematics is here. Kinematics describes all possible motions a mechanism can have with no regard of actual forces applied. In this case the kinematics of this machine is the pivoting of one object with the one degree of freedom being the rotation angle. The rod rotates because that is the only motion allowed to do, unlike a free falling rod that has 3 degrees of freedom (two directions of motion for the center of mass and one rotation about the center of mass). Mar 9 at 19:58

Yes that's because the force from the wall is actually applying a torque on the rod. Now torque doesn't impart a linear momentum but it imparts an angular momentum.

More explicitly the wall is pushing the rod upward, since the push is not at the COM it imparts a torque on the rod. Hence the opposite side gets pushed downwards.

Why does the right end has acceleration > g?

To see this explicitly lets use the fact that "Any complicated motion can be decomposed to motion of the COM and motion about the COM".

PICTURE 1 - Motion of the COM -
Here this instant the COM is executing a simple fall under gravity. i.e. COM has a acceleration of g. To illustrate this, consider the same rod independent of pivot to be horizontal and falling under gravity, i.e every point of the rod has an acceleration equal to g.

PICTURE 2 - Motion about COM -
Consider the rod to be simply rotating about the COM and the COM is at rest. The torque that causes this rotation is equal to the torque imparted by the pivot at this instant. To illustrate this, consider our rode at rest, but rotating about the Instantaneous axis of Rotation which is COM here.

Now, SUPERIMPOSE these two pictures. Illustrating using the far right point. From Pic 1 this point has an acceleration equal to g and from pic to this point has an acceleration in the same direction due to the torque imparted by the wall. Hence the far right point has an acceleration downward that is greater than g.

• But the rod is rotating about the pivot so shouldn't we consider the torques about pivot only? Mar 8 at 12:32
• that is one torque, the torque due to gravity disappears when you are in COM frame, instead a new torque due to the wall comes to picture. Mar 8 at 12:37
• going in the COM frame is making it furthermore complicated, can u plz explain it in Ground frame? Mar 8 at 12:59
• It is correct to assert the pin reaction causes a CW rotation. But why does the end accelerate $> g$ is the question here. Mar 9 at 13:45
• Ahh okay Thanks for the comment I'll edit my post and add that as well. Mar 9 at 15:25