Question

A thin, uniform bar of length L and mass M is suspended horizontally at rest. It is suddenly released

and, at the same instant, it is struck a sharp blow vertically upwards at one end – the duration of

the impulse is negligibly short.

(a) Explain the meaning of the equation Fnet = Macom (com stands for center of mass). If we call z the vertical direction, write an equation for zCOM(t), draw a sketch of zCOM(t) vs t, and annotate it to show the initial

slope, the maximum value of zCOM(t), and the time(s) at which zCOM(t) = 0.

(b) Explain the meaning of the equation Tnet = Iα (net torque). Describe the rotational motion of the bar, by

giving an equation for the angular displacement θ(t). (The moment of inertia of a bar about

a perpendicular axis through its centre is 1/12ML^2.)

(c) By combining your answers to (a) and (b), describe in words how the bar moves after being

struck. (Note: you can try the experiment yourself, using a pen as the bar, and your finger

or another pen to impart the impulse.)

Answer 1

We are given a uniform bar of length and mass that is suspended horizontally. When it is dropped suddenly, also giving a struck at the same instant, the bar falls down under the influence of gravity, also rotating about the centre of mass.

1. We are dealing with a rigid extended object here. So, the usual Newton's law is not valid here as it is applicable only for point particles. However, the same structure of Newton's law can be reproduced for a system of particles based on the observation that for any system that traverses along with the system as if it were a particle and its trajectory as specified by Newton's law for point particles. Such a point is called the centre of mass (com) of the system. The consequence is that the com of an object thus moves like a particle that is acted upon by a force according to Newton's law.

where is the total mass of the object and is the acceleration of the com.

Thus, once we know the com (or the mass distribution) of an object, it is possible to give an account of the trajectory of the particle. However, you should understand that this is not the actual trajectory of the particle, but the trajectory of the com; it tells nothing about the overall motion of the object. For example, in this case, the motion of the centre of mass does not give any detail of the rotation of the bar about the com.

Now, let's apply (1) to our problem. We take the vertically downward direction as the positive direction of z-axis. The mass is uniformly distributed along the length of the bar. Thus, the com of the bar is at the centre. Now, the only force acting on the com is the force of gravity, acting downwards. Thus, (1) can be written as

where is the acceleration due to gravity. Thus, the com is under free fall. Now, we proceed to solve equation (2). Integrating both sides of (2),we get

where is the initial velocity (or the initial slope of the ). The bar is dropped from rest. Hence, we can take . Thus, (3) becomes

Integrating again, we can find the vertical position of the com at any instant:

where is the initial position of the com. Based on the choice of our coordinate system, we can set the initial position to be at the origin. Thus, , so that (50 reduces to

which gives the vertical position of the com at any instant. The following figure shows the graph of with .

The initial slope is nothing but the initial velocity and you can see that it is zero. The maximum value of is , the height from which the bar is dropped to the ground, assuming that the bar touches the ground horizontally (this need not be the case as we have to account for rotations also). only at the time .

Now, if the bar touches the ground making an angle with respect to the horizontal, then the maximum value of is given by .

2. It is clear that equation (1) is not the complete story as it gives only the motion of the com. The actual motion of the entire parts of the object is still unknown. So, along with the motion of the com, we should describe the motion of the bar about the com. As the bar is rotating about the centre of mass, we need the rotational analogue of (1), which gives

Equation (7) relates the net torque acting on the object about the com to the angular acceleration of the object about the com. The proportional (need not always be!) factor gives the moment of inertia of the object about the com. Just like Newton's second law, (7) indicates that the presence of an angular acceleration implies a net torque acting on the object about the same axis. Since we are analyzing the rotational motion of the bar about the com, our origin can be taken as the com.

The angular acceleration can be written as the second derivative of the angular position:

where is the angle subtended by a point on the body from the horizontal position about the com. In addition, for a rigid bar, the moment of inertia about a perpendicular axis passing through the com is given by

The question is how to find the torque? The torque is defined as

where in this case, we are finding the torque about the com and hence   is the position vector from the com to the point where the force is acting. However, we know that the only force acting on the bar is the force of gravity and it acts at the com. Thus, . This implies that the torque acting on the object is zero:

Since can't be identically zero, (11) implies that

which can be easily integrated to give the angular velocity

which is a constant and is positive as the bar will be rotating in the anticlockwise direction. Integrating again, we can find the angular position as a function of time:

Since the bar started from a horizontal configuration, . Thus, (14) becomes

which gives the angular displacement about the com as a function of time.

3. Equations (6) and (15) combined describes the overall motion of the bar. Equation (16) gives an account on the motion of the com of the bar while (16) describes the motion about the com. The com of the bar (which is the midpoint of the bar) is falling freely as if it were a particle under the influence of gravity. As it falls freely, the bar rotates about the com at constant angular velocity.

You may wonder where the struck force has gone? That force was the one that put the bar into rotation. However, the struck force acted only for an instant and after that, no force was contributing to the rotational motion. This makes perfect sense. It explains why we can't set even though it is a mathematically consistent answer for (13). But, physically it is not. The bar is struck at its end in the vertically upward direction. That creates a tangential force and thus a torque that acts only for an instant. It will put the bar int rotation. After that, since no torque is acting, the bar should continue rotating at the same angular speed.

Regards.