# Maxima and Minima, Rod Rotating on a Surface With Friction

• Phys student
In summary, the conversation discusses the problem of finding the maximum and minimum time of rotation for a uniform rod rotating around a pivot point on a rough surface. The solution involves finding the torque and angular deceleration of the rod, and using the parallel axis theorem to calculate the moment of inertia. After taking the derivative and setting it equal to 0, it is found that there is only one extremum point, and this leads to a solution of T_m = (1/2)T_M. This result can also be derived intuitively by considering the rod as two halves rotating around their respective pivot points.
Phys student

## Homework Statement

A uniform rod of mass M and length L can rotate around point P which is at position x from one end of the rod as shown in the figure. The rod is gently placed on a rough horizontal surface that has a friction coefficient μ. and at t=0 starts rotating with angular velocity ω.

The total time of rotation of the rod is T, which depends on x. Let the maximum and minimum time of rotation be T_M and T_m respectively. Find: $$\frac{T_m}{T_M}$$

## The Attempt at a Solution

We need to find the torque that the friction force exerts on the rod and from that find the angular deceleration of the rod. We need to consider the part of the rod that is above the point of rotation and the one below it. Let's take a small mass dm with length ds, we have (λ is the linear mass density of the rod) :

$$dF_f = μgdm = μgλds$$
$$dτ= μgλsds$$
The total torque is the sum of the torques of the two parts:
$$\int_{0}^{L-x} μgλs ds + \int_{0}^{-x} μgλs ds = \frac{1}{2}μgλx^2 + \frac{1}{2}μgλ(L-x)^2 = \frac{1}{2}μgλ(L^2 +2(x^2-Lx))$$
Now:
$$α=\frac{τ}{I}= \frac{\frac{1}{2}μgλ(L^2 +2(x^2-Lx))}{\frac{1}{3}ML^2 +Mx^2}= \frac{μgλ(L^2 +2(x^2-Lx))}{\frac{2}{3}ML^2 +2Mx^2}$$
$$T= \frac{ω}{α} = \frac{ω(\frac{2}{3}ML^2 +2Mx^2)}{μgλ(L^2 +2(x^2-Lx))}$$

We take the derivative with respect to x and set it equal to 0:
$$\frac{d}{dx} \frac{ω(\frac{2}{3}ML^2 +2Mx^2)}{μgλ(L^2 +2(x^2-Lx))} = \frac{4LMω(L^2+Lx-3x^2)}{3gλμ(L^2-2Lx+2x^2)} = \frac{4L^2ω(L^2+Lx-3x^2)}{3gμ(L^2-2Lx+2x^2)} = 0$$
When we solve for x we get one negative root which isn't physical.
$$x= \frac{-L+L\sqrt{13}}{-6} = \frac{L(1-\sqrt{13})}{6}$$
Can someone please tell me where's the mistake?

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Phys student said:
$$\frac{\frac{1}{2}μgλ(L^2 +2(x^2-Lx))}{\frac{1}{3}ML^2 +Mx^2}$$
How do you get that expression for I?

Phys student
haruspex said:
How do you get that expression for I?

I used the parallel axis theorem. But I think I made a mistake, because the theorem applies to the moment of inertia around the center of mass so it should have been:
$$I=\frac{1}{12}ML^2 +M(\frac{L}{2} -x)^2$$

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Phys student said:
Oh, The parallel axis theorem applies to the moment of inertia around the center of mass so it should have been:
$$I=\frac{1}{12}ML^2 +M(\frac{L}{2} -x)^2$$
Yes. Does the answer look better now?

Phys student
haruspex said:
Yes. Does the answer look better now?

We have:
$$α=\frac{μgλ(L^2 +2(x^2-Lx))}{\frac{1}{6}ML^2 +2M(\frac{L}{2} -x)^2}$$
$$T= \frac{ω}{α} = \frac{ω(\frac{1}{6}ML^2 +2M(\frac{L}{2} -x)^2)}{μgλ(L^2 +2(x^2-Lx))}$$
When we take the derivative we get:
$$\frac{d}{dx} \frac{ω(\frac{1}{6}ML^2 +2M(\frac{L}{2} -x)^2)}{μgλ(L^2 +2(x^2-Lx))} = -\frac{2L^2Mω(L-2x)}{3μgλ(L^2-2Lx+2x^2)^2}$$
Which only gives us one solution: $$x= \frac{L}{2}$$
We should have two solutions. So, there's still something wrong

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Phys student said:
##
\frac{d}{dx} \frac{ω(\frac{1}{6}ML^2 +2M(\frac{L}{2} -x)^2)}{μgλ(L^2 +2(x^2-Lx))} = -\frac{2L^2Mω(L-2x)}{3μgλ(L^2-2Lx+2x^2)^2}##
That's not what I get.
Phys student said:
Which only gives us one solution:
It is possible there is only one local extremum. There is a physical constraint on the range of x.

Phys student
haruspex said:
That's not what I get.

It is possible there is only one local extremum. There is a physical constraint on the range of x.

I guess it could be possible that there's only one extremum but it would mean that we don't have both a maximum and minimum time.

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Phys student said:
I guess it could be possible that there's only one extremum but it would mean that we don't have both a maximum and minimum time.
No, I wrote local extremum, but even that wasn't quite accurate. One end of the physical range for x may be an extremum by virtue of the fact that x cannot go any further along, but the gradient need not be zero. Indeed, the fact that there was only one d/dx = 0 point, and that the two ends of the physical range are equivalent, it follows that those end points must represent the other extreme value.

Phys student
haruspex said:
No, I wrote local extremum, but even that wasn't quite accurate. One end of the physical range for x may be an extremum by virtue of the fact that x cannot go any further along, but the gradient need not be zero. Indeed, the fact that there was only one d/dx = 0 point, and that the two ends of the physical range are equivalent, it follows that those end points must represent the other extreme value.
I see what you mean now.

If we substitute $x= \frac{L}{2}$ we get:

$$T_m = \frac{Mω}{3gλμ}$$

Now, when we substitute $x=0$ or $x=L$ we get:
$$T_M = \frac{2Mω}{3gλμ}$$

So:
$$\frac{T_m}{T_M} = \frac{1}{2}$$

Can this result be derived intuitively, without calculus?

Anyway, thank you very much for you help.

Phys student said:
Can this result be derived intuitively, without calculus?
Perhaps.
A rod length L pivoted at one end will clearly take the same time to stop as two rods of length L, and therefore the same as a single rod length 2L pivoted about its centre.
Doubling the length of the rod (for the same density) doubles the mass, multiplies the moment of inertia by 8, so multiplies the KE by 8. The frictional force doubles, so the rotational distance to stop quadruples. Under uniform acceleration, four time the distance means twice the time.
Of course, I haven't shown these are the extreme times.

Phys student

## 1. What is the concept of maxima and minima in physics?

Maxima and minima refer to the maximum and minimum values of a physical quantity in a given system or situation. In other words, they represent the highest and lowest points of a function or a graph.

## 2. How does a rod rotating on a surface with friction relate to maxima and minima?

When a rod rotates on a surface with friction, the frictional force acts as a constraint on the motion of the rod. This can lead to the presence of maxima and minima in the rotational motion of the rod, as the frictional force can cause the rod to come to a stop or change its direction.

## 3. How can we determine the maxima and minima of a rotating rod on a surface with friction?

The maxima and minima of a rotating rod on a surface with friction can be determined by analyzing the forces and torques acting on the rod. This can be done using mathematical equations and principles of mechanics, such as Newton's laws of motion and the concept of torque.

## 4. What factors can affect the maxima and minima of a rotating rod on a surface with friction?

The maxima and minima of a rotating rod on a surface with friction can be affected by various factors, such as the coefficient of friction between the rod and the surface, the shape and mass distribution of the rod, and the initial conditions of the motion (e.g. initial velocity and angular velocity).

## 5. How can the concept of maxima and minima be applied in real-life situations?

The concept of maxima and minima is widely used in various fields of physics, engineering, and other sciences. For example, it can be applied in the design of machines and structures to optimize their performance and minimize frictional losses. It is also used in the analysis of motion and forces in mechanical systems, such as in the study of rotational motion of gears and pulleys.

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