A uniform rod of length L oscillates as a pendulum about a pivot ...

Question

A uniform rod of length L oscillates as a pendulum about a pivot that is a distance x from the center.a. For what value of x, in terms of L, is the oscillation period a minimum?

Accepted Answer

Hey, everyone in this problem, a regular piece of wood with a length L is pivoted about a position X measured from the center of mass of the piece of wood. Calculate the value of X that gives the lowest aci period express X. In terms of L, we're given four answer choices. Option AC and D all start with X equals the square root of one divided by multiplied by L for option A. That's it for that function. For option C, there's an extra minus one at the end for option D, there's plus one at the end and option B is X is equal to the square root of one divided by 12 L squared. So we have this regular piece of wood that's gonna be pivoting. And so what we want to think about is a physical pendulum. OK. This is going to be a physical pendulum and what we have physical pendulum. And we're looking at the period we're called that the period T is equal to two pi multiplied by the square root of the moment of inertia eye divided by M G D where M is a mass G is the acceleration due to gravity and D is the distance from the rotational axis to the center of mass. Now, in this case, we're told that the distance to the center of mass is X and so our period T is actually going to be equal to two pi multiplied by the square root of I divided by M G Pets. Now, we want the value of X that gives the lowest period. What we need first is to find this moment of inertia eye, we need to put in the mass and the gravitational acceleration so that we can simplify this. So the moment of inertia eye I recall is going to be equal to the moment of inertia about that center of mass plus M X squared. And again, X is that distance from the center of mass to the piece of wood. So we have this regular piece of wood, you can think of this as like a rectangle like a slab or block. And so the moment of inertia I above the center of mass is going to be 1/12 multiplied by M multiplied by L square. And that's an expression that you can find in a table in your textbook or that your professor has provided. So our moment of inertia, I is gonna be 1 12 M L squared plus M X squared. And we've used capital M here just to indicate the mass of this particular piece of wood instead of just a general mass with a little M. So if we go back to our equation for the period, now we have that the period T is equal to two pi multiplied by the square root of 1/12 M L squared plus M X squared divided by M G. That's, and now we can see that there's an M in every single one of these terms. So we can divide the numerator and the denominator inside our square root by M and those will all divide up. So we're left with our period T is equal to two pi multiplied by the square root of 1 L squared plus X squared divided by G. Now, the goal of this problem is to find the value of X that gives the lowest Aci toy period. OK. So what we wanna do is we want to minimize T are period. Now recall in order to minimize T we wanna find the critical points and those are gonna occur when D T divided by D X is equal to zero. OK. So when the derivative of the period, with respect to that variable, we're interested in X is equal to zero. Let's start by finding the derivative of our period T, with respect to X D T by D X, we have this two pi out front, that's a constant. And so we're just gonna multiply it out in front and leave it where it is. We have this entire expression inside of a square root, we can think of the square root as an exponent of one half, then we can use our exponent rule. So we're gonna bring the exponent down in front. So we have one half, then we're gonna take the entire expression and raise it to the exponent that it had minus one. Now, the exponent we had was one half. When we subtract one, we get negative one half. So we get 1 12 elsewhere plus X squared divided by G X to the exponent negative one half. OK. So we've done the first part of our derivative now be by the chain rule, we need to multiply by the derivative of what was inside of that square root. Now what's inside that square root is a quotient. So we want to use our quotient rule for division. So we're gonna take the derivative of the numerator first now with respect to X 1 12 L L squared is just a constant. So the derivative is going to be zero. And then we have X squared. When we take the derivative, we bring the exponent down in front and then reduce the exponent by one. And so we get Q X and I'm gonna multiply that by the denominator. OK. So the derivative of the numerator multiplied by the denominator minus, then we're gonna do the opposite. We're gonna leave the numerator alone 1/12 L squared plus X squared. And we're gonna multiply by the derivative of the denominator. Now, the denominator in this case is G X. So the derivative is just G and we're gonna divide all of that by the denominator squared. And so we get G X squared in the denominator. All right. So we found our derivative that looks really messy. Let's go ahead and simplify where we can, we have D T by D X is equal to, we have two pi divided by two. And that just gives us pi we have 1 L squared plus X squared divided by G X to the exponent negative half. OK. So when we have a negative exponent recall that we can move everything to the denominator and make that a positive exponent. OK. So we can do that. And then our exponent of one half again, we can write as a square root. So this is gonna be equal to the square root of G X divided by 1 12 L squared plus X squared. OK. So the numerator and the denominator have reversed since we moved that negative exponent into the denominator in order to make it positive. OK. All right. And then we are still multiplying by this other big term and we can simplify a little bit. We have two X multiplied by G X. So we get two X squared G, we have minus 1, 12 L squared plus X squared multiplied by G which gives us minus 1 12 L square G minus X squared G and this is all divided by G squared X squared. Simplifying a little bit more. In that last big term we had, we had D T by D X is equal to pi multiplied by the square root of G X divided by 1/ L squared plus X squared multiplied by. Now, we have a G in every single term here. So we can divide the numerator and denominator by G. When we do that in the numerator, we have no more G terms left. And when we do that in the denominator, we had an exponent two and that's gonna reduce to exponent one. Now we have two X squared minus X squared in the numerator. So we get X squared and then we have also have minus 12 L squared and all of this is divided by G X squared. OK. We're getting there. Now, remember we're looking for the critical points. We want to minimize this ciliary period T. So we wanna set it equal to zero, this derivative equal to zero. When we set the derivative equal to zero, we get pi multiplied by G or sorry, multiplied by the square root of G X divided by 1 12 L squared plus X squared, multiplied by X squared minus 1 12 L squared divided by G X squared. Now we have these three terms multiplied together. So in order for this entire thing to be equal to zero, we just need any one of those terms to be equal to zero. So the first thing we have is pie, that's never gonna be equal to zero. That's just a constant. So we can ignore that this second term, the square root of G X divided by 1 12 L squared plus X squared. Now this will be equal to zero if A G multiplied by X is equal to zero, which tells us that X is equal to zero. However, we have to be careful if we have X equals zero. Then in the last term, we get G multiplied by X squared in the denominator, we're gonna have zero in the denominator. And so this is actually not going to exist there. Yeah, we, we're gonna eliminate that as an option. OK? Because we have um a restriction there. Now, the second part of our equation, we could have that equal to zero as well. We get that X squared minus 1 12 L squared divided by G X squared is equal to zero. In order for this to be equal to zero, we just need the numerator to be equal to zero. So we have X squared minus 1 12 L squared is equal to zero, which tells us that X is going to be equal to the square root of 1 12 L squared. And the square root of L squared is just L and so we can write this as the square root of 1 12 multiplied by L and that is the X value that is going to minimize the oscillatory period, which is what we wanna do. And so X value we want is X is equal to the square root of one divided by 12 multiplied by L. That's gonna give us the lowest asci period that corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

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A uniform rod of length L oscillates as a pendulum about a pivot that is a distance x from the center.a. For what value of x, in terms of L, is the oscillation period a minimum?

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