A uniform rod of mass M and length L swings as a pendulum on a pi...

Question

A uniform rod of mass M and length L swings as a pendulum on a pivot at distance L/4 from one end of the rod. Find an expression for the frequency f of small-angle oscillations.

Accepted Answer

Hey everyone. Welcome back in this problem. An L meter long rectangular wooden bar of mass M and uniform composition is pivoted using a nail at a 0.0 point three L as measured from one end of the bar. For the bar oscillating at small angles. We're asked to derive an expression for the frequency of the oscillation. We're given four answer traces. Option A two divided by pi multiplied by the square root of 15 G divided by 37 L. Option B one divided by pi multiplied by the square root of 60 G divided by 37 L. Option C one divided by four pi multiplied by the square root of 37 G divided by 15 L and option D one divided by pi multiplied by the square root of 15 G divided by 37 L. We're gonna start by drawing out what we have. So we have an L meter long rectangular wooden bar. So the entire length of this rectangular wooden bar is L it has a mass um in a uniform composition. If it has a uniform composition, that means the center of mass is going to be in the center of the bar. So the point I've drawn right in the middle of the bar is going to be the center of mass. We're gonna say C M in the distance from that center of mass to the end of the bar is going to be L over two or one half L. And because this is uniform composition, the center of the mask is gonna be at the center of that bar. So it's gonna be at a length of half. OK. Now, for our pivot, I'm gonna draw our pivot in blue and we're told that our pivot is 0.3 L from one end. So if we imagine it on the left hand side, it doesn't matter what end, we draw it on the distance from the end to this pivot point is 0.3 L, which means that the distance from that pivot to the center of mass is going to be 0.2 L. Since this entire half has a length of 0.5 L. Now we needed to do this because we needed to know the distance from the center of maps to our pit. Now, we're looking for frequency, let's recall the relationship between period when we have a physical pendulum like this, we have that the period T is equal to two pi multiplied by I, the moment of inertia divided by M G D where M is the mass G is the acceleration due to gravity and D is the distance from the pivot to the center of mass. OK. So that's why we needed to draw out our diagram and figure out that value which we found to be 0.2 L. OK. That was for that D value in our period equation. Now, this is great. We've written the frequent er the period, but what we're interested in is the frequency. So let's recall that the period is related to the frequency by T is equal to one divided by F. And we're gonna come back to that. Now, in our equation for the period, we needed to know the moment of inertia. OK. So for I recall that this is gonna be equal to the moment of inertia a about the center of mass plus the mass multiplied by D squared, that distance from the center of mass to the pivot. Now, we have a rectangular wooden bar. So you can look in a table in your textbook or that your professor provided and you'll find that the moment of inertia about the center of mass is going to be 1/ multiplied by M multiplied by L squared. And so I is equal to that 1 12 M squared plus M multiplied by 0.2 L all squared. OK. Again, that's that distance from the pivot to our center of mass. And if we simplify this, we get that M is equal to 1 12 or sorry, not M I is equal to 1 12 M L squared plus 0.4 M. OK. So we have our moment of inertia eye. Let's get back to our period. T now our period T again, two pi multiplied by the moment of inertia I, and this is under the square root. So two pi multiplied by the square root of I which we found to be 1/12 M L squared plus 0.5 M L squared divided by the mass M multiplied by the acceleration due to gravity. G multiplied by D which is 0.2 L. Now we have an L in every single one of these terms, numerator and denominator. So let's divide both the numerator and denominator by L in the denominator. That term of L is gonna go away in the numerator. We have L squared terms. And so those are gonna turn into just L terms. OK. The exponent is gonna reduce by one. We also have mass in every one of these terms. And so we can divide numerator and denominator by the mass M as well. And so all of those masses are going to divide out. So after we do that, we're left with two pi multiplied by the square root of 12 L plus 0.4 L divided by G multiplied by 0.2. And we can write that as 1/5. OK. So we're getting there, we want to do some simplifying. Now, we're dividing by 1/5. This is the same as multiplying by five. So we can write this as two pi multiplied by the square root of five divided by 12 L plus 1/5. L OK. We had 0.4 L we multiplied by five, which gives us 0.2 L which we can write as 1/5. And all of this is divided by. Now, we have 5/12 L plus 1/5 L in the numerator. Here, we want to find a common denominator so that we can simplify and add those together. Now, the lowest common denominator here is going to be 60. So we get T is equal to two pi multiplied by the square root of, we have 5 12 L. In order to get the denominator to be 60 we need to multiply it by five. What we do to the denominator we need to do to the numerator. And so we're gonna multiply it both by five and we get divided by 60 L plus. Same thing in the next term, we have 1/5 L. In order to get that common denominator of 60 we need to multiply the denominator by 12. If we do it to the denominator, we do it to the numerator. And so we end up with 12 divided by 60 L and this is all still divided by G little bit more simplifying to go. We have T is equal to two pi multiplied by the square root 25 divided by 60 L plus 12, divided by 60 L gives us 37 divided by 60 L. And we can write this as 37 L divided by 60 G. All right. So we've simplified quite a bit our period. T but remember that's not what we're looking for. We're looking for the frequency, we wrote that the frequency is equal to one divided by the period. So that's gonna be equal to one divided by this entire expression. We found two pi multiplied by the square root of 37 L divided by six G. And we can write this as one divided by two pi multiplied by the square root of six G divided by 37 L. Now, at this point, you might stop and think that this is the final answer and you could compare with the answer choices and you will see that none of the answer choices match what we have here. However, we can do one more thing if we want, we have this factor divided by two out in front of our square root, we can bring that into our square root to simplify some of those numbers. So we know that two can be written as the square root of four. So we can write this as one divided by pi multiplied by the square root of 60 G divided by four times 37 L. OK. So we had a two in the denominator, we kept it in the denominator, but we wrote it as a square root of four, OK. And brought it into that square root symbol. And now we have 60 divided by four, which is going to give us 15. And so we can write this as one divided by pi multiplied by the square root of 15 G divided by L. OK. And so there was a lot of simplification that we needed to do there. You just have to work at it one step at a time and pay attention to what's in the new writer, what's in the denominator and just make sure you're not mixing up any of those terms. And so we found that the frequency of the oscillation is given by F is equal to one divided by pi multiplied by the square root of 15 G divided by 37 L which corresponds with answer choice. D. Thanks everyone for watching. I hope this video helped see you in the next one.

A uniform rod of mass M and length L swings as a pendulum on a pivot at distance L/4 from one end of the rod. Find an expression for the frequency f of small-angle oscillations.

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