The three 200 g masses in FIGURE EX12.11 are connected by massles...

Question

The three 200 g masses in FIGURE EX12.11 are connected by massless, rigid rods. (b) What is the triangle's kinetic energy if it rotates about the axis at 5.0 rev/s?

Accepted Answer

Hey, everyone. In this problem, we're asked to consider three masses of 150 g each that are joined to each other with massless rods that form a circular ring, which is shown in the figure, we're asked to determine the circular rings kinetic energy if the rotation speed is six revolutions per second about the central axis. So we're given our diagram here, we have this circular ring with our three masses. We're told that the radius is 45 centimeters and the axis of rotation is right in that central axis, we have four answer choices all in jewels. Option A 129 option B 64.8, option C 34.3. And option D 43.3, we asked for the kinetic energy, OK. This is rotating, our ring is rotating. And so this is gonna be the rotational kinetic energy. And let's recall that the rotational kinetic energy K R is given by one half multiplied by I multiplied by omega squared. And where I is a moment of inertia and omega is the angular speed. So let's start with Omega. OK? Because we are told in this problem that we have this speed of six revolutions per second. And so we know that Omega is going to be equal to six revolutions per second. But we want to convert this to our standard unit of radiance per second. OK. So that we can work this out to get the unit of jewels that we want in the answer choice. Now we're gonna multiply, we're gonna multiply by two pi radiant divided by one revolution. Since we know that there are two pi radiant in every revolution. OK. You can think about going around a circle one entire time. That's one revolution. And we know there are two pi in a circle, the unit of revolution will divide out. We're left with six multiplied by two pi radiance divided by second, which gives us 12 pi radiance for a second. So we have our omega value. Now we need the moment of inertia. I, in order to calculate our kinetic energy K R, we have these three masses. OK? We can consider those as point masses. We have massless rots. So the moment of inertia is only gonna be made up of the moment of inertia of these three MA. OK. And let's recall that the moment of inertia in this case is going to be equal to the sum of the mass multiplied by the radius squared. And in this case, because we have three masses, it's gonna be the sum of I equals 1 to 3 of M I R I squared mhm Expanding this, we can write it as M one R one squared plus M two R two squared plus M three R three squared or the subscript just indicates which mass we're talking about. Now, in this case, all three of the masses are exactly the same, they have the same uh not and they have the same weight and they're the same distance from the central axis and they're the same distance from our axis of rotation because it's directly in the center of this ring. And so we can simplify this and just write it as three M R squared. OK. Again, because all of the masses and all of the radiuses are the same substituting in our values, we have three multiplied by the mass. So we are told that these are 150 g masses. We want to convert this into kilograms so that we can get our unit of jewel. At the end, we're gonna multiply by one kg divided by g, the unit of gram divides out and we're left with a unit of kilograms. OK. So we're essentially dividing by 1000 to go from grams to kilograms because we know there are 1000 g in every kilogram, then we multiply by the radius R which is 45 centimeters. And again, we want to convert this into our standard unit of meters so that we can get that unit of jewel. At the end, we're gonna multiply by one m divided by 100 centimeters because we know that there are 100 centimeters in every meter. The unit of centimeter divides out, we're essentially dividing by 100 this R value is squared. So simplifying these units, we have three multiplied by 0.15 kg, multiplied by 0.45 m all squared. And this gives us a moment of inertia of 0. kilogram meters squared. So now we have our moment of inertia. I we have our angular speed omega and we can calculate the kinetic energy the rotational kinetic energy. In this case of this ring, we have that K R is equal to one half I omega squared K, just like we had written above, we can substitute in our values. Now this is gonna be equal to one half multiplied by 0. kilogram meter squared multiplied by 12 pi radiance per second all squared. If we work all of this out on our calculator, we get the kinetic rotational energy is going to be worse 0.7545 Jules. OK. We have kilogram meter squared per second squared, which is equivalent to a jule. And so the kinetic energy we were looking for which in this case is rotational kinetic energy is 64.8 Jes rounding to one decimal place. Which corresponds with answer choice. B thanks everyone for watching. I hope this video helped see you in the next one.

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The three 200 g masses in FIGURE EX12.11 are connected by massless, rigid rods. (b) What is the triangle's kinetic energy if it rotates about the axis at 5.0 rev/s?

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