Hey, guys. So in this video, I want to talk about a torque producing an acceleration on a point mass. You may remember that in all these rotation questions, we can have either point masses, which are tiny objects with no radius and no volume, or we can have rigid bodies or shapes that have a radius, have volume. Let's check out how torques and acceleration works for point masses. So, real quick, I want to point out that most torque problems are actually going to involve shapes or rigid bodies. But I want to quickly just do one with point masses because it works just the same. Most of the time you're going to be here, but this works just the same. Cool?
So, a quick example: you spin a small rock. A small rock is an indication that this is going to be a point mass because they say it's small and I don't give you the shape of the rock. The mass is 2 kilograms. You do this at the end of a light string of length 3 meters. If you spin a rock or any object at the end of a string, the length of the string will be the distance to the center. Okay. So it will be your little r. So little r equals the length of the string, which is 3 meters, and the mass is 2 kilograms. Okay. We want to know what net torque is needed to give the rock an acceleration of 4. So if you want to have an acceleration of 4, what is the torque you need? And I want to remind you that net torque is the same thing as the sum of all torques. Okay. So check this out. I am asking for the sum of all torques, and I'm giving you alpha. So I hope that you immediately thought of using ∑τ = Iα (sum of all torques equals I multiplied by alpha). This is what we're looking for. Okay?
Now, to calculate this, we just have to figure out these two guys. Remember, point masses have a moment of inertia. The moment of inertia of point masses is given by m r2, where r is the distance to the center. Okay. So we're going to replace this with m r2 alpha, and we have all of these numbers. The mass is 2. The r is the distance, which is going to be 3, and we have to square that. Then alpha is 4. Okay. And if you multiply all of this, you get 72 Newton meters. Remember, torque has units of Newton meters. That's it for Part A. Got that done.
For Part B, it's asking us to find the tangential acceleration while it has that speed. Remember, the tangential acceleration, atan, is related to alpha by this equation: atan = r alpha. So all we got to do is multiply. R is the distance here, which is 3, and alpha is 4. So that means that at that point when you have an alpha of 4, your tangential acceleration is 12. This is an acceleration a. It's a linear acceleration. So it's going to be 12 meters per second squared. Cool?
These two are somewhat unrelated. This is new stuff. This is just plugging it back into a different kind of acceleration, some old stuff bringing that back, putting it all together. Alright? So that's it for this one. Let me know if you have any questions and let's keep going.
