A rock rotates around a light, 4-m long string. The rock is initially at rest, but reaches 150 RPM in 3 seconds. Calculate its tangential acceleration after 3 s. BONUS: Calculate its tangential speed after 3 s.

All right. So here we have a rock that rotates around a light 4 meter long string. So if you rotate a rock at the end of a string, it goes in a circular path, and the length of the string is the radius of the circle you form. So you got a little object here and it goes like that. The radius of the circle will be the length. So the distance, the radial distance of the rock, little r, is the same as the length, which in this case it is, let's put a 4 here, 4 meters. Actually, I'm just gonna put this here. Little r is the length, which is 4 meters. It forms a circle of a radius of 4. The rock is initially at rest, so mega initially equals 0, but reaches, in just 3 seconds, it reaches a 150 RPM. So RPM final is 150 and this happens in just 3 seconds. I want to calculate its tangential acceleration. Tangential acceleration is a t, it's not alpha. Alpha is angular acceleration or rotational acceleration. Okay. Now, the equation we have for a t is that a t equals r alpha. We got r, r is 4, but we don't have alpha. So all we gotta do is go find alpha. Alpha is one of our motion equations so I can use one of our one of our 5 motion variables so I can use one of the 4 motion equations. Okay. Here, I have 4 of them already. Now, this is going to have to be converted into omega final before we can solve it. And the 5th one that's missing is delta theta. So I'm going to put a little sad face. So first things first, let's convert this because you can't plug RPM into a motion equation. It has to be in terms of w. So I already showed you this a bunch of times, but RPM goes into frequency, which then goes into omega. Frequency is RPM over 60 and omega is 2 pi f. I can combine these 2. Omega is 2pi RPM over 60. RPM is 150. So this is going to be 2pi 150 divided by 60. And when you do all of this, you get an omega of 15 point 7 radians per second. Okay? So let me put this here, 15.7, a little messy. I'm sorry about that. The ignored equation is, the ignored variable is this one. We're going to use the only equation that doesn't have delta theta, which is the first one. So I have omega final equals omega initial plus alpha t, and I'm looking for alpha. This is 0. So alpha will be omega final divided by time or delta t and it is 15.7 divided by 3 And the answer is, I have it here, 5.23 radians per second squared. But that's not what we want. What we want is at, tangential acceleration. The r is 4 and the 5.23 goes here. And if you multiply these 2, you get 20.9. Tangential acceleration is a linear acceleration. So I get meters per second squared. Okay. Alright. So for the bonus question here, we want to calculate the tangential speed after 3 seconds. Tangential speed is v t and I want to know this after the same 3 seconds. Well, vt is r omega. And if I want to know the initial zero, so I want to know v t after 3 seconds, after means it's final. So this is v t final omega final. I know the r, but I don't know omega final. But I could get it. Okay. I can get it. Actually, I do know omega final. Omega final is simply 15.7. We got that. We just plug in b t equals r, which is 4 times 15.7. And this is going to be 62.8 meters per second. Okay. So just more plug it into the equations that you already know how to use. Alright. That's it for this one. Let me know if you have anything.

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A rock rotates around a light, 4-m long string. The rock is initially at rest, but reaches 150 RPM in 3 seconds. Calculate its tangential acceleration after 3 s. BONUS: Calculate its tangential speed after 3 s.

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A rock rotates around a light, 4-m long string. The rock is initially at rest, but reaches 150 RPM in 3 seconds. Calculate its tangential acceleration after 3 s.
BONUS: Calculate its tangential speed after 3 s.