A 2.80-kg grinding wheel is in the form of a solid cylinder of ra...

Question

A 2.80-kg grinding wheel is in the form of a solid cylinder of radius 0.100 m. (a) What constant torque will bring it from rest to an angular speed of 1200 rev/min in 2.5 s?

Accepted Answer

Welcome back everybody. We have a flywheel and we are told a couple different things about this flywheel. We're told that it has a mass of 15 kg and we are also told that has a radius of eight cm or zero meters. Now we are asked to find the amount of torque required to accelerate this flywheel from rest, meaning starting out with an initial angular velocity of zero to a rotational speed of 3600 rpm. In a matter of 3.2 seconds we mentioned acceleration. We need to find tour the main automatic formula that pops out to me is that torque is equal to our moment of inertia for this solid disc times our angular acceleration. Now the moment of inertia for a solid disc is given by this formula here. Our moment of inertia is equal to one half times our mass times our radius squared. Well before moving on, I'm just plug in our values here and just find this super quick. We have one half times our mass of 15 times our radius of 150.8 squared Gives us a moment of inertia of 0. kilograms meters. Where great. So we have a moment of inertia. But what about our acceleration? What is our acceleration equal to? Well, we are actually going to use these values over here along with our rotational kid. Matic formulas, The first one that catches my eyes. This one, it says that our final angular velocity equal to our initial angular velocity plus our angular acceleration times time. Now I'm actually gonna isolate this term right here. So this is what's gonna look like attracting our initial omega from both sides. And dividing by t. We get that our angular acceleration is equal to our final angular velocity minus our initial angular velocity, all divided by our time. Now, before we can plug in our values, we have to make sure that everything is in the right unit here. So It looks like everything's in the right unit. Except this guy, we don't need revolutions per minute, rather we need radiance per second. So I'm gonna go ahead and convert this. We are given 3600 RR for one minute In one minute there is 60 seconds and in one revolution there is two. I radiance. These units cancel out and these units cancel out. And when you plug this into your calculator we get that our final angular velocity is 100 and 20 pi radiance per second. Great. So now we have this, we have that our initial angular velocity is zero and we have our time in seconds. So let's go ahead and plug in our values now. So our angular acceleration is equal to 1 20 pi minus zero, All divided by 3.2. Which when you plug into your calculator that our angular acceleration is 117.8 Radiance per second squared to find our torque. Now that we have the moment inertia and the angular acceleration. Let's go ahead and plug into our torque equation. Once again, our torque is equal to our moment of inertia times our angular acceleration plugging in our values, we have that moment of inertia is 0. kg meters. Where'd Times are angular acceleration of 117.8 radiance per second squared? Giving us a final answer of 5.65 Newton meters. Sorry, meters corresponding to our final answer choice of B. Thank you guys so much for watching. Hope this video helped. We will see you all in the next one.

A 2.80-kg grinding wheel is in the form of a solid cylinder of radius 0.100 m. (a) What constant torque will bring it from rest to an angular speed of 1200 rev/min in 2.5 s?

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