A 1.0 kg ball and a 2.0 kg ball are connected by a 1.0-m-long rig...

Question

A 1.0 kg ball and a 2.0 kg ball are connected by a 1.0-m-long rigid, massless rod. The rod is rotating cw about its center of mass at 20 rpm. What net torque will bring the balls to a halt in 5.0 s?

Accepted Answer

Hey, everyone in this problem, a very light bar of length, 0.8 m has two cubes, a mass, 0.25 kg and 0.75 kg welded to its two ends, the bar attached at its center of mass to a fan motor revolves counterclockwise at a steady angular speed of 45 R PM. Now we're gonna come back to the rest of this question, but let's draw this out first. So we have our very light bar and it has a length of 0.8 m on one end of our bar. And we're gonna draw it on the left end, we have a cube with a mass which we'll call M one which is equal to 0.25 kg. And on the other end, we have a cube with a mass M two which is 0.75 kg. Now, we're told that this thing is rotating counterclockwise, counterclockwise, it could be kind of this direction at 45 R PM. So we have omega equals to 45 R PM, the angular speed. OK. Back to this problem, we're told that a torque is applied that causes the this to stop rotating in 12 seconds. And we're asked to calculate that torque tell we're given four answer choices. Option A 0.47 Newton meters clockwise. Option B 0.24 Newton meters counterclockwise. Option C 0.58 Newton meters counterclockwise and option D 1.4 Newton meters applied clockwise. Now, if we think about this conceptually, we can already eliminate two answer choices. OK. If this thing is rotating counterclockwise, and we want to apply a torque that's gonna stop it from rotating, we're gonna have to apply that torque to oppose the motion, that it means that that torque is gonna have to be applied clockwise. OK. So we already expect that the answer is gonna be either A or D, right? Because we need that opposing direction to stop this from rotating. Now, we're looking for a torque we're called the torque to can be written as the moment of inertia. I and that's the moment of inertia about the center of mass because that's where we're rotating about, they multiplied by alpha the angular acceleration. Now, we're looking for the moment of inertia of the center of mass in order to find that we first want to calculate what that center of mass is. OK. So we're gonna do that first and we're gonna find X C M OK. That position of the center of mass. Now, recall at the center of mass, we have two objects here. OK. This bar we're told is very, very light. So we're going to assume that its mass is negligible. So the two masses we have to worry about are the two cubes. And so the position of the center of mass is gonna be given by M one X one plus M two X two divided by M one plus M two. No, let's choose to calculate this from the left end. If we choose to calculate this from the left end, we get the mass of cube one which is 0.25 kg multiplied by X one, which is the distance from that mass to whatever our reference point is. Now, if we're taking the reference point to be the left end, then the distance is just gonna be zero m. OK? So that first part is not gonna contribute based on our reference point. The second term we have the mass and in this case, the mass of that second cube, 0.75 kg multiplied by the distance to the reference point. Now the reference point is at the far left end, this cube is at the far right end. The bar has a length of 0.8 m. And so the distance is gonna be 0.8 m. All right, we're gonna take all of this and divide it by the sum of those two masses. 0.25 kg plus 0. kg. Now, in the denominator, we're just gonna get one kg. OK? So the unit of kilogram will cancel out or divide out through the entire equation. And then we're just left with 0.75, multiplied by 0.8 m, which is gonna give us 0.6 m. OK? So the position of the center of mass is gonna be 0.6 m from the left end. I'm gonna draw that in red. This is our X C M position at the center of mass and it is 0.6 m from the left-hand end, right? If it's 0.6 m from the left-hand end, that means it's 0.2 m from the right hand end because the entire thing is 0.8 m. Now that we know where our center of mass is, we can calculate the moment of inertia about that center of mass. So the moment of inertia about that center of mass is gonna be made up of the moment of inertia of the bar. Pull off the moment of inertia of Q one, pull us the moment of inertia of cube two. Now again, this bar is very light, we're gonna assume that it has negligible weight. And so the moment of inertia of the bar is actually just going to be zero. OK. So we're down to two moment of inertia we need to be concerned about, we aren't told much about these cubes. Let's go ahead and assume that they are point masses. If we assume that these are point masses, then we're called that the moment of inertia for each is gonna be MD squared. So we have MC one DC one squared plus MC two DC two squared. OK. Where M is the mass and D is the distance from that particular cube to the center of mass where we're rotating about and substituting in our values here, the mass of Q 10.25 kg, the distance from Q one which is on the left to the center of mass. Well, we just found that that's 0.6 m. OK. This is why it was important to find that position of the center of mass because we needed these distances that's gonna be 0.6 m squared. And then we add the mass of cube 20. kg multiplied by the distance of cube two to the center of mass. A cube two is on the right. And we said that that distance is gonna be 0.2 m. So we get 0.2 m squared. Simplifying we have 0. kilogram meter squared plus 0.3 kilogram meter squared, which gives us a total moment of inertia about that center of mass of 0.12 kilogram meters squared. And I'm going to put a ball box around this. So we don't lose track of it. That is one of those values we need in order to calculate our torque. Mhm. All right. So we've done a lot of calculations already. Let's remind ourselves of what, what we're trying to do. We want to calculate the torque to, we can do that by taking the moment of inertia at the center of mass and multiplying it by the angular acceleration alpha. We've calculated the moment of inertia about the center of mass IC M already. So now we need to calculate this alpha value. OK. This acceleration, let's write out the information that we have. We know that the initial angular speed omega knot is equal to 45 R P F, right? This is a positive value because we're told that it's rotating counter clockwise. Now we can write this as 45 revolutions per minute. OK? We wanna convert to our standard unit of radiance per second. We're gonna multiply by two pi radiant divided by one revolution. OK? Because we know that in every revolution there are two py radians. You can imagine going around a circle. What's equates to two py radiant, the unit of revolution will cancel. Then we're gonna multiply again and we're gonna multiply by one minute divided by 60 seconds. Since there are 60 seconds in every minute, the unit of minute will divide out. We're gonna be left with 45 multiplied by two pi radians divided by 60 seconds which we can simplify to three pi divided by two radiant per second. Yeah. So that is our omega knot value omega F we know is gonna be equal to zero radiance per second. OK. We're told that this torque is applied in order to stop the bar from rotating. So we start at this speed of 45 R PM and we end up being stopped, zero radiance per second and it takes 12 seconds to stop. So we know that T is equal to 12 secondss. We're looking for alpha. OK. So we can use our rotational kinematic equations. These are very similar to the linear case and the conditions are the same, we have three known values and one thing we want to find. So that's enough to find this angular acceleration. We're gonna choose the equation that does not include delta data because we don't have information about that. And that's not what we're looking for. OK. What we get is that Omega F is equal to Omega knot plus alpha T substituting in our values zero radiance per second is equal to three pi divided by two radiant per second plus alpha multiplied by 12 seconds. We're gonna move this alpha multiplied by 12 seconds to the left hand side. We want a isolate for alpha. We get negative alpha multiplied by 12 seconds is equal to three pi divided by two radiant per second. We're gonna divide by negative 12 seconds. We get that alpha is equal to negative three pi divided by 24 radiance per second squared. Now because we divided by that extra unit of seconds and we can approximate this to negative 0. radiance per second squared. OK. So this is that other value we needed for our torque calculation. And it makes sense that we found a negative angular acceleration. OK? Because this is gonna be slowing that bar down to a stop. OK. So that makes sense we have this negative acceleration, it's gonna be slowing the motion. So now we can get back to that final calculation that we have been working towards. And that is that torque calculation, the torque tau is equal to the moment of inertia about the center of mass multiplied by the angular acceleration alpha. This is equal to 0.12 kg meter squared multiplied by negative 0.3927 radiance per second squared. OK? We have kilogram meter squared per second squared. We call that a Newton is a kilogram meter per second squared. So our unit is gonna be equivalent to a Newton meter which is exactly what we want for torque. And we get that this is equal to negative 0. Newton meters. And that is the final answer we were looking for. Now we have a negative, we need to interpret the direction of that. OK. When we have a negative torque that indicates clockwise, OK. That's a clockwise torque and that's exactly what we were expecting. Ok. That makes sense. We have this bar rotating counterclockwise. We want to apply a torque to stop it. That means that the torque needs to apply it, be applied in the opposite direction to oppose that motion and stop it. Ok. So we have a negative torque which indicates the clockwise direction and rounding to two significant digits. We see that the correct answer is option a 0.47 Newton meters applied clockwise. Thanks everyone for watching. I hope this video helped see you in the next one.

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A 1.0 kg ball and a 2.0 kg ball are connected by a 1.0-m-long rigid, massless rod. The rod is rotating cw about its center of mass at 20 rpm. What net torque will bring the balls to a halt in 5.0 s?

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