FIGURE P8.54 shows two small 1.0 kg masses connected by massless but rigid 1.0-m-long rods. What is the tension in the rod that connects to the pivot if the masses rotate at 30 rpm in a horizontal circle?

Question

Diagram showing two 1.0 kg masses connected by rods, rotating at 30 rpm around a rotor.

Accepted Answer

Welcome back, everybody. We are making observations about this system right here. We have two balls both with equal mass, but I'll name this ball ball two and I will name this ball ball one. They're connected by sticks of, of equal length to this rotor right here. Now, we are told a couple of different things about this system. We are already told that they have the same mass of 100 g or 0.1 kg. And that each of the sticks have a length of 75 centimeters or 0.75 m. We are also told, let me go ahead and move this over just a little bit here. We are also told that they have a constant angular velocity or the rotor has a constant angular velocity of 22 revolutions every one minute. And we are tasked with finding what the tension is in this first stick right here. So this tension that's going to be facing towards the rotor. We need to find T one. Now, before diving into this, I want to highlight a couple of other tensions as well. We are going to have the tension in this second stick acting in words, and this will be T two. Both of these are equal. It's just from the perspective of whichever ball we're looking at here. So what we are going to do is we are going to apply Newton's second law for each particular sphere. So first, let's start out with the, some of all forces in the radial direction around this second mass right here is just going to be equal to the second mass times the radial acceleration of that mass. So what does this equal? We're looking at our forces present T two is the only one attached to it and it's in the positive radial direction. So we're gonna have that T two is equal to M two times. I'm going to sub in B squared over R for our radial acceleration. So this will be the velocity of our second mass squared divided by the distance to the rotor which is going to be two L but we can sub in for L two L omega weird and then everything else will remain the same. So this will be two L times M two. So what does this give us here? We have then that our T two, let me write this over just a little bit. We have that T two is equal to two L times our mass of that particular sphere times our angular velocity squared. Now is everything in the right units before we can plug in. Uh I'm looking over here and it doesn't look like it is. We need this in radiant per second. Well, we know that per revolution, there are two pi radiant and every one minute there are 60 seconds, this is going to give us an angular velocity of 2.304. Now, let's go ahead and plug in all of our values to find our tension in the second stick. This is going to be two times 0.75 times 0.1 times 2.304 squared, giving us a tension in the second stick of 0.8 students. Great. So now what we are going to do is we are going to use Newton's second law again. But this time we are going to do it around the first sphere right here. This is just going to be equal to M one times the radial acceleration of our first sphere. So let's go and plug in some values this time, we have two forces in play in the negative radial direction is T two and in the positive radial direction is T one, this is going to be equal to M one times. I'm once again going to sub in V squared over our distance to the rotor. So this will be V one squared divided by this time, just L and once again, we can sub in for RV one, this is just going to be two times the length times our angular velocity squared, which gives us L times M one times our angular velocity squared. Now, what I am going to do, let's see here, I'm going to add T two to both sides of our equation. And you will see that T one is then equal to LM one Omega squared plus T two. So let's go and plug in our values. We have 0.75 times 0.1 times 2.304 squared plus our T two of 0.8 giving us a tension in stick, one of 1.2 newtons corresponding to our final answer. Choice of D. Thank you all so much for watching. I hope this video helped. We'll see you all in the next one.

FIGURE P8.54 shows two small 1.0 kg masses connected by massless but rigid 1.0-m-long rods. What is the tension in the rod that connects to the pivot if the masses rotate at 30 rpm in a horizontal circle?

Verified Solution

Key Concepts

Centripetal Force

Tangential Velocity

Rotational Dynamics