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20) A uniform sphere of mass 1 kg and radius 5 cm is welded to the end of a thin uniform rod of length 50 cm and mass 0.5 kg, as shown in the figure below. The sphe...

Question

20) A uniform sphere of mass 1 kg and radius 5 cm is welded to the end of a thin uniform rod of length 50 cm and mass 0.5 kg, as shown in the figure below. The sphere/rod system will balance horizontally on a knife edge placed a distance x from the left end of the rod: What is x?5 cmA 25 cm B. 45 cm C 40 cm D. 48 cm E. 52 cm50 cm

20) A uniform sphere of mass 1 kg and radius 5 cm is welded to the end of a thin uniform rod of length 50 cm and mass 0.5 kg, as shown in the figure below. The sphere/rod system will balance horizontally on a knife edge placed a distance x from the left end of the rod: What is x? 5 cm A 25 cm B. 45 cm C 40 cm D. 48 cm E. 52 cm 50 cm



Answers

Two spheres each of mass $M$ and radius $R / 2$ are connected with a massless rod of length $2 R$ as shown in the figure.
The moment of inertia of the system about an axis passing through the centre of one of the spheres and perpendicular to the rod is (a) $\frac{21}{5} M R^{2}$ (b) $\frac{2}{5} M R^{2}$ (c) $\frac{5}{2} M R^{2}$ (d) $\frac{5}{21} M R^{2}$

This problem, we have to find which of the following object has the greatest rotational energia one kg solid wall with a radius of five centimeters. Masses option paste masses one kg. The radius is five centimeters. So I will write it as a small lot. So we know that moment of inertia of a solid wall is a cult two by five Mm Hardest quick. Not an option we we have given this mass is one kg. So this will be equal to 55 Hardest square in option B a one kg hollow ball with radius of five centimeters. So we know that moment of inertia of olive all formulas is two by three and into our square. Their mass is one kg and this is for option we and masses one kg. So this will be called to buy three Artist Claire, you know, for option it is to buy five bars square the options See says mhm five kg solid wall with radius five centimeters. So this will be equal to two by five. Mm artists. Where now? This time masses, five KT. So this will be called to buy 55 and 22 by five how to square all weekends. It is two r squared now in part D. We have given five kg hall of all with radius five centimeters. So this will be equal to two by three masses five k d two r squared. So this is various is same for all the objects. So I will right here 10 by three are the square. Now we will see the greatest in us here. This is equal to around 0.4 I'll square and it is around 0.6 r squared an option we and it is to our square and it is around more than three r squared. So if we see, this is for C and this is far or Sunday Now we have to find the greatest moment of an L C here. So this will be for option D five kg hall of all with radius five centimeters. This is the answer. Option is correct. Mhm. Thank you

Rotational inertia. It's proportional to salutation. Inertia rise proportional to the muscle for an object and also proportional to the distance from a tradition of access. So Option Deke is the right answer because of the high marks and the greater distance from excess of tradition.

All right. So this brushing me are after a both the area on a curse. Um, so in question, they were asked with the area under the curve between time with zero and time in calls for five. So I mean, we consider this to be your time, because five, uh, and we can kind of put this into three areas. So we got two rectangles. They're trying. So the first rec time goal is to buy two. So the area is gonna be, um, four plus on mrs. To this, rectangles will be to buy three go five, uh, plus six. Plus. Now, this triangle the the height. Ah. Is there three and the, um yeah, I just three in length is three. So, um, the area is gonna be 1/2 times nine, which is 4.5. The area is 14.5. So in part B were asked about the area between zero or sore between five and 10. So between five and 10 again are basically going to do the same thing. We got a little triangle, appear we've got a nice rectangle and another nice rectangle. So the smaller rectangle it is won by five. So area, it's areas five The triangle It's one by one. So it's area his 0.5 by one by one. I mean base and I have one. And then the big rectangle it is four Ah, by six. So the area is 24 so total area is 20 9.5. So now we and by so yes, O c So total from 0 to 10 which just add the 055 10 up. So that is gonna give us 44 and then lastly, in part D that ask us about the center of mass. So, um, the what this is really asking us is, um what, uh, X value is, um, half. Um, I want at up to what? X value is half of the total area under the curve. So, um, just by looking, um so Okay, it's a total area under a curve is 44 so x. So the center of mass is gonna be where up where the total, um, area under the curve above and below. It is 22. So, um, that if we we know going up to exceed six the area of the service 20 so and going up to seven, um, the area under the curve A's, um, you're gonna be Yes, we know that, um, it's gonna be between six. And seven. So now, um, we're gonna So yes. So now we're gonna get the ratio between six. And seven. So, um, we know, Yeah.

Let's start with party from from the graph, we can see the graph. The area under the curve is divided into little cubes, and we can try to find the area of each of these cubes. So for each cube, its height is one grams per centimeter. What's its length is one centimeter. So the area off this cube represents one grand. Okay, now, to find the mess off the rod from 0 to 5, we need to count how many of these cubes there are. So when you look at the graph, you can count 13 whole cubes, plus three halves off a cube. Now, one thing to notice. Two halves make one cube, so this turns out to be three cubes plus two halves, plus 1/2 this two halves. Make one cube. So have 13 cube plus one Q plus 1/2 and so then we have 14 cubes, plus 1/2 of a cube. So then, since we have since we know that one Cube represents one gram, we know that this is 14 grams and we have half a gram. So it's 0.5 Gramps. So that so then the mass of the Rod from 0 to 5 is 14.5 grams. Next we have Oops, Not not zero. We have 5 to 10. We can count that There are 29 cubes from 5 to 10. Plus, we have 1/2 of a cube. So substituting what we know about each cube, you know, this represents 29 grams plus 0.5 grams, so get 29.5 grams. Now, the mass of the entire rod should be the mass off the length 0 to 5, plus the mass off the length 5 to 10. So the total mass should be equal to 14 plus 14.5 plus 29.5 and you should get 44 grams in order to find the sent off mass off our the location off our center of mass on the road. Let's find what half of out mess is. So since our total mess from what we calculated before is 44 then our sent up then that half of this mass in 1/2 mess should be 22. Now what we want to do is now we want to find a point between zero and 10 such that both sides have 22 grams or in terms of the graph, we have 22 squares and when you look at look at the graph, you can't count. Roughly the point of 22 squares should be between between six and seven between six and seven, you can count roughly. Both sides has 22 grams slash square groups, number 22 Gramps, and that's it between six centimeters and seven centimeters.


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