u r a real saviour sir.. i was not good in ed but after listening to ur lecture videos i scored really good and got the correct imagination also. thank u so much sir
Hy can you answer qsn 1:Determine the position of centroid[3] 2:second moment of area about the vertical centroidal axis[6] 3:Radius of gyration about the vertical centroidal axes[3]
Thanks for your efforts sir....please upload some videos on machine assembly....your lectures on engineering drawing helped me lot to pass my exams...please upload videos on machine drawing..🙏🙏
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sir your videos are awesome to learn from and i love it,but can you do something bout the ads , they are irrtating and disturbing while learning something from you
Sir can you explain in depth problems very depth problems, i exam questions are asking in depth sir, in this centroid and moment of inertia models sir.....
my question I just saw that on the other video when calc x for semi semi circle you subtract the radius from 4r/3pi and on this one you just used 4r/3pi help me understand sir
Distance has to be taken from the axes and then measurements are done. Think about it and i would suggest you observe both the solutions again. There is a subtle difference.
Hi therr.... look when you are taking distances you need to consider an axis and here we are taking vertical distances okay or the y cordinates so you have to take all of these distances with respect to the x axis. Area 1 happens to be a rectangle and its height is 30 so 30 by 2 will be 15 and this distance directly comes wrt x axis, but Area 2 is a right angle triangle whose height is 20 so 20 by 3 is the value from the base of the triangle But we want a distance from the x axis that is why I added 30 to it so the distance of the centroid of area 2 from the exactis will be 30 + 20 / 3
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in this problem for calculating the area of A1 why not 750-(pi*225)/(2) because material whose area is about half of the semicircle has been removed from the area of rectangle
Since the diameter of the circle is same with the height of the rectangle, the diameter is 30. So we get the radius 30÷2=15. 4r÷3π is the centroid formula for semicircles.
Make video on Question Example:- Two blocks of masses M1 and M2 are connected by a string as shown in Figure 5. below Assuming the coefficient of friction between block Mị and the horizontal surface to be u if the system is released from rest, find velocity of the block A after it has moved a distance of 1 m Assume Ml00kg.and M2-150kg and u0.20.
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u r a real saviour sir.. i was not good in ed but after listening to ur lecture videos i scored really good and got the correct imagination also. thank u so much sir
right❤
didi ED ya EM
Your a savior Manish ji. I understood exactly what you said from your video. Thank you!
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Hy can you answer qsn
1:Determine the position of centroid[3]
2:second moment of area about the vertical centroidal axis[6]
3:Radius of gyration about the vertical centroidal axes[3]
Sir u r expect in cad and cam... U r the best explanation for engineering drawing and engineering mechanics
Very very good technique s and teaching methods
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Thanks for your efforts sir....please upload some videos on machine assembly....your lectures on engineering drawing helped me lot to pass my exams...please upload videos on machine drawing..🙏🙏
Thank you sir
Good explanation
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Nice explanation sir👌
Thanks for your lecture
Keep on lecturing sir. .
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plz upload more videos on this concept...
Nice presentation Sir. May I know the software you used during the presentation?
My calculus exam is tomorrow. Thank you for this sir!
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Hi Manas - Would I get the same answer if I would have computed area 1 as the {area of the rectangle - area of the semi circle} ?
sir when do we use 2/3 in calculating x or y coordinate
Your lectures are awesome sir keep it up
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Maan I understand this guy better than my professor
sir your videos are awesome to learn from and i love it,but can you do something bout the ads , they are irrtating and disturbing while learning something from you
thank you for the video , i was asking for the semicircle y Did u use 4r/3π
Same doubt for me also
If we are not given the x and y axis like this question, can we place the axis anywhere along the structure?
Nice sir thanks it is to useful
Thank you so much sir , You are our god 🙏 You bless you . (God bless you 😁)
Sir... Please upload on how to draw nut and bolts...
Sir please upload how to take strip while finding centroid of objects??
Great....
excellent explaination
Thanks Qaiser and enjoy the videos
sir i would like to request you to make more videos on numericals based on dynamics
Thsy have already been uploaded. Search it on my channel page.
And more videos will be uploaded in future as well.
thank you sir
Nice expalinations
Thank you sir for this explaination.
wow
Sir i wanted to know how to calculate that area
Sir plz solve some ques on basis of moment of inertia
Sir can you explain in depth problems very depth problems, i exam questions are asking in depth sir, in this centroid and moment of inertia models sir.....
Thank you sir🎉🎉🎉
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thank u sir
helped a lot
Excellent
Thanks Chandresh....
What did you study ?
sir, centroid for semicircle along Y -axis is 0. right?
and along X-axis is (4R/3pie)
how can we take 15 as Y in semicircle.?
TnQ sir
Why did you use minus sign for the semi circle
Sir u just rock m blessed that I listen you 👉sir can make a video in friction chapter
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my question I just saw that on the other video when calc x for semi semi circle you subtract the radius from 4r/3pi and on this one you just used 4r/3pi help me understand sir
Distance has to be taken from the axes and then measurements are done. Think about it and i would suggest you observe both the solutions again. There is a subtle difference.
Great
Sir please tell us how to locate the position of the centroid
Why you didn't negative the unshaded areas?
Thank you so much sir
Sir tell about the force in a space
Sir can you please slove problems on Moment of inertia
What textbook are the questions from?
Thank you sir
Good
The centre of semi circle ⭕ can it be 15/2=7.5
Super
Sir when you find y2 you take 30+20/3
But when you find y1 you directly take 30/2 not 20+30/2
Why?
Hi therr....
look when you are taking distances you need to consider an axis and here we are taking vertical distances okay or the y cordinates
so you have to take all of these distances with respect to the x axis.
Area 1 happens to be a rectangle and its height is 30 so 30 by 2 will be 15 and this distance directly comes wrt x axis, but Area 2 is a right angle triangle whose height is 20 so 20 by 3 is the value from the base of the triangle
But we want a distance from the x axis that is why I added 30 to it so the distance of the centroid of area 2 from the exactis will be 30 + 20 / 3
Bro you need to eliminate the hemisphere from rectangle for A1
The area of the rectangle should be (bh- pi*r(sqr)/2). I think.
sir y 20/3 and 25/3 in case of rectangle pls explain
sir please upload rolling numerical
Why in x3 you use 4r/3bay I think you must use R which D/2 !!
I'm learning but also send for me the centroid of circles enclosed together, because I'm having a problem in finding the Iyy and Ixx
Kibet.......Can u please rephrase your question. Be very clear and concise in what u ask
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Upload circle questions bro
in this problem for calculating the area of A1 why not 750-(pi*225)/(2) because material whose area is about half of the semicircle has been removed from the area of rectangle
Yes exactly
Thank you sir❤️❤️
in semicircle r is 30 na
Use camtesia studio
Sir why did you subtracted the A3
Because it is a hole. It has negative mass.
I mean it is not included in the solid.. i hope you understand what im trying to say.
In 4r÷3pie what is the value of r??
Since the diameter of the circle is same with the height of the rectangle, the diameter is 30. So we get the radius 30÷2=15. 4r÷3π is the centroid formula for semicircles.
Sorry, I commended too early. You did it at the end.
I don't know how to apply the formula
X3 Kash nikala sir nahi aaya
Someone please calrify that how to take X3 calcualtion
the equation for xi for a semicircle is 4r/3π that is what sir used here
Please sir
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Question Example:-
Two blocks of masses M1 and M2 are connected by a string as shown in Figure 5.
below Assuming the coefficient of friction between block Mị and the horizontal
surface to be u if the system is released from rest, find velocity of the block A
after it has moved a distance of 1 m Assume Ml00kg.and M2-150kg and
u0.20.
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Y* is wrong
31.0597
That is current
Sir semi circil to cut hua hai na
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Good day
Can I ask how you got the >> 4r/3.pi
thank u sir
helped a lot
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