Recent content by Pi-is-3

Thank you everyone, this problem has been solved.

Got it! Since their is no translational motion, the normal reaction from ground+ frictional on the left wall= mg. And, friction on ground = Normal reaction from the left wall. Done!

Oh wait, maybe the normal reaction at ground isn't mg. I have had an idea, let me try it out first.

Ignore the minus sign behind mgk. Their has to be some involvement of centripetal force in friction, but I am not sure how, because it is directed towards the center, and not towards the ground. I know I have probably done it very wrong. I'm sorry if it's too bad.

I am unable to find the frictional force on the left wall.

Hello, I'm stuck in this rotational motion problem (advanced high school level). Source: Problems in General Physics- IE Irodov My attempt(s): First I tried using work done by the moment of friction (mgkR) and equated it with change in KE. I got the answer as ## \frac{R (\omega_0)^2}{8 \pi...

You can simplify the formulas of n=odd. For n and m both odd the formula can be simplified to ## \frac{mn+1}{2} ## . For n odd, m even, the formula can be simplified to ## \frac{mn}{2} ## . This way both the cases of n feel interconnected.

So much better than what I was going to post. I just expanded ## e^x ## upto first 3 terms to get the second greater than first.

Sorry for late reply. I was not able to come online a lot since a long time. Anyway, I tried to prove it, but I couldn't. Hence my proof is wrong. Sorry.

Sorry, I thought it was sufficient to leave it at that. EDIT: I made one more mistake, I didn't mention that z is not equal to 0. If z=0, then the answers come to be (0,0,0) which is excluded.

One wrong assumption, x=a is not the global maximal for all a. If a is negative or an even natural number, then it is just a critical point. But the answer is correct because it is the global maxima of ##e^{-2}##.

In this method, ##a^2=9x^2, b^2=9y^2## . This implies ##c^2## is a multiple of 3. Then you can continue with the same method I did. Using contradiction to prove that solutions don't exist. However, I'm not sure how to prove that d divides c (if I'm not wrong, d is the gcd of a,b).

I didn't write the answer for this one, because I didn't really come up with the solution. It was an example question in my physics textbook, and it had 2 methods of solving it. One of them was similar to yours and the other was similar to that of @nuuskur gave.

I'm sorry. I advised it because when I came to this site, I learned that copy pasting latex causes some problems in this website. So I found out alternative ways. It also had the replace function, so I would type shortcuts, and then find and replace them. It also preserved the formatting. You...