Recent content by adeelrizvi

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    Competitive exam Question: Work done to shorten string

    I really appreciate your help. I hope we are correct but it will be good to know if someone else here can verify our solution.
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    Competitive exam Question: Work done to shorten string

    Solution: dW= F.dr dW= {(m*v^2)/r}dr (v=k/mr) dW= {(k^2/m)*(1*r^3)}dr Integrate this with limits Ro to R W = [(k^2)/(2m)]*[(1/Ro^2) - (1/R^2)] <----Answer Please check and let me know if this seems correct to you? Thanks
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    Competitive exam Question: Work done to shorten string

    I guess you are right. I was taking the wrong assumption. If angular momentum is conserved then velocity increases if radius decreases. Now when you have corrected me I think my solution to the problem was not correct either. Can you please verify that for me as well? Thanks
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    Competitive exam Question: Work done to shorten string

    You think my solution is correct? Can someone else also please confirm this. v=rw, w is constant. r decreases so v decreases as well. Correct me if I am wrong. Angular momentum I believe is conserved.
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    Competitive exam Question: Work done to shorten string

    Q) A particle of mass 'm' attached to a string rotates with velocity Vo when the length of the string is Ro. How much work is done in shortening the string to R? One way I thought about doing this was: W= {(m*r*w^2) * r}dr and integrate this from R to Ro But I am not sure if that is...
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