Menu Close

Question-33214




Question Number 33214 by Tinkutara last updated on 13/Apr/18
Commented by Tinkutara last updated on 13/Apr/18
Commented by Tinkutara last updated on 15/Apr/18
But if we consider the torque of mg instead of ma it still gives approximately the same answer. Which one is conceptually correct?
Commented by ajfour last updated on 14/Apr/18
Commented by ajfour last updated on 14/Apr/18
mg−T=ma  TR=I((a/R))  ⇒   mg=a[m+(I/R^2 )]    ...(1)  s=(1/2)at^2    ⇒   a=((2s)/t^2 )  using in (1)  mg=((2s)/t^2 )(m+(I/R^2 ))  or    I = (((mgt^2 )/(2s))−m)R^2             =(((0.5×10×16)/(2×1.5))−0.5)(0.15)^2           =(((80)/3)−0.5)(0.0225)kg.m^2          =(((80)/3)−(1/2))((9/(400))) kg.m^2          =((157)/6)×(9/(400)) =((157×3)/(800)) = ((4.71)/8)        = 0.59 kg.m^2   .
mgT=maTR=I(aR)mg=a[m+IR2](1)s=12at2a=2st2usingin(1)mg=2st2(m+IR2)orI=(mgt22sm)R2=(0.5×10×162×1.50.5)(0.15)2=(8030.5)(0.0225)kg.m2=(80312)(9400)kg.m2=1576×9400=157×3800=4.718=0.59kg.m2.
Commented by ajfour last updated on 15/Apr/18
if torque of mg is taken  net  moment of inertia of system  about fixed axis of pulley is  I_(system) =I+mR^2   mgR=(I+mR^2 )((a/R))  .
iftorqueofmgistakennetmomentofinertiaofsystemaboutfixedaxisofpulleyisIsystem=I+mR2mgR=(I+mR2)(aR).
Commented by Tinkutara last updated on 15/Apr/18
Both are same in value approximately. Which is correct then?
Commented by ajfour last updated on 15/Apr/18
have same value exactly. how approximately?
havesamevalueexactly.howapproximately?
Commented by Tinkutara last updated on 15/Apr/18
Thank you very much Sir! I got the answer. ��������

Leave a Reply

Your email address will not be published. Required fields are marked *