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Differential Calculus
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8 Mar 2012 21:54:40 IST
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The moment of inertia of the circular disc through the central axis(I) :mr^2/2. -------- eq(1)
torque(T)=r x F --------eq(2)
F=moment of inertia x angular acceleration-----------eq(3)
angular acceleration =r x angular velocity-----------eq(4)
take angular velocity to be w.
substitutin eq(4) in eq(3),we get
force F=mr^2/2 x r x w.
=mr^3 x w/2.---------eq(5)
substituting eq(5) in eq(2)
we get
torque T=(r x mr^3 x w)/2
=(mr^4 x w )/2
ans:(C)
thus torque T is proportional to r^4
9 Mar 2012 00:35:59 IST
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how do u give the ans. guys i dont know!! u dont know urself how blindly u r trying hard to justify the ans...arjun virmani sir pls give me ewwplanation why disc needs centripital force...sir i think if disc p rotates every particle get centripital force d by w9 sorrunding particals cgi being s rigid body. the work of ewwternal torque is to work against frictionajl tarque so that net torque is zero and body rmotates wid constant angular velocity. my ans is b.




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The Friction is responsible to provide the centripetal acc. and the friction rises from the force applied.
T = r x F........eq1
F = 1/2 I w2.............eq2
I = k r2..........eq3
From eq2 and eq3
F is proportional to r2
T is thrfr proportional to r3
Ans!!