Question

(a) the tension of each thread and the angular acceleration of the cylinder.

(b) the time dependence of the instantaneous power developed by the gravitational force.

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Solution

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i.e. from equation $F_{x}=mw_{c}$ and $N_{z}=I_{c}β_{x}$.

$mg−2T=mw_{c};2TR=2mR_{2} β$

As there is no slipping of thread on the cylinder

$w_{c}=βR$

From these three equations

$T=6mg =13N,β=52 Rg =5×10_{2}rad/s_{2}$.

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