Question #269506

A 3.56 kg block resting on a smooth horizontal surface is attached to a string that passes over a frictionless pulley as shown in the figure. The pulley has a radius of 0.28 m. A force of 7.5 N pulls the other end of the string. As a result, the block moves a distance of 0.2 m starting from rest. Find the (a) acceleration of the block, (b) angular acceleration of the pulley, and (c) moment of inertia of the pulley. 




Expert's answer

Assume that the moving time is 1.5 s.


(a) acceleration

s=at22a=2st2s=\frac{at^2}{2}\to a=\frac{2s}{t^2}

a=2×0.21.52=0.18 (m/s2)a=\frac{2×0.2}{1.5^2}=0.18\ (m/s^2)


(b) angular acceleration

ϵ=ar=0.180.28=0.64(rad/s2)ϵ=\frac{a}{r}=\frac{0.18}{0.28}=0.64 (rad/s^2)


(c) moment of inertia

T=ma

FrTr=IarFr-Tr=\frac{Ia}{r}

Frmar=IarI=r(Frmar)aFr-mar=\frac{Ia}{r}\to I=\frac{r(Fr-mar)}{a}

I=0.28[(7.5×0.28)(3.56×0.18×0.28)]0.18=3 (kg/m2)I=\frac{0.28[(7.5×0.28)-(3.56×0.18×0.28)]}{0.18}=3\ (kg/ m^2)



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