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. 




1
Expert's answer
2021-11-22T17:51:00-0500

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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