# Chapter 8 - Rotational Motion and Equilibrium - Learning Path Questions and Exercises - Exercises - Page 307: 41

$I=I_{CM}+M(\frac{1}{2})^{2}=\frac{1}{12}M(1)^{2}+\frac{1}{4}M=\frac{1}{3}M$ $T=\frac{Mg}{2}$ Thus, angular acceleration =$\frac{T}{I}=\frac{3g}{2}$ a). Tangential acceleration of the 100 cm position = $1\times \frac{3g}{2}=1.5g$ Yes, the result is surprising as it is more than g. b). $a_{t}=g$ $g=r \times \frac{3g}{2}$ Thus, $r=\frac{200}{3} cm=66.67cm$. Hence, tangential acceleration is g at position 66.67 cm.

#### Work Step by Step

$I=I_{CM}+M(\frac{1}{2})^{2}=\frac{1}{12}M(1)^{2}+\frac{1}{4}M=\frac{1}{3}M$ $T=\frac{Mg}{2}$ Thus, angular acceleration =$\frac{T}{I}=\frac{3g}{2}$ a). Tangential acceleration of the 100 cm position = $1\times \frac{3g}{2}=1.5g$ Yes, the result is surprising as it is more than g. b). $a_{t}=g$ $g=r \times \frac{3g}{2}$ Thus, $r=\frac{200}{3} cm=66.67cm$. Hence, tangential acceleration is g at position 66.67 cm.

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