A block of 200 g mass moves with a uniform speed in a horizontal circular groove, with vertical side walls of radius 20 cm. If the block takes 40 s to complete one round, the normal force by the side walls of the groove is :
Solution
Normal force will provide the necessary centripetal force.<br><br>$\Rightarrow$ N = m$\omega$<sup>2</sup>R<br><br>Also, $\omega$ = ${{2\pi } \over t}$<br><br>N = (0.2)$\left( {{{4{\pi ^2}} \over {{T^2}}}} \right)$(0.2)<br><br>$\Rightarrow$ N = 0.2 $\times$ ${{4 \times {{(3.14)}^2}} \over {{{(40)}^2}}}$ $\times$ 0.2<br><br>$\therefore$ N = 9.859 $\times$ 10<sup>$-$4</sup> N
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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