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Coln:\begin{array}{l}m=40 \mathrm{~kg} \\D=18 \mathrm{~m}, \quad R=9 \mathrm{~m} \\\end{array}\begin{array}{l}t=\frac{60}{4}=15 \mathrm{sec} \text { for } 1 \mathrm{rev} . \\v=\frac{2 \pi R}{T}=3.77 \mathrm{~m} / \mathrm{s}\end{array}a.) a_{c}=\frac{V^{2}}{R}=\frac{(3.77)^{2}}{8}.\begin{array}{c}a_{c}=1.77 \mathrm{~m} / \mathrm{s}^{2} \text { towards the } \\\text { center }\end{array}b.) at lowest point\begin{array}{l}N=m\left(g+a_{c}\right)=40(9.81+1.77) \\N=463.2 \mathrm{~N} \\\text { direction = upward } \therefore \text { angle }=0^{\circ}\end{array}c) at highest point\begin{array}{l}N=m\left(g-a_{c}\right)=40(9.81-1.77) \\N=321.6 \mathrm{~N} \quad \text { direchion = upward } \\\therefore \theta=0^{\circ}\end{array}d) at halfway\begin{aligned}N_{x} & =40(1.77)=70.8 \mathrm{~N} \\N_{y} & =40(9.81)=392.4 \mathrm{~N} \\N & =\sqrt{N_{x}{ }^{2}+N_{y}^{2}} \\& =398.73 \mathrm{~N}\end{aligned}\begin{array}{c}\&=\tan ^{-1}\left(\frac{N x}{N_{y}}\right)=\tan ^{-1}\left(\frac{70.8}{392.4}\right) \\\theta=10.227^{\circ}\end{array} ...