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Question Number 28905 by tawa tawa last updated on 01/Feb/18

$$\mathrm{A}\mathrm{body}\mathrm{rolls}\mathrm{down}\mathrm{a}\mathrm{slope}\mathrm{from}\mathrm{a}\mathrm{height}\mathrm{of}100\mathrm{m}.\mathrm{the}\mathrm{velocity}\mathrm{at}\mathrm{the}\mathrm{foot}\mathrm{of}\mathrm{the}$$$$\mathrm{slope}\mathrm{is}20\mathrm{m}/\mathrm{s}.\mathrm{What}\mathrm{percentage}\mathrm{of}\mathrm{the}\mathbf{P}.\mathbf{E}\mathrm{is}\mathrm{converted}\mathrm{in}\mathbf{K}.\mathbf{E}?$$$$$$$$\mathbf{A}\mathrm{nswer}:20\mathrm{\%}$$

Answered by mrW2 last updated on 01/Feb/18

$$PE=mgh$$$$KE=\frac{1}{2}{mv}^2+\frac{1}{2}I\frac{v^2}{R^2}=\frac{m}{2}(1+\frac{I}{{mR}^2})v^2$$$$\alpha =\frac{KE}{PE}=\frac{v^2}{2gh}(1+\frac{I}{{mR}^2})$$$$$$$$Forasolidball:I=\frac{2}{5}{mR}^2$$$$\Rightarrow \alpha =\frac{v^2}{2gh}(1+\frac{2}{5})=\frac{{20}^2}{2\times 10\times 100}\times 1.4=0.2\times 1.4=0.28=28\mathrm{\%}$$$$$$$$Forahollowball:I=\frac{2}{3}{mR}^2$$$$\Rightarrow \alpha =\frac{v^2}{2gh}(1+\frac{2}{3})=0.2\times 1.67=0.34=34\mathrm{\%}$$$$$$$$Foradisc\left(cylinder\right):I=\frac{1}{2}{mR}^2$$$$\Rightarrow \alpha =\frac{v^2}{2gh}(1+\frac{1}{2})=0.2\times 1.5=0.30=30\mathrm{\%}$$$$$$$$Foraslidingobject\left(withoutrotation\right):I=0$$$$\Rightarrow \alpha =\frac{v^2}{2gh}(1+0)=0.2\times 1.0=0.20=20\mathrm{\%}$$

Commented by tawa tawa last updated on 01/Feb/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{i}\mathrm{really}\mathrm{appreciate}.$$

Commented by NECx last updated on 01/Feb/18

$$\mathrm{T}hisis\mathbb{W}\mathit{o}\mathit{n}\mathit{d}\mathit{e}\mathit{r}\mathit{f}\mathit{u}\mathit{l}.$$

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