NEET-XII-Physics
07: Circular Motion
- #27-athe normal force by the wall on the block, (b) the frictional force by wall and (c) the tangential acceleration of the block. (d) Integrate the tangential acceleration
dvdt=vdvdsto obtain the speed of the block after one revolution. (b) the frictional force by wall and (c) the tangential acceleration of the block. (d) Integrate the tangential acceleration
dvdt=vdvdsto obtain the speed of the block after one revolution.Ans : Normal reaction by the wall on the block = N = `` \frac{m{v}^{2}}{R}`` (b) Force of frictional by the wall = `` \mu N=\frac{\mu m{v}^{2}}{R}`` (c) Let at be the tangential acceleration of the block.
From figure, we get:
`` -\frac{\mu m{v}^{2}}{R}=m{a}_{t}``
`` \Rightarrow {a}_{t}=-\frac{\mu {v}^{2}}{R}`` (d) `` \,\mathrm{\,On \,}\,\mathrm{\,using \,}a=\frac{dv}{dt}=v\frac{dv}{ds},\,\mathrm{\,we \,}\,\mathrm{\,get \,}:``
`` v\frac{dv}{ds}=\frac{\mu {v}^{2}}{R}``
`` \Rightarrow ds=-\frac{R}{\mu }\frac{dv}{v}``
`` \,\mathrm{\,Integrating \,}\,\mathrm{\,both \,}\,\mathrm{\,side \,},\,\mathrm{\,we \,}\,\mathrm{\,get \,}:``
`` s=-\frac{R}{\mu }\,\mathrm{\,In \,}v+c``
`` \,\mathrm{\,At \,},s=0,v={v}_{0}``
`` So,c=\frac{R}{\mu }\,\mathrm{\,In \,}{v}_{0}``
`` \Rightarrow s=-\frac{R}{\mu }\,\mathrm{\,In \,}\frac{v}{{v}_{0}}``
`` \Rightarrow \frac{v}{{v}_{0}}={e}^{\mathit{-}\frac{\mu s}{R}}``
`` \mathit{\Rightarrow }v={\,\mathrm{\,v \,}}_{0}{e}^{\mathit{-}\frac{\mathit{\mu s}}{R}}``
`` \,\mathrm{\,For \,}\,\mathrm{\,one \,}\,\mathrm{\,rotation \,},\,\mathrm{\,we \,}\,\mathrm{\,have \,}:``
`` s=2\pi r``
`` \therefore v={v}_{0}{e}^{-2\pi \mu }``
`` ``
`` ``
Page No 116: (b) Force of frictional by the wall = `` \mu N=\frac{\mu m{v}^{2}}{R}`` (c) Let at be the tangential acceleration of the block.
From figure, we get:
`` -\frac{\mu m{v}^{2}}{R}=m{a}_{t}``
`` \Rightarrow {a}_{t}=-\frac{\mu {v}^{2}}{R}`` (d) `` \,\mathrm{\,On \,}\,\mathrm{\,using \,}a=\frac{dv}{dt}=v\frac{dv}{ds},\,\mathrm{\,we \,}\,\mathrm{\,get \,}:``
`` v\frac{dv}{ds}=\frac{\mu {v}^{2}}{R}``
`` \Rightarrow ds=-\frac{R}{\mu }\frac{dv}{v}``
`` \,\mathrm{\,Integrating \,}\,\mathrm{\,both \,}\,\mathrm{\,side \,},\,\mathrm{\,we \,}\,\mathrm{\,get \,}:``
`` s=-\frac{R}{\mu }\,\mathrm{\,In \,}v+c``
`` \,\mathrm{\,At \,},s=0,v={v}_{0}``
`` So,c=\frac{R}{\mu }\,\mathrm{\,In \,}{v}_{0}``
`` \Rightarrow s=-\frac{R}{\mu }\,\mathrm{\,In \,}\frac{v}{{v}_{0}}``
`` \Rightarrow \frac{v}{{v}_{0}}={e}^{\mathit{-}\frac{\mu s}{R}}``
`` \mathit{\Rightarrow }v={\,\mathrm{\,v \,}}_{0}{e}^{\mathit{-}\frac{\mathit{\mu s}}{R}}``
`` \,\mathrm{\,For \,}\,\mathrm{\,one \,}\,\mathrm{\,rotation \,},\,\mathrm{\,we \,}\,\mathrm{\,have \,}:``
`` s=2\pi r``
`` \therefore v={v}_{0}{e}^{-2\pi \mu }``
`` ``
`` ``
Page No 116:
- #27-bthe frictional force by wall andAns : Force of frictional by the wall = `` \mu N=\frac{\mu m{v}^{2}}{R}``
- #27-cthe tangential acceleration of the block.Ans : Let at be the tangential acceleration of the block.
From figure, we get:
`` -\frac{\mu m{v}^{2}}{R}=m{a}_{t}``
`` \Rightarrow {a}_{t}=-\frac{\mu {v}^{2}}{R}``
- #27-dIntegrate the tangential acceleration
dvdt=vdvdsto obtain the speed of the block after one revolution.Ans : `` \,\mathrm{\,On \,}\,\mathrm{\,using \,}a=\frac{dv}{dt}=v\frac{dv}{ds},\,\mathrm{\,we \,}\,\mathrm{\,get \,}:``
`` v\frac{dv}{ds}=\frac{\mu {v}^{2}}{R}``
`` \Rightarrow ds=-\frac{R}{\mu }\frac{dv}{v}``
`` \,\mathrm{\,Integrating \,}\,\mathrm{\,both \,}\,\mathrm{\,side \,},\,\mathrm{\,we \,}\,\mathrm{\,get \,}:``
`` s=-\frac{R}{\mu }\,\mathrm{\,In \,}v+c``
`` \,\mathrm{\,At \,},s=0,v={v}_{0}``
`` So,c=\frac{R}{\mu }\,\mathrm{\,In \,}{v}_{0}``
`` \Rightarrow s=-\frac{R}{\mu }\,\mathrm{\,In \,}\frac{v}{{v}_{0}}``
`` \Rightarrow \frac{v}{{v}_{0}}={e}^{\mathit{-}\frac{\mu s}{R}}``
`` \mathit{\Rightarrow }v={\,\mathrm{\,v \,}}_{0}{e}^{\mathit{-}\frac{\mathit{\mu s}}{R}}``
`` \,\mathrm{\,For \,}\,\mathrm{\,one \,}\,\mathrm{\,rotation \,},\,\mathrm{\,we \,}\,\mathrm{\,have \,}:``
`` s=2\pi r``
`` \therefore v={v}_{0}{e}^{-2\pi \mu }``
`` ``
`` ``
Page No 116: