Qstnid -Iv-19-a Motorcycle Has To Move With A Constant Speed On An Over Bridge Which Is In The Form Of A Circular Arc Of Radius R And Has A Total Length L. Suppose The Motorcycle Starts From The Highest Point. (A) What Can Its Maximum Velocity Be For Which The Contact With The Road Is Not Broken At The Highest Point? (A) What Can Its Maximum Velocity Be For Which The Contact With The Road Is Not Broken At The Highest Point? - 07: Circular Motion

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NEET-XII-Physics

07: Circular Motion

with Solutions - page 5

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#19
A motorcycle has to move with a constant speed on an over bridge which is in the form of a circular arc of radius R and has a total length L. Suppose the motorcycle starts from the highest point. (a) What can its maximum velocity be for which the contact with the road is not broken at the highest point? (a) What can its maximum velocity be for which the contact with the road is not broken at the highest point?
Ans : R = Radius of the bridge
L = Total length of the over bridge (a) At the highest point:
Let m be the mass of the motorcycle and v be the required velocity.

`` mg=\frac{m{v}^{2}}{R}``
`` \Rightarrow {v}^{2}=Rg``
`` \Rightarrow v=\sqrt{Rrg}``
`` ``
`` \left(\,\mathrm{\,b \,}\right)\,\mathrm{\,Given \,}:``
`` v=\left(\frac{1}{\sqrt{2}}\right)\sqrt{Rg}``

Suppose it loses contact at B.
`` \,\mathrm{\,At \,}\,\mathrm{\,point \,}\,\mathrm{\,B \,},\,\mathrm{\,we \,}\,\mathrm{\,get \,}:``
`` mg\,\mathrm{\,cos \,}\theta =\frac{m{v}^{2}}{R}``
`` \Rightarrow {v}^{2}=Rg\,\mathrm{\,cos \,}\theta ``
`` \,\mathrm{\,Putting \,}\,\mathrm{\,the \,}\,\mathrm{\,value \,}\,\mathrm{\,of \,}v,\,\mathrm{\,we \,}\,\mathrm{\,get \,}:``
`` \sqrt{{\left(\frac{Rg}{2}\right)}^{2}}=Rg\,\mathrm{\,cos \,}\theta ``
`` \Rightarrow \frac{Rg}{2}=Rg\,\mathrm{\,cos \,}\theta ``
`` \Rightarrow \,\mathrm{\,cos \,}\theta =\frac{1}{2}``
`` \Rightarrow \,\mathrm{\,\theta \,}=60°=\frac{\,\mathrm{\,\pi \,}}{3}``
`` \because \theta =\frac{L}{R}``
`` \therefore L=R\theta =\frac{\pi R}{3}``
So, it will lose contact at a distance `` \frac{\pi R}{3}`` from the highest point.
(c) Let the uniform speed on the bridge be v. The chances of losing contact is maximum at the end bridge. We have:

`` \alpha =\frac{L}{2R}``
`` \,\mathrm{\,So \,},\frac{m{v}^{2}}{R}=mg\,\mathrm{\,cos \,}\alpha ``
`` \Rightarrow v=\sqrt{g\,\mathrm{\,Rcos \,}\left(\frac{L}{2R}\right)}``
Page No 115: (a) At the highest point:
Let m be the mass of the motorcycle and v be the required velocity.

`` mg=\frac{m{v}^{2}}{R}``
`` \Rightarrow {v}^{2}=Rg``
`` \Rightarrow v=\sqrt{Rrg}``
`` ``
`` \left(\,\mathrm{\,b \,}\right)\,\mathrm{\,Given \,}:``
`` v=\left(\frac{1}{\sqrt{2}}\right)\sqrt{Rg}``

Suppose it loses contact at B.
`` \,\mathrm{\,At \,}\,\mathrm{\,point \,}\,\mathrm{\,B \,},\,\mathrm{\,we \,}\,\mathrm{\,get \,}:``
`` mg\,\mathrm{\,cos \,}\theta =\frac{m{v}^{2}}{R}``
`` \Rightarrow {v}^{2}=Rg\,\mathrm{\,cos \,}\theta ``
`` \,\mathrm{\,Putting \,}\,\mathrm{\,the \,}\,\mathrm{\,value \,}\,\mathrm{\,of \,}v,\,\mathrm{\,we \,}\,\mathrm{\,get \,}:``
`` \sqrt{{\left(\frac{Rg}{2}\right)}^{2}}=Rg\,\mathrm{\,cos \,}\theta ``
`` \Rightarrow \frac{Rg}{2}=Rg\,\mathrm{\,cos \,}\theta ``
`` \Rightarrow \,\mathrm{\,cos \,}\theta =\frac{1}{2}``
`` \Rightarrow \,\mathrm{\,\theta \,}=60°=\frac{\,\mathrm{\,\pi \,}}{3}``
`` \because \theta =\frac{L}{R}``
`` \therefore L=R\theta =\frac{\pi R}{3}``
So, it will lose contact at a distance `` \frac{\pi R}{3}`` from the highest point.
(c) Let the uniform speed on the bridge be v. The chances of losing contact is maximum at the end bridge. We have:

`` \alpha =\frac{L}{2R}``
`` \,\mathrm{\,So \,},\frac{m{v}^{2}}{R}=mg\,\mathrm{\,cos \,}\alpha ``
`` \Rightarrow v=\sqrt{g\,\mathrm{\,Rcos \,}\left(\frac{L}{2R}\right)}``
Page No 115:

#19-a
What can its maximum velocity be for which the contact with the road is not broken at the highest point?
Ans : At the highest point:
Let m be the mass of the motorcycle and v be the required velocity.

`` mg=\frac{m{v}^{2}}{R}``
`` \Rightarrow {v}^{2}=Rg``
`` \Rightarrow v=\sqrt{Rrg}``
`` ``
`` \left(\,\mathrm{\,b \,}\right)\,\mathrm{\,Given \,}:``
`` v=\left(\frac{1}{\sqrt{2}}\right)\sqrt{Rg}``

Suppose it loses contact at B.
`` \,\mathrm{\,At \,}\,\mathrm{\,point \,}\,\mathrm{\,B \,},\,\mathrm{\,we \,}\,\mathrm{\,get \,}:``
`` mg\,\mathrm{\,cos \,}\theta =\frac{m{v}^{2}}{R}``
`` \Rightarrow {v}^{2}=Rg\,\mathrm{\,cos \,}\theta ``
`` \,\mathrm{\,Putting \,}\,\mathrm{\,the \,}\,\mathrm{\,value \,}\,\mathrm{\,of \,}v,\,\mathrm{\,we \,}\,\mathrm{\,get \,}:``
`` \sqrt{{\left(\frac{Rg}{2}\right)}^{2}}=Rg\,\mathrm{\,cos \,}\theta ``
`` \Rightarrow \frac{Rg}{2}=Rg\,\mathrm{\,cos \,}\theta ``
`` \Rightarrow \,\mathrm{\,cos \,}\theta =\frac{1}{2}``
`` \Rightarrow \,\mathrm{\,\theta \,}=60°=\frac{\,\mathrm{\,\pi \,}}{3}``
`` \because \theta =\frac{L}{R}``
`` \therefore L=R\theta =\frac{\pi R}{3}``
So, it will lose contact at a distance `` \frac{\pi R}{3}`` from the highest point.
(c) Let the uniform speed on the bridge be v. The chances of losing contact is maximum at the end bridge. We have:

`` \alpha =\frac{L}{2R}``
`` \,\mathrm{\,So \,},\frac{m{v}^{2}}{R}=mg\,\mathrm{\,cos \,}\alpha ``
`` \Rightarrow v=\sqrt{g\,\mathrm{\,Rcos \,}\left(\frac{L}{2R}\right)}``
Page No 115: