We have to find the height at which boy loses contact?
We have to find the height at which boy loses contact?
Suyash Kumar Gupta , 6 Years ago
Grade 12
Mechanics> We have to find the height at which boy l...
1 Answers
Arun
As the boy slides along the surface of the hemispherical mound, the component of gravity normal to the surface provides a centripetal force which keeps him moving in a circle. The further down he slides, the weaker this component becomes; at the same time his speed increases, which requires an ever-larger centripetal force to keep him moving in a circle of radius R. When the decreasing component of gravity is no longer strong enough to provide the increasing centripetal force required for his speed, the boy is on the point of losing contact with the hemisphere.
Let A be the angle between the vertical through the centre and top of the hemisphere and a radius from the centre to the boy's position at some time t. The height lost by the boy at this point is :
R - RcosA = R(1-cosA).
Since there is no work done against friction, the PE lost = the KE gained by the boy, ie.
(1/2)mv^2 = mgh = mgR(1-cosA)
mv^2/R = 2mg(1-cosA)
where v is the boy's speed at this point (or angle).
At this point the component of the boy's weight normal to the hemispherical surface is mgcosA. The centripetal force required to keep him moving in a circle of radius R with speed v is mv^2/R. When these two forces are equal (ie when the component of gravity is used up in providing centripetal force) we have -
mgcosA = mv^2/R = 2mg(1-cosA)
cosA = 2 - 2cosA
3cosA = 2
cosA = 2/3
So when the boy loses contact with the ice mound, his height above the centre of the hemisphere is RcosA
Let A be the angle between the vertical through the centre and top of the hemisphere and a radius from the centre to the boy's position at some time t. The height lost by the boy at this point is :
R - RcosA = R(1-cosA).
Since there is no work done against friction, the PE lost = the KE gained by the boy, ie.
(1/2)mv^2 = mgh = mgR(1-cosA)
mv^2/R = 2mg(1-cosA)
where v is the boy's speed at this point (or angle).
At this point the component of the boy's weight normal to the hemispherical surface is mgcosA. The centripetal force required to keep him moving in a circle of radius R with speed v is mv^2/R. When these two forces are equal (ie when the component of gravity is used up in providing centripetal force) we have -
mgcosA = mv^2/R = 2mg(1-cosA)
cosA = 2 - 2cosA
3cosA = 2
cosA = 2/3
So when the boy loses contact with the ice mound, his height above the centre of the hemisphere is RcosA
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