A small block is freely sliding down from top of a smooth inclined plane the block reaches bottom of inclined plane then the block describes vertical circle of radius 0.5m along smooth track the minimum vertical height of inclined plane should be
Ans:1.25m
To find the minimum vertical height of the inclined plane, we need to consider the conservation of energy.
Let's assume the initial position of the block on the inclined plane is at height 'h' above the bottom. The gravitational potential energy at this position will be mgh, where m is the mass of the block and g is the acceleration due to gravity.
As the block slides down the inclined plane, it will gain kinetic energy equal to (1/2)mv^2, where v is the speed of the block at the bottom of the inclined plane.
At the bottom of the inclined plane, the block should have enough kinetic energy to complete the vertical circle of radius 0.5m along the smooth track. This means that the total kinetic energy at the bottom should be equal to the potential energy at the initial position.
So, we have the equation:
mgh = (1/2)mv^2
The mass cancels out, leaving us with:
gh = (1/2)v^2
We can express the speed of the block at the bottom in terms of the radius of the vertical circle:
v = √(gr)
Substituting this into the equation above, we get:
gh = (1/2)(gr)^2
Simplifying the equation:
2h = r^2
Given that the radius of the vertical circle is 0.5m, we can substitute this value into the equation:
2h = (0.5)^2
Solving for h:
2h = 0.25
h = 0.25/2
h = 0.125m
Therefore, the minimum vertical height of the inclined plane should be 0.125m or 1.25m.
radius = R = .5 m
at top, m g = m v^2/R for contact force = 0
sso
v^2/R = g
so v = sqrt (g R) at top of circle which is height 2 R
potential energy at top of plane = m g h
potential energy at 2 R = 2 m g R
loss of potential =
m g h - 2 m g R = mg (h-2R)
= Ke at top of circle = (1/2) m v^2
so
(1/2) m v^2 = m g (h-2R)
but
we know v^2 = g R
so
(1/2) m g R = m g h - 2 m g R
2.5 m g R = m g h
h = 2.5 R = 2.5 (.5) = 1.25 meters