When a body slides along an inclined plane of height 'h' and describes a vertical circle of radius r on reaching the bottom
Ans:h =5R/2
PE starting point= PE at top + Ke at top
mgh=mg(2r)+1/2 m v^2
but mg=mv^2/r or v^2=rg
mgh=2r*mg+ 1/2 m rg
h= 5r/2
Thank you
Please can you explain the derivation of the formula used to solve the problem?
hi how are you i am fine
To answer this question, we can use the concept of conservation of energy and gravitational potential energy.
When the body is at the top of the inclined plane, it possesses gravitational potential energy given by mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the inclined plane.
As the body slides down the inclined plane, it gains kinetic energy equal to the loss in potential energy. At the bottom of the inclined plane, the body will have converted all its potential energy into kinetic energy.
Now, when the body reaches the bottom and begins to describe a vertical circle of radius r, it will have centripetal force acting on it to keep it in circular motion. The centripetal force is provided by the tension or normal force acting on the body.
The minimum force required to keep an object moving in a vertical circle is the weight of the object. Therefore, at the bottom of the inclined plane, the tension or normal force acting on the body is equal to its weight, which is mg.
The centripetal force required is provided by the vertical component of tension or normal force. Therefore, we can equate the weight of the object (mg) to the vertical component of tension or normal force (mg cosθ), where θ is the angle between the inclined plane and the horizontal.
Simplifying the equation, we get:
mg = mg cosθ
Canceling out mass and g from both sides, we get:
1 = cosθ
As the body is describing a vertical circle, the angle θ is the angle between the vertical line and the radius of the circle. Therefore, the angle θ is equal to 90 degrees.
Substituting θ = 90 degrees into the equation, we get:
1 = cos(90)
Taking the cos inverse of both sides, we get:
θ = cos^(-1)(1)
θ = 0 degrees
Since the angle θ is 0 degrees, the inclined plane is horizontal. In this case, the height of the inclined plane (h) will be equal to 0.
Therefore, the answer to the question is h = 0.