Respuesta :
Answer:
[tex]H = \frac{5}{2}R[/tex]
Explanation:
As we know that if the block will complete the circular motion of the path then the speed at the bottom most part of the path must be equal to
[tex]v = \sqrt{5Rg}[/tex]
now we know that
velocity at the bottom of the path is due to conversion of potential energy to kinetic energy
so we can say it is given as
[tex]U = KE[/tex]
[tex]mgH = \frac{1}{2}mv^2[/tex]
now we have
[tex]mgH = \frac{1}{2}m(5Rg)[/tex]
[tex]H = \frac{5}{2}R[/tex]
The minimum height that the block must start to make it around the loop without falling off is;
h = 5R/2
We are told that the block slides around the inside of a circular loop. This means we are dealing with critical velocity in vertical circular motion.
In critical velocity in vertical circular motion, the formula for the minimum velocity at the lowest Point of the vertical circle required to get the critical velocity at the highest point is given as;
v = √(5gR)
Now, from conservation of energy we know that;
K.E = P.E
Thus;
½mv² = mgh
Where;
h is the minimum height that the block must start to make it around the loop without falling off.
g is acceleration due to gravity
Thus;
m will cancel out to give;
½v² = gh
Plugging in the relevant values gives;
½ × [√(5gR)]² = gh
Simplify to get;
5gR = 2gh
g will cancel out to get;
5R = 2h
h = 5R/2
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