Question

Review. A bead slides without friction around a loop-the loop (Fig. P8.5). The bead is released from rest at a height h=3.50R (a) What is the speed at point (A)?

Answer 1

The speed of the beed at point A is $\sqrt{3gR}$

We are familiar with the energy conservation law, which states that energy can't be destroyed or created, In simple terms relating to this system we can say that the sum of the kinetic and potential energies are the same at all points along the path.

So the equation for the conservation law we can refer to is,

(1/2)mv₁ + mgh₁ = (1/2)mv₂² + mgh₂,

The left-hand side refers to the energy of the bead before sliding and the right-hand side refers to the energy sum after the bead slides without friction.

Since, the initial point of the bead was at rest so,

v₁ will be 0  and h₁ will be 3.5R

Similarly, v₂( the velocity at point A which is unknown )

h₂=2R

Now, substituting the required value in the equation we get,

(1/2)mv₁ + mgh₁ = (1/2)mv₂² + mgh₂

⇒ 0 + mg(3.5) = (1/2)v₂² + mg(2R)

⇒ mg(3.5) - mg(2R) = (1/2)v₂²

⇒mg(1.5R)= (1/2)v₂²

⇒V₂ =√3gR

To know more about conservation law refer to the link brainly.com/question/20971995?referrer=searchResults

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