Question

can you please answer the 23 rd question ?

Answer 1

Ans is 2 sqrt gR cos theta
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Answer 2

11th/Physics

Kinematics

Answer :

Initial velocity of bed = zero

Radius of circle = R

We have to find final velocity of bed when it reaches to point $.$

⧪ Distance covered by bead (AB) = 2Rcosθ

Horizontal component of gravitational acceleration = gcosθ

Now, let's apply third equation of kinematics!

• v² - u² = 2as

v denotes final velocity

u denotes initial velocity

a denotes acceleration

s denotes distance

ATQ, initial velocity (u) = zero

➙ v² - 0² = 2 (gcosθ)(2Rcosθ)

➙ v² = 4gRcos²θ

➙ v = √(4gRcos²θ)

➙ v = 2√(gR) cosθ

Cheers!