Question

This is in need of an hour. Please help with Rotational Mechanics!
Correct answers would be appreciated :) ​

Answer 1

Answer:

The tension in the string at lowest point is 3mg

Explanation:

Length of the pendulum =I

mass of bob =m

let tge velocity at the lowest position is v

The bob is released from the horizontal position then its potential energy will be converted to kinetic energy. at tge lowest position.

mgl=1/2mv square

v square =2gl

The centrifugal acceleration at the lowest point

m v square /l

m 2gl/l

The tension in the string = weight of the bob+ centre due to centrifugal force

=mg+2mg

=3mg.

Hope it helps u

Mark it as BRAINLIEAST please i request

Answer 2

Answer= ${tan}^{ - 1} ( \frac{tanθ}{2} )$

On applying energy conservation taking the bottom line as reference,

Potential Energy at top = (Potential Energy + Kinetic Energy) at given innsttant

mgl=mgl(1-cosθ) +$\frac{1}{2} {mv}^{2}$

mglcosθ=$( \frac{mv^2}{2} )$

so v = $\sqrt{2glcosθ}$

tangential acceleration of bob is $a_{T}$=gsinθ

Radial acceleration$a_{R} = \frac{ {v}^{2} }{1} = 2gcosθ$

acceleration vector angle with string is

$tan a = \frac{ a_{T} }{ a_{R}} = \frac{gsinθ}{2gcosθ}$

a= ${tan}^{ - 1} ( \frac{tanθ}{2} )$

hope it's helpful for you