Grade Select GradeMechanics

A particle is hanging from a fixed point O by means of a string of length a. There is a small nail Q in the same horizontal line with O at a distance of a/3 from O. Find the minimum velocity with which the particle should be projected so that it may make a complete revolution around the nail without being slackened.

Aman Mishra

8 Years agoGrade Select Grade

2 Answers

Sumeet Gupta

8 Years ago

Hi,

Here we have to find the velocity of particle at the lowest position required to complete the constrained motion arround the nail

Firstly we know that the velocity required to complete 90° is sqrt(2ga)

Total kinetic energy decrease to complete 90 of the circle of radius a K1 = 1/2mv²

now the radius of particle is decreased to 2a/3

so , we know that the velocity required to complete a circle with radius R is /{sqrt{5gr}}

so here R =2a/3 v at

Lets take it as a completely rotational motion about point O and Q.

by energy conservation(w_b representa angular velocity at the bottom (ZPL) and w_T represents the velocity at the top)
(1/2)I₁w_b²= mg(a+2a/3) + (1/2)I₂w_T^{2 ...(1)}

Simplify it to get

(1/2)(ma²)w_b²= mg(a+2a/3) + (1/2)(m(2a/3)²⁾w_T²

put values I₁= ma² ; I₂=m(2a/3)² .

here we need to find w_b , so we will eleminate w_T:

at the top the Tension just tends to 0

Therefore, mw_T²(2a/3) = mg

Now, substitute the value of mw_T(2a/3) in equation 1 :

(1/2)(ma²)w_b²= mg(5a/3)+(1/2)(mg(2a/3))

Solve this to get :

$Wb=2{\sqrt{g/a}}$

So the velocity at bottom will be

$Vb = 2{\sqrt{ga}}$

Please Let me know whether its correct or not..........

$Hi, Here we have to find the velocity of particle at the lowest position required to complete the constrained motion arround the nail Firstly we know that the velocity required to complete 90° is {\sqrt{2ga}}} and total kinetic energy is 1/2mv^{2}$