IIT JEE , AIEEE Forum - Physics , Mechanics

raman_shadow

A uniform rod of mass m is pivoted on top of the cieling as shown.

It is fixed to a wall with a spring of spring const. k

if it is slightly pushed in the direction shown find its time period.

aankurverma

SORRY I SOLVE IT IN DIFFERENT MANNER N BIT CONFUSED N I DON'T WANT TO CONFUSE U 2 SO I HAVE EDITED MY METHOD

aditya_arora04

Hi,

Look,

Young's modulus of rod, Y = FL / A

L

So, F = (YA / L)

L

i.e the spring constant for the rod is k₁ = YA / L

Now, this rod and the spring are in series.

So, net spring constant, k_net = k₁k₂ / (k₁ + k₂ )

= kYA / L(k + YA/L )

So, the time period = 2(pie)( m / k_net )^1/2

Here, m = mass of rod

pinnaka

That is a simple Question
Here u need to consider the torques exerted by two different forces
SO when the Stick or teh rod makes an angle theta
Then torques due to two forces act on the rod
One is due to garvity and the other is due to the restoring force exerted by the spring and by the earth

NOw lets us consider the situation when the stick has made an angle theta with the vertical. Then we have the torque due to gravity to be
tau_g = -mgsin( theta)*l/2
the force due to garvity acts through the center of mass therfore the lever arm is l/2 sin(theta)
Now the torque due to the restoring force exerted by the spring is equal to -kx sin(90-theta) *l = tau_spring
the angle is deduced from geometry
The torques due to forces point in the same direction. therefore they can be added them up
tau_net = -mgsin(theta)*l/2- kxcos(theta)*l
where l is the lenght of the rod which should have been given in the question
for very small angles we have cos(theta) = 1
sin (theta) = theta
these results follows from talyor series
tau_net = -mg(theta)*l/2 - kx*l
now we have theta = x/l so x =theta*l
tau_net = - (mg*l/2-kl^2) theta
now we have the torque acting as a function of the angle rotated
therefore omega = sqrt{ ( mg*l/2+kl^2)/(1/3 ml^2)}= sqrt{3 mg/2l + 6k/2m}
the rod rotates about the pivot
T = 2 pie /omega
which gives u
T = 2 pie/omega
Now lets us test this result when u r spring constant is equal to zero
which means that no spring is present
then u have T to be a expected result
therfore the preiod
T = 2(pi) * sqrt{2ml/(6Kl+3mg}

----Hope that Helps

nitin62225

ans is 2(pi) [ 2ml/(6Kl+3mg)]^1/2

raman_shadow

thnx ppl

shivamk

Is the answer to the question T = 2pie( (m) (YA+KL)/YAK)^1/2. If yes plz reply . I will post my explaination along with it .

raman_shadow

yaar i dont know the answer.
but yes pls post ur explanation

shivamk

Okay here is my explaination . Youngs modulus of a wire is given by Y= F/A*l/L where L is the change in lenght of the spring.So if we consider the spring to be cut into 2 halves one having a spring constant k₁& other having spring constant kthen k_eq= k₁*k/k₁+k. Substituting value of k₁which is YA/L*K/YA/L+K = YAK/YA+KI. hence substituting this values iget my final answer.

shivamk

I HAVE TYPED THIS ANSWER IN A BIT OF HURRY . IF U WANT ANY EXPLAINATIONS PLZ ASK.

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