Mechanics> Using 2 light cords a bob is held in equi...

Using 2 light cords a bob is held in equilibrium at A. The horizontal cord is cut and the bob is allowed to swing as a pendulum. Calculate the ratio of the tension of the supporting cord in position B to that in position A.

Raisinten , 10 Years ago

Grade 11

7 Answers

Neeti

Finally.

See in the first case when the cord has no been cut, and the bob is in equilibrium, therefore, T_acos $\Theta$ = mg (equilibium in vertical direction ) and T_a sin $\Theta$ = F ( equilibirum in horizontal direction) .

Squaring and adding the two equations we get T_a = (F² + (mg)²)^1/2

Now when the string is cut and the bob starts oscillating, at the extreme positing B ( assuming it to be exteme since $\Theta$ same for positions A and B and and oscillation starts at A so it is considered to be one extreme position so B is the other extreme)

For position B, there is no force is radial direction since body is momentarily at rest, So only centripetal = mv²/l = T_b -mgcos $\Theta$

So T_b = mgcos $\Theta$ + mv²/l

But i’m not sure if this is the correct but this is the only answer i could get.

But try solving it, try to simplify it a little. That might help.

Last Activity: 10 Years ago

Raisinten

I liked your working but the answer is (cos@)^2

Last Activity: 10 Years ago

Neeti

I’m sure you’ll somehow get it from the equations i derived. Because there’s no other way to solve it i guess. sorry i couldn’t get the answer, it just needs to be simplied.

Last Activity: 10 Years ago

Raisinten

Do you know anyone who might be able to solve

Last Activity: 10 Years ago

Neeti

Well you can wait, some other askiitian tutor might.

But what about your physics teacher? In school or coaching?

Last Activity: 10 Years ago

Raisinten

I have no tutor and my physics teacher won’t answer it as they haven’t started the chapter.

Last Activity: 10 Years ago

Neeti

Well that is a weird physics teacher. Do one thing, post the question again, you may have to wait for some time , you’ll hopefully get another answer from another tutor :) I’ll try this question again, i just hope i’m able to solve it and i’ll get back to you ASAP :)

Last Activity: 10 Years ago