1tryyyyyyy it conceptual one
A homogeneous rod AB of length L=1.8m and mass M is pivoted at the center o such that it can rotate freely in the vertical plane . The rod is initially in the horizontal position. A insect S of the same mass M falls vertically on the point C , Midway B/W points O&B with speed v. Immediately after falling, The insects moves towards the end B such that the rod rotates with constant angular velocity w,. The insect reaches the end B when the rod has turned an angle of 90°.
1.) The angular velocity in terms of v&L
2.) Speed V of the insect when it reaches end B.
3.) The final angular velocity when angle is 90°
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9 Answers
39well, this involves 3 steps
1] conserve angular momentum about the pooint o
2]conserve linear momentum for rod and insect in vertical direction
3]conserve kineteic energy of the rod and the insect
now u will get the answer
- rahul chaudhary how can we conserve angular momentum about O in this case as there will be external torque or impulse acting about O due to weight of insect.Upvote·0· Reply ·2018-03-06 19:07:43
3after turning also ang vel. is same .....[see question].............
MgL/2 = 1/2[ML^2/12 + ML^2/4]W^2 ................
find W again ..........
NOW EQUATE THE Ws GOT BY THE DIFF METHODS TO GET 'v' ....................
and b555555 u cant conserve liniar mom. along vertical because of piviot force actin .......
3speed v when it is falling is req. i hope found this ques. in IIT paper uve totally made it ur own way ...................
66@iitimcomin,
you cannot conserve energy as you have done in #8 for the reason that there will be some negative work done by friction. The basic idea is that the rate of change in angular momentum (of the system rod + insect about the pivot) brought about by the motion of the insect will be equal to the torque of the force of gravity acting on the insect, since the force of gravity on the rod passes through the pivot.
If x be the distance of the insect from the pivot along the rod, then we get (I have taken the length of the rod as 2L)
Mgx\cos\omega t=\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{ML^2}{3}+Mx^2\right)\omega
Here, \omega is the angular speed calculated in part i (#7). Therefore, we get
\dfrac{\mathrm{d}x}{\mathrm{d}t}=\dfrac{g}{2\omega}\cos\omega t
This gives the velocity of the insect w.r.t. the rod. Further, integrating this equation and using the initial condition that at t=0, x = L/2, we get the position of the insect as
x(t)=\dfrac{L}{2}+\dfrac{g}{2\omega^2}\sin\omega t
Since when the insect reached the other end, it has turned by 90°, we get \omega=\sqrt{g/L}
Eliminating t from the above two equations, we obtain
v_x=\dfrac{\mathrm{d}x}{\mathrm{d}t}=\omega\sqrt{\dfrac{g^2}{4\omega^4}-\left(x-\dfrac{L}{2}\right)^2}
which gives the velocity (with respect to the rod) of the insect as function of x. When the insect reaches the other end, we get its velocity w.r.t. the rod as
V=\omega \sqrt{\dfrac{g^2}{4\omega^4}-\dfrac{L^2}{4}}=0
@ramlal,
what do you mean when you say in part (iii) that "Find the final angular velocity when angle is 90°"? . Wasn't it already specified that the angular velocity was a constant.