106not in the options..
options r now given in the ques
Ill be posting some of the questions of Aakash AITS which i cudnt manage to do
I have been messing up Physics very very badly, at Aakash also as well as in BT
Q1. A rod of mass m (hinged at one of its ends at O) has a particle of mass m attached at its other end. The rod is released in horizontal orientation to fall under gravity. Tension at the midpoint of the rod when it becomes vertical is _________
(a) 17mg/16 (b) 27mg/16 (c) 31mg/16 (d) 147mg/32
Q2. Paragraph
Consider the figure shown. A uniform rod PQ of mass m and length l strikes rigid frictionless support A elastically such that distance between A and the end Q of the rod is l/3. The rod is rotating with an angular velocity \omega (clockwise) while its centre of mass has zero velocity just before the collision
(i) Angular speed of end P of the rod wrt end Q, immediately after the impact is
(a) 0.25\omega (b) 0.5\omega (c) 0.75\omega (d) zero
(ii) Speed of the end Q of the rod, immediately after the impact is
(a) 0.25\omega l (b) 0.75\omega l (c) 1.25\omega l (d) zero
(iii) Angular impulse exerted on the rod by the support at A is
(a) m\omega l^2/6 (b) m\omega l^2/12 (c)m\omega l^2/24 (d) Zero
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UP 0 DOWN 0 0 12
12 Answers
106dunno post it (the test was today) (and they dont give sols)
3yep ... i got it ...... didnt see the stupid mass attached to the other end!!!
3@xyz simple bro .......
imagine the two parts are attached by a massless string!!!!!!! ... now the tension on this is only because of the lower part!!!!
let it have W at bottommost position .....
we get W =9g/l ... simple conc of engy!!!!
now ........
T =(m/2)(W^2)(l/2+l/4) + m(W^2)l + mg/2 + mg ...
= 147mg/32
3typo!!!
second paragraph ....
1st answe is 0.5W .... rest of it follows frm the same ....
method [takin AQ=l/3 cant read ur latex ]......
let an impluse J act upwards ....
so we have ..... J=mv .............(1)
also -J(l/6)= (ml^2/12)(W-W0) ...(2)
now at the point of impact v=0 ...
or v = Wl/6 !!!!!!!!!!
or J =mWl/6
solvin we get W =3/4W0
now vel. of Q ...
Wl/2 - v = Wl/2 - Wl/6 = W0l/4 !!!!!!
ANG IMP . = Jl/6 = mWl^2/36 ...... but W=3/4W0
so = mWl^2/48 !!!! .....
pls veryfy ans.... cac errors possible ....... by da way nice questions ...........