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Ask iit jee aieee pet cbse icse state board experts Expert Question: need sol of HC VERMA CHAPTER ROTATION Qno-37,63,50,69,60
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deepesh28 (12)

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waterdemon (5140)

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Here is the solution:
Assuming mass of pulley as "M"
mass of block given as "m = 1 kg"

First we will find the mass of pulley:

I = MR2/2
Given:
I = 0.2 kg-m2
R = 0.2m.

(0.2) = M(0.2)2/2
M = 0.4/0.04
M = 10 kg.

Now writing equations of motion for block:
T1 = ma .........(1)
m = 1
T1 = a ...........(1)

Now supposing that the block accelerates with "a" then:
From (b) FBD pulley will accelerate with "a/2".

Now writing equations of motion for "pulley":

Mg - T1 - T2 = M(a/2)

(10*9.8) - (T1+T2) = 5a

98 - 
(T1+T2) = 5a ............(2)

Now for linear motion of the block the pulley is going down
and hence the pulley is rotating:
 = I@
T2R - T1R = I@
T2(0.2) - T1(0.2) = (0.2)@
T2 - T1 = @

We know that @ = a/R
Therefore we get,
@ = (a/2)0.2
as we know pulley is accelerating down with "a/2"
@ = 5a/2

T2 - T1 = 5a/2...........(3)

Now from eq.(1) w know T1 = a.
So we have,
T2 - a = 5a/2
T2 = 7a/2

Substituting the values of "T2" and "T1" in equation (2) :
98 - (T1+T2) = 5a
5a + 7a/2 + a = 98
6a + 7a/2 = 98
19a/2 = 98
19a = 196
a = 196/19
 10 m/s2.

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<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>







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budku007 (396)

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50.Torque of the given force about centre of the rod=Fr=5*(.25)Nm
     we know Torque=I@
       I is moment of inertia of system through centre of rod perpendicular to plane
      I=Mr2+Mr2
       
I=2Mr2
       
M is mass of each ball and r is distance between the from center
      I=2*.5*(.25)2
       
On using the two equations we get @=20rad/s2
        
so use =0+@t
        t=.1
        we get =12rad/s
       Answer
        Rate me if satisfied

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Anyone who has never made a mistake has never tried anything new
You can never solve a problem on the level on which it was created.
When you are courting a nice girl an hour seems like a second. When you sit on a red-hot cinder a second seems like an hour. That's relativity

Purpose of solving a problem is not simply to get the answer(the answer is only an evidence) but to develop your thinking ability
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budku007 (396)

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60
Apply Law of conservation of angular momentum about centre
we get
mvL/2=ML2,/12,as particle stops after collision
we get =6mv/ML             M is mass of rod,m is mass of particle,v is velocity of particle
as angular velocity is constant
apply theta=t
theta=/2
we get t=ML/12mv
now apply law of conservation of linear momentum
we get
mv=MV    V is velocity of centre of rod
V=mv/M
use s=Vt
we get s=mv/M*ML/12mv
we get s=L/12
answer
Rate me if satisfied
Nudge me for dought  

<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>





Anyone who has never made a mistake has never tried anything new
You can never solve a problem on the level on which it was created.
When you are courting a nice girl an hour seems like a second. When you sit on a red-hot cinder a second seems like an hour. That's relativity

Purpose of solving a problem is not simply to get the answer(the answer is only an evidence) but to develop your thinking ability
 this reply: 5 points  (with Olaaa!! Perrrfect answer.   in 1 votes )   [?]
 
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deepesh28 (12)

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wt about other qestions mainly 68,6763
 
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anandghegde (1707)

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why dont you search for it in the forum first man??Mad

"I a universe of atoms.......an atom in the universe"
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