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 a 10 m/s2.
Hope u find it useful. Rate me if useful.
Cheers!!!!!!!!!!@@@@!!!!!!!!! :)
Always available for help !
But Remember Don't hesitate to ask a good Question but
Be damn serious for Questioning a weak one.
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
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
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
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
Moderation Team
![[Post New]](/templates/default/images/icon_minipost_new.gif){kind=icon}
![[Up]](/templates/default/images/icon_up.gif){kind=icon}
2
= I@
10 m/s2.
![[Thumb - FBD.jpg]](http://www.goiit.com/upload/2007/12/14/958998777ac0084dfb087808d39b8e91_20170.jpg_thumb)
=
/2
