This is simple problem. The difficulty is conceptualization.
First, note that the upper rod will rotate around the hinge, and the lower rod will initially rotate around the center of mass (this is because only torque will be supplied to it -- no motion either up or down). Therefore, knowing the mutual endpoint, the lower rod will rotate at twice the rate of the upper.
α.u = 1/2 α.b ("u" is for upper and "b" is for the lower or bottom rod).
I.u = 1/3 m L² ... (it rotates about hinge)
I.b = 1/12 m L² ... (it rotates about com)
Now, we need to set the torques around the upper rod equal to zero.
I.u * α.u = τ - τ.b
τ = the torque supplied by the force, i.e., = F x d F = 45 N d = 1/2 m
τ.b = the reactionary torque from the lower rod. (Newton's 3rd law)
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ReplyDeleteI am getting 45/7
ReplyDeleteyes 45/7 assuming 45N acts on the mid point of the upper rod
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDelete@sambhav:i am not able to solve.....pls tell hw u solved..
ReplyDeletei'll tell once the answer is confirmed
ReplyDeleteThis is simple problem. The difficulty is conceptualization.
ReplyDeleteFirst, note that the upper rod will rotate around the hinge, and the lower rod will initially rotate around the center of mass (this is because only torque will be supplied to it -- no motion either up or down). Therefore, knowing the mutual endpoint, the lower rod will rotate at twice the rate of the upper.
α.u = 1/2 α.b
("u" is for upper and "b" is for the lower or bottom rod).
I.u = 1/3 m L² ... (it rotates about hinge)
I.b = 1/12 m L² ... (it rotates about com)
Now, we need to set the torques around the upper rod equal to zero.
I.u * α.u = τ - τ.b
τ = the torque supplied by the force, i.e., = F x d
F = 45 N
d = 1/2 m
τ.b = the reactionary torque from the lower rod. (Newton's 3rd law)
I.u * α.u = τ - I.b * α.b
τ = I.u * α.u + I.b * α.b
τ = 1/3 m L² * 1/2 α.b + 1/12 m L² α.b
α.b = 4 * τ / ( m L² )
α.b = 4 * ( F x d ) / ( m L² )
α.b = 10 rads/s
α.b = 10 rads/s²
ReplyDeleteMade a mistake:
ReplyDeleteI.u * α.u = τ - τ.b
τ = the torque supplied by the force, i.e., = F x d
F = 45 N
d = 1/2 m
τ.b = the reactionary torque from the lower rod. (Newton's 3rd law) = F.b x L
L = 1 m = 2 d
I.u * α.u = τ - F.b x L
The torque on the lower bar from this force will be
&tau.b' = F.b x 1/2L = I.b * α.b
τ = I.u * α.u + 2 I.b * α.b
τ = 1/3 m L² * 1/2 α.b + 2 * 1/12 m L² α.b
α.b = 3 * τ / ( m L² )
α.b = 3 * ( F x d ) / ( m L² )
α.b = 7.5 rads/s²
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ReplyDeletei am getting 10.
ReplyDeleteSAMBHAV (or someone else who knows the reason), please reply to this
http://anujkalia.blogspot.com/2010/08/kings-of-s.html
thanks
The answer's 45/7.. (just looked it up from the textbook).
ReplyDeletePeople able to solve this will do a bit better than me in their Applied Mechanics courses.
nw someone pls gv d solution....
ReplyDeletenot able to solve this ,please give the solution.
ReplyDeletemake FBD of rods separately. Equate the accn of hinge from both the rods.
ReplyDeletestill not able to get it.....pls give the eqns of fbd....
ReplyDeletepratik
ReplyDeleteassume upper rod to be rotating with (alpha1)about hinge
lower rod with angular accl (alpha2)about COM and Acm
since both the rods are joined their accelerations at the point of joint will be same
alpha1 L = Acm + alpha2 L/2
On the upper rod there will be 3 forces
1 from the hinge at top
2 is 45 N
3 is the force from lower hinge (towards left)
On the lower rod there will be 1 force only (opp to force '3' on the upper block )
make FBD and solve,answer will be 45/7
delhite check
ReplyDeletehttp://anujkalia.blogspot.com/2010/08/kings-of-s.html
i have answered it