Question

A uniform rod of mass M = 250.0 g and length L = 50.0 cm stands vertically on a horizontal table. It is released from rest to fall. Calculate the angular speed of the rod as it makes an angle θ = 45° with respect to the vertical.

Answer 1

we have concentrated the mass of the rod on the centre of mass of the rod for finding the work done by gravity
if we break the position vectors , we get that the centre of mass(com) has a decrease in the height of \[25\div \sqrt{2}\]cm
so by the work energy theorem,(for rotation, as there is no tranalational motion i.e. there is no straight line or any other kind of motion other than rotaional motion)
mgh(work done by gravity)=(I (omegaf)^2)/2 - (I(omegai)^2)/2
here, m is the mass of the rod and g is the accn. due to gravity and h is the decrease in the height of the com and I is the moment of inertia of the rod about the contace point (mr^2 /3) (our ref. pointis the contace point)and omegai and omegaf are initial and final angular velocities of the rod and the initial omega=0 as it is released so mgh=(I(omegaf)^2)/2
so by putting the values given in the question into the equation, we get,
omegaf=7.445 rad/s
and remember to convert the given values into s.i. units
hope this helps you

Answer 2

the answer is 4.152 rad/s. what i dont understand is that dont we count normal kinetic energy as well as rotational kinetic energy?

Answer 3

the normal kinetic energy is counted if there is any kind of translational motion i.e. the body shoud be moving and not only be rotating but here , the body is just rotating so we take onl the rotational kinetic energy in account

Answer 4

yes man you are right the answer is 4.152 rad / sec
i made a calculation mistake
sorry

Answer 5

sir, i'll upload a picture of my calculations from what you told me. can you check where i do wrong?

Answer 6

yeah sure and i'm not any kind of sir (sorry for that)
but i am in the 11th class

Answer 7

sir as i respect your knowledge of physics :) this is what i did

Answer 8

okay you are up to the mark
but there's something wrong
like you should take the height as 0.25/squrt2 m and not 25/squrt2 cm
next i've specified in the first post that the initial angular velocity is 0 as it is released from rest
and rest of the things are CORRECT!

Answer 9

thanks for your help :) still i m making some kind of error and i get wrong results i'll have to work on this more :D

Answer 10

even i am getting some other answer so i'll work on it too and sorry that i've done something wrong

Answer 11

thanks for the medal

Answer 12

came to my mind that shouldn't be 0.25-(0.25/root2) since the initial position of com was 0.25 and final was 0.25/root2 and we need to find delta(h) dont we?

Answer 13

YES i found it YES

Answer 14

i got it now even i was thinking about that and it came into my mind and you posted it right now
and the height should be 0.25-(0.25/root2)
thanks for making it right:)

Answer 15

Thanks for YOU you showed the method answer is just a number. Impotant things is to understand the path that leads to answer. I got one more question about torque that i'll post now. I would be very glad if you could take a look at that as well! THANKS

Answer 16

i'll surely help you and your thinking is just so positively awesome