Question

A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass 'm' and radius 'R' . Assuming pulley to be a perfect circular disc, the acceleration of mass if the string doesn't slip on the pulley is
a)g/3
b)3g/2
c)g
d)2g/3

Answer 1

K just wait

Answer 2

Should i upload the direct answer or explain u step by step ?

Answer 3

isn't this of AIEEE 2011 ?

Answer 4

explain!

Answer 5

\[\large{\frac{mg}{m+\frac{1}{R^2}}=\frac{mg}{m+\frac{m}{2}}=\frac{2g}{3}}\]

Answer 6

d) 2g/3 is the correct answer

Answer 7

Ok. How do you get it? I didn't understand!

Answer 8

\[\frac{I}{R^2}\]

Answer 9

I dont want to cheat any1 ; http://brilliantiitjee.com/docs/aieee-solution-2011.pdf

if it does not help then u can ask me

Answer 10

http://www.narayanagroup.com/pdf/AIEEE_PAPERCODE_2011.pdf

Answer 11

I still don't get it. Can you explain a bit?

Answer 12

Do you know that mg-T=ma ?

Answer 13

Ok, tell me what is \[\frac{1}{2}mR^2 \alpha\]??
To be precise, tell me what is \(\alpha\) ??

Answer 14

Awesome question that is what i am thinking about wait

Answer 15

Ok!

Answer 16

alpha is angular acceleration

Answer 17

\[\large{\frac{1}{2}mR^2\alpha}\]is related to momentum of inertia

Answer 18

got it now ?

Answer 19

I had read moment of inertia a year back. An i didn't enjoy it much! so, I have forgotten almost everything related to moment of inertia. :(

Answer 20

no problem search it on internet u will get everything there

Answer 21

Thanks!

Answer 22

Is \[I=\frac{1}{2}mR^2\]??
Is it the moment of inertia of disc?

Answer 23

yes

Answer 24

How do they get it? And isn't moment of inertia = mR^2 ?? Where does (1/2) come from?

Answer 25

\[I=\int\limits r^2 dm\]
for a thin disc
\[I= \int\limits_{0}^{R} \int\limits_{0}^{2\pi} r^2 \rho \ r \ dr \ d \theta= \frac{R^4}{4} 2 \pi \rho= \frac{1}{2} M R^2 \\ since \\ M=\rho 2 \pi R^2 \]

Answer 26

sorry
M=rho pi R^2 *******

Answer 27

\(\huge \tau\large=I\:\alpha\)
is the equivalent of

\(\Large F\large=m\:a\)

for rotating bodies.

I think you have to memorise the moment of inertia of a solid disc/cylinder rotating about its axis:

\(\Large I=\frac12\large mR^2\)

Answer 28

so the T in mg - T = ma is torque and not tension?

Answer 29

Its tension! TR is the torque.

Answer 30

i thought that was tension*radius! i was wondering why we're taking that!

Answer 31

Tension \(\times\) radius= torque.

Answer 32

oh yes! T = r x F....right! now i get it. thanks!

Answer 33

:)