circular path

A uniform disc of mass m and radius r is rolling on a circle by virtue of torque Ï„ given to rod(mass M length L) pivoted to disc.Find angular accelrations of rod and disc.

47 Answers

24
eureka123 ·

actually you are very close.....
here are the answers::

Ï„/L2(2M+9m) and Ï„/Lr(2M+9m)

33
Abhishek Priyam ·

no i am far far away from the answer [2]

sorry i can't help u here [2]

24
eureka123 ·

Ok..............But can u tell me how u got ang acc of disc= 3Ï„
Lr(M+3m)

33
Abhishek Priyam ·

Leave it..

It is wrong...
:(

and u will get confused bekar me..

its not worth solving..
[3](dur ke angur khatte hai isliye)

24
eureka123 ·

Okzzzzzzzzzz maybe nishant sir can help me........

24
eureka123 ·

PLZZZZ HELP meeeeeee

24
eureka123 ·

nops....

62
Lokesh Verma ·

There are 2 rotations involved..
in is along the axis of the rod.
the other is by the center..

(Does this hint work enuf for finding the moment of inertia?)

24
eureka123 ·

I have been scratching my head for over a week.............plzzzz send the solution........I am getting mad.......

24
eureka123 ·

I have been scratching my head for over a week.............plzzzz send the solution........I am getting mad.......

1
g4utanmoy ·

I know the ans is diffcult but iwill find it out Mr. u rekal.
give me some time

62
Lokesh Verma ·

inertia along the axis of the rod is mr2/2

inertia along the center will be ML2/3 + mr2/4+mL2

Just check that this is correct!

1
Philip Calvert ·

this is correct but how to get the ans. bhaiyya
i wrote some eqns but could not get the ans eureka has given

can you write the eqns. so that i correct mine

62
Lokesh Verma ·

I will try to give some more hints... (I din solve it myslef... but see if this helps!)

Take friction ... call it F

and then find the total torque due to friction and applied torque... at the central rod...

THen use torque = I alpha

again there is torque along the central axis of the rod.. and there is rotation or angular acceleration ... again due to friction alone

also there is a relation between these two torque....

Use the moment of inertias that i put above... i guess that should help :)

1
Philip Calvert ·

wil the net Ï„ about the central axis be equal to (tau) - FL or will it be something else?

1
Philip Calvert ·

ok here are my equations please point out the mistake in them

(tau) - FL= Iα (1)

(tau)/L - F = ma where a =αL (2)

Fr =(mr2/2)a/r (3)

but this does not give the ans
[2] what's wrong
maybe 2nd eqn is doubtful

62
Lokesh Verma ·

philip we are dealing with 2 different angular accelerations..

i think in the second equation u have used the same angular acceleration as the first ....

Your Answer

Close [X]