1. A solid cylinder of mass " M " and radius " R " and height " R " has a helical groove cut on it's surface with a helix
angle of tan^{-1}\frac{1}{2\pi}. It can freely rotate about a vertical
axis passing through the center as shown in figure. The moment of Inertia of cylinder is " KMR2 "
where K < 1.
A small particle of mass m = K' M is gently released at the top of the groove and it slides down the groove to the end.
Find angular velocity of cylinder when particle reaches the bottom of the groove.
Hint: This sum is very very easy in concepts.
49haan sahi hai... this is a hinged system.. thus my solution is wrong! :(
71@Subho Bhaiya..
You too have almost the same as I had did. Then please see the qwerty's post.
49yaar bahut late ho gaya hai toh baaki ke answers parhne ka kuch patience nahi bacha... so I ll "try to" post the theory and the solution that I feel is correct...sorry if it is a repitition of a previous correct solution!!
So, first of all the thing is the given path is in the form of a "screw" (ghar me toh dekha hi hoga__ see there is a marked similarity between ghar-wala screw and question wala path..)
Secondly, screw is a modification of an inclined plane (just usko ghuma diya gaya hai bachchho ko chondu banane k liye asal me hai wahi purani cheez..)
Now let us draw the new revised and simplified version of same problem...
yawnnnn...:( fir se paint kholna parhega! :'(
yaar bura mat manna..apna man chahi values le liya hoon...
Abhi dekhne ka baat ye hai ki isme important concepts kya hai:
Pura dono ko agar system maane to phir the linear momentum can be conserved in the horizontal direction...
and: ALWAYS energy+work+losses in a mechanical system is constant.
And aise kuch bhi diya huwa nahi hai to phir paine soch hi liya ki there is no friction anywhere nahi to sum ghaatak ban jayega mere liye! :P
Anyway, so, let at the time the particle reaches the base of the wedge, its velocity be v downwards with respect to the wedge (as shown by arrow) and wedge's velocity be v' (as shown by arrow).
Since initial momentum of system in horizontal direction is 0, the final momentum must also be 0.
So, m(vcosθ-v') - Mv' = 0
another equation is,
mgH = 12m(v2+v'2-2vv'cosθ) + 12Mv'2
from the above two equations you would need to find out v' (i.e. the velocity of the wedge in horizontal direction.)
Now, that the wedge is formed into a screw and even here there is no resistance, even here the wedge should be moving with a same velocity v'.
Thus, the angular velocity ω, should be given by, ω=v'R
And that would give the answer.
The above solution is subject to falacy-risks please trust it only if validated by experts...and validation is a responsibility of me as well as all my fellow Targetiitians! :)
Cheers!!
49yaar height nahi diya huwa??
mujhe toh solution batane k liye height chahiye!! :-/
I am considering it to be H.
21A bit explanation :
1)Gravitational force is acting in vertically downward direction. So Torque produced by the force is acting horizontally with respect to the axis(vertical) of the cylinder.
2)The centripetal force does not produce any torque at all.
So ,indeed there is not external torque in vertical direction. So total angular momentum of the system in vertical direction remains constant(ie xero)
71I don't think your Momentum conservation equation is correct! I may be wrong though
Cause, The gravittional force is acting on that ball.
1please confirm if the answer or method is right qwerty ,or shubhodip,or shubomoy
1mvr*cos(tan-1(12π))=Iω0
here ω0 is the angular velocity of the cylinder
71Post your angular momentum conservation equations please.
The final answer as given in the book is :
\omega = \sqrt{\frac{2K'g}{KR\left\{1+ \frac{K}{K'}\left(1+ \frac{1}{4\pi ^{2}} \right) \right\}}}
1is it somethung like
w(0)^2=(2mg/MR)*(k^2(M/m)(1+pi^2/4)+k)
???
1remember but ,you have to tak ethe angulkar momentum of the particle along the vertical direction,as it is not revolving abiut the the horizontal plane :)
1it is simple angular momentum conservation as told by qwerty
23how can u conserve momentum of the system ( pt B + the particle ) ?
the points of cylinder adjacent to pt B will exert some forces on B and hence this system is not free of external forces
rather select the cylinder + particle as the system
now angular momentum of this system about the vertical axis should be conserved ..now u can do it easily
71I am posting my attempt. However, I'm at my wits end at the end of the solution:
Let the velocity of the ball when it reaches the bottom of the groove (i.e., at point B) is " v ". Also, the angular velocity of the cylindrical drum be " w ".
Now Since,
i) Ei = Ef = mgR ...... [ i ]
ii) Pi = Pf = 0 ...... [ ii ]
Solving for [ i ] :
mgR = mv22 + Iω22
=> 2mgR = mv2 + Iω2
Also,
m(v sinθ - v0) + Mv0 = 0 ...... Where v0 is Linear Vel. of Point B
But How Do I related the variables!
v0 = rω will suffice? I don't think so
23wo aits wala paragraph ? i remember that one , usme 2 questions to bina solve kiye ho gaye mere LoL
