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Hot goIITian

Joined: 18 Jan 2009
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4 Mar 2009 00:13:23 IST
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Rotational - Rolling
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See the figure.

There is a big block of mass M=1kg and radius R=62.5cm placed on a smooth surface and a cylinder of mass m=0.5kg and radius r=10cm is first suspended by a string so that it just touches the surface of the block. Now the thread is burnt and the cylinder is allowed to roll down. Friction is sufficient to cause roling. When the cylinder reaches the bottom, find the velocity of the cylinder and that of block.

Ans given is :-  Velocity of cylinder =2 m/s and that of block=1.5 m/s

 

Please give the explanation also.

 


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Blazing goIITian

Joined: 11 Sep 2008
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4 Mar 2009 09:20:17 IST
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this question is from arihant---------mechanics----part 1--------chapter of collisions and c.o.m-----level 2!!!!!!!!!!! better refer to the hints and solutions in the book itself............the explanation given there is good!!!!!!!


Blazing goIITian

Joined: 11 Sep 2008
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4 Mar 2009 09:20:26 IST
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this question is from arihant---------mechanics----part 1--------chapter of collisions and c.o.m-----level 2!!!!!!!!!!! better refer to the hints and solutions in the book itself............the explanation given there is good!!!!!!!


Blazing goIITian

Joined: 11 Sep 2008
Posts: 1159
4 Mar 2009 09:20:42 IST
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this question is from arihant---------mechanics----part 1--------chapter of collisions and c.o.m-----level 2!!!!!!!!!!! better refer to the hints and solutions in the book itself............the explanation given there is good!!!!!!!

XYZ's Avatar

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4 Mar 2009 09:30:52 IST
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I don't have that book. So, if you could please provide the solution, it would be really helpful.

shraman  asa's Avatar

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4 Mar 2009 11:10:51 IST
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key.......

MOMENTUM  REMAINS SAME ALONG HORIZONTAL DIRECTION!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

WHY ... NO HORIZONTAL FORCE .......................................................................................................

 

CONSERVE ENERGY .............................................................................................................................

 

NOTE:ALSO INCLUDE THE RECOIL VELOCITY OF THAT BIG THING ON WHICH THE CYLINDER IS PLACED IN UR EQUATIONS ....

 

  common mistake made by every 1 .............

 

IF U STILL DONT GET IT ILL POST DETAILED SOLN.

 

 

XYZ's Avatar

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4 Mar 2009 12:35:55 IST
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Shraman! I have done all that. I know how to do these type of questions because I have done these types earlier also.

 

I did all of that- i.e. conserved momentum in the horizontal direction, after that I conserved energy taking into consideration the velocity of the block and the cylinder and the rotational velocity of the cylinder.

 

The only problem with this one is that the answer that I have given in this question is not what I'm getting. There's a slight variation in my answer. So, I posted this question to know the correct answer.

 

Can you please solve it and tell whether you are getting the same answer as given or something different.

If you are getting the same answer, then please post the solution.

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4 Mar 2009 13:09:06 IST
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I am getting the velocity of the cylinder in close proximity as 2.049m/s. first applied conservation of momentum. then for energy..........V= -mv/M ............ and mg(R-r) = 1/2 mv^2 + 1/2 MV^2 + 1/2 Iw^2. solving these two i got ans as 2.049 m/s.
XYZ's Avatar

Hot goIITian

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4 Mar 2009 14:16:58 IST
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JPS, I also did it exactly the same way as you have done and got the same answer ,i.e. , Velocity of the cylinder=2.04 m/s (approx.)

But the velocity of the block is given in the answer is 1.5 m/s. And considering the approach we have used, we'll get the velocity of the block= 1.02 m/s. The variation in this result is large.

 

Can anybody explain the above dilemma (i.e. the variation of the answers)???

OR Is the given answer incorrect???




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