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Mechanics

Forum Expert
 Joined: 18 Feb 2007 Post: 67
19 Feb 2007 12:26:06 IST
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dare to solve this problem
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Q) A uniform marble rolls down a symmetric bowl, starting from rest at the top of the left side of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction.
a) how far up the smooth side will the marble go, measured vertically from he bottom.
b) how high would it go if the both sides are rough
Using the following principle

Scorching goIITian

Joined: 7 Feb 2007
Posts: 233
19 Feb 2007 17:09:26 IST
2 people liked this

Pinnaka, I dared to solve d 1st part of ur problem but I am not too sure about the answer. I think it should be (5/7)h.
Explanation:
Let the marble be released from a vertical height h (height of the bowl).
During the downward motion, it is rolling purely. so
v=rw
Let the mass be m.
Pot. E at h = K.E. at the bottom
So,  mgh = 1/2 mv2 + 1/2 Iw2    (w is the angular speed at the bottom)
mgh = 1/2 mw2r2 + 1/2 X 2/5 mr2w2
gh = 7/10 r2w2
i.e.,   r2w2 = 10gh/7    .............................(i)

Since no external torque acts on the upward journey so w remains same.

After reaching max height H
v becomes 0.
So, K.E. possessed by the body = 1/2.2/5mr2w2
So, K.E. lost = 7/10 mr2w2 - 1/5 mr2w2  = 1/2 mr2w2
This lost energy is gained in d form of P.E.
So,
mgH= 1/2 mr2w2
So, H= 1/2 . 10/7gh    (from (i))
H = 5/7 h

Hey everyone out there, please check and tell me if I am correct or wrong.

Scorching goIITian

Joined: 7 Feb 2007
Posts: 233
26 Feb 2007 14:22:53 IST
1 people liked this

Now I am sure that the solution of the 1st part is correct and I have solved the 2nd part also. In the 2nd case,  as the ascending surface is also frictionful, so the marble will be pure rolling and both its translational and rotational velocities will become zero when it reaches its max. height. So, all the K.E. is lost thus

loss in K.E. = 7/10 mr2w2
This is gained as P.E.
So,     mgH = 7/10 mr2w2
H = 7/10 . 10h/7
So,   H= h

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