18th February 2007 - 12:33 PM
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.
how high would it go if the both sides are rough Using the following principle
please answer my question with detailed solution, I am not getting it
18th February 2007 - 03:33 PM
This is a potential energy problem. On the rough surface, the friction will make the marble roll, so some of the gravitational potential energy will be converted to rotational kinetic energy. Note that this friction is STATIC friction, so it does not lose energy to heat like kinetic friction does. Instead, the static friction force does negative work on the marble, which removes some of its translational kinetic energy. At the same time, the friction exerts a torque on the marble, which does positive work and gives the marble an equal amount of rotational kinetic energy. You can tell that the amounts are equal, because no energy is lost to the marble due to static friction. This is because the point where the marble touches the bowl has no velocity (because it is not slipping), so the friction force there cannot do any NET work on the marble. This means tha static friction TRANSFERS some of the translational kinetic energy into rotational kinetic energy, but does not change the total energy of the marble.
The marble rolls without slipping, so when it turns through an angle theta, the arc length that rolls over the bowl surface is R*theta, according to the definition of the radian. (R is the radius of the marble.) Since it does not slip, the arc length is equal to the distance it moves forward. Therefore it moves a distance R*theta when it turns through an angle theta, so when it has angular velocity omega, it must have linear velocity R*omega. This means that its rotational kinetic energy and translational kinetic energy are related, since RKE = 1/2*I*omega^2 = 1/2(2/5*mR^2)*omega^2 = 2/5*(1/2*mR^2*omega^2) = 2/5*(1/2*mv^2) = 2/5*TKE. Therefore, the rotational kinetic energy is 2/5 the size of the translational kinetic energy. (This factor of 2/5 comes from the moment of inertia, so different shapes will have different factors here. A cylinder would have 1/2 and a hoop would have 1, for example.)
Since RKE + TKE = total kinetic energy = KE, you find that 2/5*TKE + TKE = KE, so 7/5*TKE = KE. Therefore, TKE = 5/7*KE, so exactly 5/7 of the total kinetic energy is translational, and the other 2/7 of it is rotational.
This is the key to the problem. The original gravitational potential energy (GPE) is mgh, and as the ball rolls down the rough side fo the bowl, this is converted into KE, so at the bottom of the bowl, KE = mgh. When the ball comes to the frictionless side of the bowl, it is unable to make its spinning motion slow down, because it needs friction to produce the torque that changes its rotation. Therefore, 2/7 of the KE is TRAPPED in the RKE. The marble will move up the smooth side of the bowl, but it will continue spinning at a constant omega the whole time, so when it reaches its highest point, it still has RKE = 2/7*mgh. Of course, at the highest point, it has lost all of its TKE, because it stops moving up the side of the bowl. Therefore, only the TKE is converted back to gravitational potential energy. This means that mgh_final = TKE = 5/7*mgh. Cancelling factors gives h_final = 5/7*h.
Although the analysis may seem long, the basic idea is simple: GPE turns into KE, 2/7 of this is trapped in RKE, leaving 5/7 of it as TKE. This then turns back into GPE, so the final GPE is 5/7 of the initial GPE. So the final height is 5/7 of the initial height.
For part B, if the bowl is rough on both sides, then there IS friction as the marble rolls up the right half of the bowl, so v=R*omega is true there just as in the left side. Therefore the rotation must slow down as the translation slows down. There IS friction, and since it is slowing down the rotation, it transfers the RKE back into TKE, where it can turn back into GPE as the marble goes upwards. In other words, the RKE is NOT trapped now. Therefore, when the marble reaches its highest point, both its translation and its rotation stop, so ALL of its KE is converted back to GPE. Therefore, GPE_final = GPE_initial, so it comes back exactly to its initial height.
Hope this helps!
7th March 2007 - 07:35 AM
28th March 2009 - 12:24 AM
you are awesome.
3rd April 2009 - 08:35 AM
Moderator is always both impressed and happy when reading one of mr_homm's posts.
21st January 2010 - 03:30 AM
Your explanation is better than these crappy text books...GREAT JOB!!!...i can"t believe i understand it now
confused physics student
25th February 2012 - 06:41 AM
Thanks; you've made everything so clear to me!
10th April 2013 - 03:45 AM
thanks bro, go professional soon eh? but keep helping us out on the forums;)
Actually think I get it now
30th April 2013 - 12:13 AM