What is the maximum possible speed of the ball at the top of the loop?
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To get the minimum required speed to make the loop the loop, at the top of the loop we require the normal force (N) to be 0. Equating the forces at the top of the loop we have the weight of the car (mg) plus the normal force (N) equal to the centrifugal force (F), which is given by F=mv2r where r is the radius of the circle. Thus,
mv2trv2trv2tvt====N+mg,g,gr,gr−−√.
Giving us a minimum velocity at the top of the loop, vt. We now proceed with the energy argument.
The total energy at the top of the loop is equal to the potential energy at the top plus the kinetic energy at the top, respectively these are:
PEt=mgh=2mgr,
KEt=12mv2t=12mgr.
Thus the total energy at the top is:
Et=2mgr+12mgr,=52mgr.
We now find the total energy at the bottom, since there is no potential energy, this is simply the kinetic energy,
Eb=12mv2b.
Where vb is the speed we are looking for. With our assumptions, the energy at the top (Et) equals the energy at the bottom (Eb) giving
Eb12mv2bmv2bv2bvb=====Et,52mgr,5mgr,5gr,5gr−−−√.
Thus we have found the speed required to complete a loop the loop of radius r. For example, if the loop had a 4 metre diameter (2 metre radius) then the velocity required to complete the loop would be v=5×g×2−−−−−−−√=10g−−−√≈9.9.
For convience we can approximate 5gr−−−√ as
5×9.81×r−−−−−−−−−−√=49.05r−−−−−√≈7r√.
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Many thanks to Mike Barr for pointing out that the explanation from where vt=gr−−√ came from was incorrect. This has now been corrected
mv2trv2trv2tvt====N+mg,g,gr,gr−−√.
Giving us a minimum velocity at the top of the loop, vt. We now proceed with the energy argument.
The total energy at the top of the loop is equal to the potential energy at the top plus the kinetic energy at the top, respectively these are:
PEt=mgh=2mgr,
KEt=12mv2t=12mgr.
Thus the total energy at the top is:
Et=2mgr+12mgr,=52mgr.
We now find the total energy at the bottom, since there is no potential energy, this is simply the kinetic energy,
Eb=12mv2b.
Where vb is the speed we are looking for. With our assumptions, the energy at the top (Et) equals the energy at the bottom (Eb) giving
Eb12mv2bmv2bv2bvb=====Et,52mgr,5mgr,5gr,5gr−−−√.
Thus we have found the speed required to complete a loop the loop of radius r. For example, if the loop had a 4 metre diameter (2 metre radius) then the velocity required to complete the loop would be v=5×g×2−−−−−−−√=10g−−−√≈9.9.
For convience we can approximate 5gr−−−√ as
5×9.81×r−−−−−−−−−−√=49.05r−−−−−√≈7r√.
—
Many thanks to Mike Barr for pointing out that the explanation from where vt=gr−−√ came from was incorrect. This has now been corrected
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