HobieDude16
Nov7-04, 03:33 PM
ok, heres a problem that you have to work out with all variables, and i frankly have trouble doing that.... i got part a, and i did a lot of work on part b, but cant seem to come up with the right answer.... any help appreciated....
In Figure 11-32, a solid brass ball of mass m and radius r will roll without slipping along the loop-the-loop track when released from rest along the straight section. For the following answers use g for the acceleration due to gravity, and m, r, and R, as appropriate, where all quantities are in SI units.
Fig. 11-32
http://www.webassign.net/hrw/hrw7_11-32.gif
(a) From what minimum height h above the bottom of the track must the marble be released to ensure that it does not leave the track at the top of the loop? (The radius of the loop-the-loop is R. Assume R r.)
(b) If the marble is released from height 6R above the bottom of the track, what is the magnitude of the horizontal component of the force acting on it at point Q?
ok, for part a, i got 2.7R whichi is correct, so then i moved on to part b, and i was told by my TA to start with the formula mg6R = .5mv^2 + .5Iomega^2 + mgR.... so i did, and tried to solve for v. i used 2/5mR^2 for I and for omega, used v/R...... so if i do that, and solve for v, then i use a=v^2/r and then when i have that, F=ma so i have the idea, i just cant get it right. anybody have any ideas? thanks in advance
John
In Figure 11-32, a solid brass ball of mass m and radius r will roll without slipping along the loop-the-loop track when released from rest along the straight section. For the following answers use g for the acceleration due to gravity, and m, r, and R, as appropriate, where all quantities are in SI units.
Fig. 11-32
http://www.webassign.net/hrw/hrw7_11-32.gif
(a) From what minimum height h above the bottom of the track must the marble be released to ensure that it does not leave the track at the top of the loop? (The radius of the loop-the-loop is R. Assume R r.)
(b) If the marble is released from height 6R above the bottom of the track, what is the magnitude of the horizontal component of the force acting on it at point Q?
ok, for part a, i got 2.7R whichi is correct, so then i moved on to part b, and i was told by my TA to start with the formula mg6R = .5mv^2 + .5Iomega^2 + mgR.... so i did, and tried to solve for v. i used 2/5mR^2 for I and for omega, used v/R...... so if i do that, and solve for v, then i use a=v^2/r and then when i have that, F=ma so i have the idea, i just cant get it right. anybody have any ideas? thanks in advance
John