The small mass m sliding without friction along the looped trackshown in the fig

Question

The small mass m sliding without friction along the looped trackshown in the figure is to remain on the track at all times, even atthe very top of the loop of radius r. If the actual release height is 5h, calculate the normal force exerted by the track atthe bottom of the loop. If the actual release height is 2h, calculate the normal force exerted by the track atthe top of the loop. If the actual release height is 4h, calculate the normal force exerted by the trackafter the block exits the loop onto the flat section.

Explanation / Answer

Radius of loop is r Initial height is h -------------------------------------------------------------- If initial height is 5h then the velocity at the bottomof loop is                mg(5h) = (1/2)mv2                       v2 = 10 gh Then the normal force on the block at bottom of loopis                     N = mg + mv2 /r Then the normal force on the block at bottom of loopis                     N = mg + mv2 /r                         = mg + m(10gh) / r                          =mg + 10mgh /r If initial height is 2h then the velocity at the bottomof loop is                mg(2h) = (1/2)mv2                       v2 =  4 gh Now we find the velocity at the top of theloop     total energy at the bottom of theloop = total energy at the top of the loop                                 (1/2)mv2 = mg(2r) + (1/2)mu2                                           v2 = 4gr + u2                                           4gh = 4gr + u2                      then u2 =4g(h-r)                                         Then the normal force on the block at top of loopis                     N = mg  - mu2 /r                         = mg - m4g (h-r) / r                          =mg - 4mg (h-r) / r     The normal force on the falt surface isirrespective of height ( that is not depens on the initialheight)                   then N = mg                                                               mg(2h) = (1/2)mv2                       v2 =  4 gh Now we find the velocity at the top of theloop     total energy at the bottom of theloop = total energy at the top of the loop                                 (1/2)mv2 = mg(2r) + (1/2)mu2                                           v2 = 4gr + u2                                           4gh = 4gr + u2                      then u2 =4g(h-r)                                         Then the normal force on the block at top of loopis                     N = mg  - mu2 /r                         = mg - m4g (h-r) / r                          =mg - 4mg (h-r) / r     The normal force on the falt surface isirrespective of height ( that is not depens on the initialheight)                   then N = mg                                                                                          v2 = 4gr + u2                                           4gh = 4gr + u2                      then u2 =4g(h-r)                                                              then u2 =4g(h-r)                                         Then the normal force on the block at top of loopis                     N = mg  - mu2 /r                         = mg - m4g (h-r) / r                          =mg - 4mg (h-r) / r     The normal force on the falt surface isirrespective of height ( that is not depens on the initialheight)                   then N = mg                                      

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