A solid sphere marble of radius 5 cm and mass 3 g starts from rest and then rolls without slipping down a track and around the inside of a loop-the-loop of radius 45 cm. From what minimum height (in cm) must the marble have started if it is to make it around the loop-the-loop without falling off the track?

• we have,

from work energy theorem,

0.5 mv^2 + 0.5 I w^2 = mg(h-2R)
0.5 mv^2 + 0.5 x 0.4 x mr^2 (v^2/r^2) = mg(h-2R)
0.7 mv^2 = mg(h-2R)
0.7v^2 = g(h-2R)

now, also at top of hoop,
mg = mv^2/R
v^2 = Rg

0.7 x Rg = g(h-2R)
0.7 x 0.45 = h - 2 x 0.45
height = h = 1.215 meter
• Anonymous commented

• Anonymous commented

The method is right, but the R is 40 cm because the marble goes around the INSIDE of the loop. (27/10)*40cm is the answer.

• Anonymous commented

write answer after you use the new method

• Circular motion:   mg = mv^2 / r  so

v^2 = gr

total energy at top of loop = K + U =

(1/2) mv^2 + (1/2) (2/5) m r^2 * (v/r)^2   +  mg(2r)  =

= (1/2) * (7/5) mv^2  + 2mgr  =

=   0.7 m (gr)  +  2mgr  =

=  2.7 mgr

initial energy  =  final energy

mgh  =  2.7 mgr

h  =  2.7 r  = 2.7 * 9.80 * 0.45  =  1.215 meters  =   121.5 cm   (watch the units... it might want cm)
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