v^2/R = g or it loses contact
m g(h-D) = .5 mv^2
v^2 = g R = .5 g D
so
m g(h-D) = .5 m *.5 g D =.25 mgD
h-D = .25 D
h = 1.25 D
a body is allowed to slide down a frictionless track from rest under gravity . The track end in a circular loop of diameter D . What should be the minimum height of the body in terms of D so that the body successfully complete the loop.
2 answers
Energy is conserved so mgh =mgD
D is the maximum height
m and g are constant and the ony variable is h and h = D
all the PE at the start from height h is converted back to PE at height D
D is the maximum height
m and g are constant and the ony variable is h and h = D
all the PE at the start from height h is converted back to PE at height D