Four objects are placed in a row at the same height near the top of a ramp and a

Question

Four objects are placed in a row at the same height near
the top of a ramp and are released from rest at the same
time. The objects are (i) a 1-kg solid sphere; (ii) a 1-kg
hollow sphere; (iii) a 2-kg solid sphere; and (iv) a 1-kg
thin hoop. All four objects have the same diameter, and
the hoop has a width that is one-quarter its diameter. The
time it takes the objects to reach the finish line near the
bottom of the ramp is recorded. The moment of inertia
for an axis passing through its center of mass for a solid
sphere is ; for a hollow sphere it is ; and for
a hoop it is . Rank the four objects from fastest (shortest time) down the ramp to slowest.
Explain your reasoning.

Explanation / Answer

for solid sphere

Apply Enrgy conservation

m*g*h = 0.5*m*v^2 + 0.5*I*w^2

m*g*h = 0.5*m*v^2 + 0.5*(2/5)*m*r^2*w^2

m*g*h = 0.5*m*v^2 + 0.2*m*v^2

m*g*h = 0.7*m*v^2

==> v(solid sphere) = sqrt(g*h/0.7)

for solid hollow sphere

Apply Enrgy conservation

m*g*h = 0.5*m*v^2 + 0.5*I*w^2

m*g*h = 0.5*m*v^2 + 0.5*(2/3)*m*r^2*w^2

m*g*h = 0.5*m*v^2 + 0.333*m*v^2

m*g*h = 0.833*m*v^2

==> v(hollow sphere) = sqrt(g*h/0.833)

for hoop

Apply Enrgy conservation

m*g*h = 0.5*m*v^2 + 0.5*I*w^2

m*g*h = 0.5*m*v^2 + 0.5*m*r^2*w^2

m*g*h = 0.5*m*v^2 + 0.5*m*v^2

m*g*h = m*v^2

==> v(hoop) = sqrt(g*h)

clearly

v(solid sphere) > v(hollow sphere) > v(hoop)

both sollid spheres reach first.

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