Question

A. A sphere and a cylinder roll down an inclined plane, started from rest at the same time. They have the same radii, but not the same mass. Does one reach the bottom first? If so, which? Explain your reasoning.

B. Two spheres roll down an inclined plane, starting from rest at the same time. Sphere A has radius R and mass m. Sphere 2 has radius 2R and mass m. Does one reach the bottom first? If so, which? Explain your reasoning.

Answer 1

1)

for sphere,

intial mechanical enrgy = final mechanical enrgy

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 = sqrt(g*h/0.7)

for cyllinder,

intial mechanical enrgy = final mechanical enrgy

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

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

m*g*h = 0.5*m*v^2 + 0.25*m*v^2

m*g*h = 0.75*m*v^2

==> v = sqrt(g*h/0.75)

clearly V(cyllinder) < V(sphere)

so sphere reaches the bottom first.

here v does not depend on mass

B) for sphere1,

intial mechanical enrgy = final mechanical enrgy

m*g*h = 0.5*m*v1^2 + 0.5*I*w1^2

m*g*h = 0.5*m*v1^2 + 0.5*(2/5*m*R1^2)*w1^2

m*g*h = 0.5*m*v1^2 + 0.2*m*v1^2

m*g*h = 0.7*m*v1^2

==> v1 = sqrt(g*h/0.7)

for sphere2,

intial mechanical enrgy = final mechanical enrgy

m*g*h = 0.5*m*v2^2 + 0.5*I*w2^2

m*g*h = 0.5*m*v2^2 + 0.5*(2/5*m*R2^2)*w2^2

m*g*h = 0.5*m*v2^2 + 0.2*m*v2^2

m*g*h = 0.7*m*v2^2

==> v2 = sqrt(g*h/0.7)

clearly V1 = V2

so both spheres reach the bottom at the same time