A solid cylinder of mass 3.0 kg and radius 0.2 m starts from rest at the top of a ramp, inclined 15°, and rolls to the bottom without slipping. (For a cylinder I = MR2.) The upper end of the ramp is 1.5 m higher than the lower end. Find the linear speed of the cylinder when it reaches the bottom of the ramp. (g = 9.8 m/s2)

v = 4.4271 m/s

Explanation:

Given

m = 3 Kg

R = 0.2 m

∅ = 15°

h = 1.5 m

g = 9.8 m/s²

v = ?

Ignoring frictional losses, at the bottom of the plane

Total kinetic energy is = Potential Energy at the top of plane

Using Law of conservation of energy we have

U = Kt + Kr

m*g*h = 0.5*m*v² + 0.5*I*ω²

knowing that

Icylinder = 0.5*m*R²

ω = v/R

we have

m*g*h = 0.5*m*v² + 0.5 * (0.5*mR²) * (v/R) ² = 0.75*m*v²

⇒ v = √ (g*h/0.75) = √ (9.8 m/s²*1.5 m/0.75)

⇒ v = 4.4271 m/s