Answer:
2/R*sqrt (g*s*sin(θ)) = w
Explanation:
Assume:
- The cylinder with mass m
- The radius of cylinder R
- Distance traveled down the slope is s
- The angular speed at bottom of slope w
- The slope of the plane θ
- Frictionless surface.
Solution:
- Using energy principle at top and bottom of the slope. The exchange of gravitational potential energy at height h, and kinetic energy at the bottom of slope.
ΔPE = ΔKE
- The change in gravitational potential energy is given as m*g*h.
- The kinetic energy of the cylinder at the bottom is given as rotational motion: 0.5*I*w^2
- Where I is the moment of inertia of the cylinder I = 0.5*m*R^2
We have:
m*g*s*sin(θ) = 0.25*m*R^2*w^2
2/R*sqrt (g*s*sin(θ)) = w
- The angular velocity depends on plane geometry θ , distance travelled down slope s, Radius of the cylinder R , and gravitational acceleration g
