Question

A solid cylinder of radius 10.0 cm rolls down an incline with slipping. The angle of the incline is 30°. The coefficient of kinetic friction on the surface is 0.400. What is the angular acceleration of the solid cylinder? What is the linear acceleration?

a) α = 2.2 rad/s², a = 1.8 m/s²
b) α = 3.0 rad/s², a = 2.5 m/s²
c) α = 4.5 rad/s², a = 3.6 m/s²
d) α = 5.8 rad/s², a = 4.2 m/s²

Answer 1

The angular acceleration of the solid cylinder is 2.2 rad/s² and the linear acceleration is 0.22 m/s².

Step-by-step explanation:

The angular acceleration of a solid cylinder rolling down an incline with slipping can be calculated using the formula:

α = (2/5) * g * sin(θ) * R / (R * R + (k * k))

Where α is the angular acceleration, g is the acceleration due to gravity, θ is the angle of the incline, R is the radius of the cylinder, and k is the coefficient of kinetic friction. Plugging in the given values:

α = (2/5) * 9.8 m/s² * sin(30°) * 0.1 m / (0.1 m * 0.1 m + (0.4 * 0.4))

α = 2.2 rad/s²

The linear acceleration can be calculated using the formula:

a = α * R

Where a is the linear acceleration and R is the radius of the cylinder. Plugging in the given values:

a = 2.2 rad/s² * 0.1 m

a = 0.22 m/s²