A cylinder with a diameter of 16 rolls with an angular speed of 0.055 on a level surface.

If the cylinder experiences a uniform tangential acceleration of 0.020 without slipping until its angular speed is 1.0 , through how many complete revolutions does the cylinder rotate during the time it accelerates?

To determine the number of complete revolutions the cylinder rotates during the time it accelerates, we can use the kinematic equation relating angular acceleration and angular speed:

ω^2 = ω₀^2 + 2αθ

Where:
ω = final angular speed
ω₀ = initial angular speed
α = tangential acceleration
θ = angular displacement

In this case, the initial angular speed (ω₀) is 0.055, the final angular speed (ω) is 1.0, and the tangential acceleration (α) is 0.020.

Rearranging the equation to solve for θ, we have:

θ = (ω^2 - ω₀^2) / (2α)

Plugging in the values, we get:

θ = (1.0^2 - 0.055^2) / (2 * 0.020)
= (1 - 0.003025) / 0.04
= 0.996975 / 0.04
= 24.924375

Therefore, the angular displacement is approximately 24.924375 radians.

To convert this to revolutions, we need to know how many radians are in one revolution. Since one revolution is equal to 2π radians, we can calculate the number of revolutions as:

Number of revolutions = θ / (2π)

Plugging in the value of θ, we get:

Number of revolutions = 24.924375 / (2π)
≈ 3.96782

Therefore, the cylinder rotates approximately 3.96782 complete revolutions during the time it accelerates.