The drum of a mixer is accelerated uniformly from rest by a driving torque of 12 Nm androtates through 50 revolutions in the first 30s. The acceleration continues for a further 20s toreach the steady operating speed of the mixer.Determine:a) the angular acceleration

To solve this problem, we first need to understand the relationship between linear motion and rotational motion. The key equations are similar to those of linear motion, but with rotational equivalents.

a) To find the angular acceleration, we first need to find the angular displacement.

The angular displacement (θ) is given by the number of revolutions multiplied by 2π (since there are 2π radians in one revolution). So, θ = 50 * 2π = 100π radians.

We know that the drum was initially at rest, so the initial angular velocity (ω0) is 0. We also know that the drum was accelerating uniformly for 30 seconds (t).

We can use the equation of motion θ = ω0t + 0.5α*t^2 to find the angular acceleration (α).

Substituting the known values, we get:

100π = 030 + 0.5α*30^2

Solving for α, we get:

α = (100π) / (0.5*30^2) = 100π / 450 = 20π/9 rad/s^2

So, the angular acceleration of the drum is 20π/9 rad/s^2.

To solve this problem, we need to use the equations of rotational motion.

a) To find the angular acceleration, we first need to convert the number of revolutions to radians.

1 revolution = 2π radians So, 50 revolutions = 50 * 2π = 100π radians

We know that the equation of motion in terms of angular displacement (θ), initial angular velocity (ω0), angular acceleration (α), and time (t) is:

θ = ω0t + 0.5α*t^2

In this case, the drum starts from rest, so the initial angular velocity (ω0) is 0. Substituting the given values into the equation, we get:

100π = 0.5α(30)^2

Solving for α, we get:

α = (100π) / (0.5*(30)^2) = 0.70 rad/s^2

So, the angular acceleration of the drum is 0.70 rad/s^2.