Question

A particle accelerates uniformly from rest to 60 rev/min in 10 seconds. Determine the magnitude of angular acceleration and angle rounded by the particle, and the number of revolutions for the time of motion.

Answer 1

The magnitude of angular acceleration is 6 rev/min/s, the angle rounded by the particle is 300 rev, and the number of revolutions for the time of motion is also 300 rev.

Step-by-step explanation:

To determine the magnitude of angular acceleration, we can use the formula:

angular acceleration = (final angular velocity - initial angular velocity) / time

Given that the initial angular velocity is 0 rev/min, the final angular velocity is 60 rev/min, and the time is 10 seconds, we can substitute these values into the formula to find the magnitude of angular acceleration.

angular acceleration = (60 rev/min - 0 rev/min) / 10 s = 6 rev/min/s

To find the angle rounded by the particle, we can use the formula:

angle = initial angular velocity x time + (1/2) x angular acceleration x time^2

Substituting the values, we get:

angle = 0 rev/min x 10 s + (1/2) x 6 rev/min/s x (10 s)^2 = 300 rev

Lastly, to find the number of revolutions for the time of motion, we can use the formula:

number of revolutions = initial angular velocity x time + (1/2) x angular acceleration x time^2

Substituting the values, we get:

number of revolutions = 0 rev/min x 10 s + (1/2) x 6 rev/min/s x (10 s)^2 = 300 rev