Rotational acceleration on merry go round

[mike2007]

The rotational velocity of a merry go round increases at a constant rate from 0.6rad/s to 1.8rad/s in a time of four seconds. What is the rotational acceleration of the merry go round?

This is what i have done so far but not too sure


Time = 4seconds
Change in velocity = 1.8rad/s – 0.6rad/s
= 1.2rad/s
Acceleration = ∆ω/t
= 1.2rad/s/4seconds
Answer = 0.3rad/s2

 

[CaptainZappo]

As far as I can tell, everything is correct.

 

[m2003]

Your calculations are correct. The rotational acceleration of the merry go round is 0.3rad/s2. This means that the rotational velocity is increasing by 0.3rad/s every second. To put it in simpler terms, the merry go round is spinning faster and faster as time goes on, and this change in speed is occurring at a constant rate. This information can be useful in understanding the dynamics and mechanics of rotating objects, and can also be applied in other scientific fields such as physics and engineering.