Rotational motion problems/angular acceleration

Question

A barrel ride at an amusement park

starts from rest and speeds up to

0.520 rev/sec in 7.26 s. What is

the angular acceleration during

that time?

could someone plz answer this thourgly?

Answer 1

Hi Jordan!

So angular acceleration can be defined as the change in angular velocity over the change in time:

α = Δω/Δt

We know that the ride starts at rest (ω = 0 rev/s) and speeds up to 0.520 rev/s. Therefore Δω in the equation above is 0.520 rev/s. This speeding up takes place over 7.26 s, so Δt = 7.26 s.

Therefore, using the equation above:

α = Δω/Δt

α = (0.520 rev/s) / (7.26 s)

α = 0.0716 rev/s²

Note: one revolution (rev) is 2π radians. We can convert the answer above to units of rad/s²:

α = (0.0716 rev/s²) × (2π rad / rev)

α = 0.450 rad/s²

Hope that helps!