How Long Does It Take for a Flywheel to Reach 125 Rev/Min from Rest?

[MMCS]

A flywheel 1.2 m in diameter is uniformly accelerated from rest and revolves completely 68 times in reaching a speed of 125 rev/min. Find the time taken to reach this speed

i have been given the correct answer as 65.29 seconds

Wht i have up to now

ω1 = 0
ω2 = 125 r/min = 125/60 = 2.083 r/s
theta = 68*2∏ = 427.26 rads
t = ?

I have tried to use theta = ((ω1+ω2)*t)/2 rearranged to get t = 2theta/ω1+ω2 however my answer doesn't come close

Any suggestions would be appreciated

 

[TxRationalist]

you need the equation of motion for rotation. the flywheel is undergoing uniform acceleration; you need to take that into consideration.

 

[Doc Al]

ω1 = 0
ω2 = 125 r/min = 125/60 = 2.083 r/s

Make sure ω is in rads/sec.

 

[MMCS]

That sorted it, silly mistake, Thanks Doc Al

 

[rezeew]

Your approach is on the right track, but there are a few issues with your calculations. First, you should convert the diameter of the flywheel to its radius, which is 0.6 m. Then, you should convert the revolutions per minute to radians per second by multiplying by 2π/60. This gives you an angular velocity of 0.2185 rad/s.

Next, you should use the rotational kinematics equation θ = ω1t + (1/2)αt^2, where θ is the total angle covered, ω1 is the initial angular velocity, α is the angular acceleration, and t is the time. Since the flywheel starts from rest, ω1 = 0, and we can rearrange the equation to solve for t:

t = √(2θ/α)

Substituting in the values we have, we get:

t = √(2*427.26/0.2185) = 65.29 seconds

So your answer is correct! It's important to note that in this case, the angular acceleration is constant, so we can use the simplified equation above. However, if the acceleration was not constant, we would need to use the more general equation θ = ω1t + (1/2)αt^2 + (1/6)ω1t^3, where the last term takes into account the change in angular acceleration over time. But in this problem, we can assume the acceleration is constant and use the simplified equation.