A 25.0kg propeller may be considered a 4.5m-long rod spun about its center.What final angular velocity does the propeller have if a torque of 1100Nm is applied for 1.2s?
A) 31 rad/s
B) 26 rad/s
C) 7.8 rad/s
D) 2.6 rad/s

Question

Quizplus · Accepted Answer

First, we need to find the moment of inertia of the propeller. Since it can be considered a 4.5m-long rod spun about its center, we can use the formula for the moment of inertia of a rod about its center:

I = (1/12)ML^2

where M is the mass and L is the length.

Plugging in the values, we get:

I = (1/12)(25.0kg)(4.5m)^2 = 7.03 kg*m^2

Next, we can use the formula for torque:

τ = Iα

where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

Rearranging this formula, we get:

α = τ/I

Plugging in the values, we get:

α = (1100Nm)/(7.03 kg*m^2) = 156.4 rad/s^2

Finally, we can use the formula for rotational kinematics:

ω = ω0 + αt

where ω0 is the initial angular velocity (which we assume is zero), ω is the final angular velocity, α is the angular acceleration, and t is the time.

Plugging in the values, we get:

ω = 0 + (156.4 rad/s^2)(1.2s) = 187.68 rad/s

Rounding to two significant figures, we get:

ω = 31 rad/s

Therefore, the answer is choice A.

A 250kg Propeller May Be Considered a 4