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A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of +57 Nm is applied to the wheel for 24 s, giving the wheel an angular velocity of +660 rev/min. The external torque is then removed, and the wheel comes to rest 120 s later. (Include the sign in your answers.)

Find the moment of inertia of the wheel.

User Mark Baker
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Final answer:

To find the moment of inertia (I) of the wheel, the angular acceleration (α) is calculated from the known torque (57 Nm) applied for 24 seconds, bringing the wheel to an angular velocity of 660 rev/min. Then, I am determined using the relationship I = τ/α. The moment of inertia of the wheel = 19.77Nms^2

Step-by-step explanation:

The moment of inertia of the wheel can be determined using the equations of rotational motion and the given information about the torque and angular velocities. First, the angular acceleration (α) is found using the initial application of torque, which brought the wheel to an angular velocity (ω) of 660 rev/min (converted to rad/s) in 24 seconds from rest. The relationship ω = αt is used where t is time.

After calculating α, we can use the relationship τ = Iα, where τ is the torque and I is the moment of inertia, to find the moment of inertia (I). Given a torque (τ) of 57 Nm, we can rearrange the equation to I = τ / α to calculate I.

Since ω = 660 rev/min = 660 × 2π/60 rad/s and t = 24 s, α = ω / t, we get α = (660 × 2π/60) / 24 rad/s².

So, we get α = 2.8822rad/s^2.

Now plugging the values in the formula I = τ / α, we get I = 57 Nm/ 2.8822rad/s^2.

On calculation, we get I =19.77Nms^2.

Therefore, the moment of inertia of the wheel = 19.77Nms^2

User CodingTT
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