Solve Torque Problem: Wheel (M & R) Slowing from 76.6 to 31 RPM in 4.62s

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The discussion focuses on calculating the torque exerted on a wheel that slows from 76.6 rpm to 31 rpm in 4.62 seconds. The initial and final angular velocities are converted to radians per second for accurate calculations. The angular acceleration is derived using the change in angular velocity over time. The torque is then calculated using the moment of inertia and angular acceleration, resulting in an approximate value of -2.43, indicating the direction of the torque is opposite to the wheel's motion. Proper unit inclusion is emphasized for the final answer.
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Homework Statement



A wheel (M = 8.64 kg, R = radius 0.738 m) in the shape of a disk is rotating at fo = 76.6 rpm when a tool is pressed against the edge of the wheel, slowing it down at a constant rate to ff = 31 rpm in time t = 4.62 seconds. Find:

τ, the magnitude of the torque exerted by the tool on the wheel

Homework Equations



Used the kinematic equations and the um.. torque formula.

The Attempt at a Solution



Here is my approach...

ω_0 = 76.6 * 2π/60
ω_f = 31 * 2π/60

Then...

α = (ω_f - ω_0)/t

So I believe I get...

τ = Iα
= mr²/2 * (ω_f - ω_0)/t
= 8.64 * 0.738²/2 * (3.25 - 8.02)/4.62
≈ -2.43?
 
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Looks good to me (if proper units are included in the answer). :smile:
 
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