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

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SUMMARY

The discussion focuses on calculating the torque exerted on a rotating wheel with a mass of 8.64 kg and a radius of 0.738 m, which slows from 76.6 RPM to 31 RPM in 4.62 seconds. The initial and final angular velocities are converted to radians per second, yielding ω_0 = 8.02 rad/s and ω_f = 3.25 rad/s. The angular acceleration (α) is calculated as -1.58 rad/s², leading to the torque (τ) being determined using the formula τ = Iα, resulting in a torque of approximately -2.43 N·m.

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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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