A 35 cm -diameter potter\'s wheel with a mass of 24 kg is spinning at 180 rpm. U

Question

A 35 cm -diameter potter's wheel with a mass of 24 kg is spinning at 180 rpm. Using her hands, a potter forms a pot, centered on the wheel, with a 14 cm diameter. Her hands apply a net friction force of 1.3 N to the edge of the pot. If the power goes out, so that the wheel's motor no longer provides any torque, how long will it take for the wheel to come to a stop in her hands? (((((((((((((((((( Express your answer to two significant figures and include the appropriate units ))))))))))))))))))))) please help I have just the last chance and I found 2.7*10^2 s, 27 s, 2.8*10^2 s, 28 s,and 2.5*10^2 all these was wrong

Explanation / Answer

= F x r = -1.3 N x 0.14 m = -0.182 Nm

Also...

= I

Where I is the moment of inertia and is the angular acceleration. Assuming the wheel is a regular cylinder, its moment of inertia is:

I = mr²/2

So the formula becomes:

= mr²/2

-0.182 Nm = (24 kg)(0.35 m)² / 2

-0.182 Nm = (1.47 kgm²)

= -0.123 rad/s²

So now we know the angular acceleration. Let's use that to find the time required for the wheel to stop.

= (f - o) / t

The original angular velocity, o, is 180 rpm. We need to change that to rad/s.

180 rpm x (2 rad / 1 rev) x (1 min / 60 s) = 18.8 rad/s

Now then...

0.123 rad/s² = (0 rad/s - 18.8 rad/s) / t

t = (-18.8 rad/s) / (-0.123 rad/s²)

t = 152.84 s

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