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A 24.0 kg wheel, essentially a thin hoop with radius 2.10 m, is rotating at 241rev/min. It must be brought to a stop in 16.0 s. (a) How nuch work must be done to stop it?

User Snmaddula
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1 Answer

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Final answer:

To stop the wheel, work must be done to bring it to a stop. The work done to stop the wheel can be calculated using the formula: Work = (1/2) * I * w^2, where I is the moment of inertia and w is the angular velocity.

Step-by-step explanation:

To stop the wheel, work must be done to bring it to a stop. The work done to stop the wheel can be calculated using the formula:

Work = (1/2) * I * w^2

where I is the moment of inertia and w is the angular velocity. In this case, the moment of inertia of the wheel can be calculated using the formula: I = m * r^2 , where m is the mass of the wheel and r is the radius.

Plugging in the values given, we can calculate the moment of inertia and the work:

mass of the wheel (m) = 24.0 kg

radius of the wheel (r) = 2.10 m

angular velocity (w) = 241 rev/min

To calculate the moment of inertia:

I = m * r^2 = 24.0 kg * (2.10 m)^2 = 105.84 kg*m^2

To convert the angular velocity to radians per second:

w = 241 rev/min * (2π rad/rev) * (1 min/60 s) = 25.32 rad/s

Now we can calculate the work:

Work = (1/2) * I * w^2 = (1/2) * 105.84 kg*m^2 * (25.32 rad/s)^2 = 33,778.65 J

Therefore, the work required to stop the wheel is 33,778.65 Joules.

User Dunbarx
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