A constant torque of 24.5 N. m is applied to a grindstone whose moment of inertia is 0.130 kg · m. Using energy principles, and neglecting friction, find the angular speed after the grindstone has made 14.0 revolutions. Hint: the angular equivalent of Wnet = FAx =. mv? -mv? is wnet = TA0 = lo} - lw?. You should convince yourself that this last relationship is correct. (Assume the grindstone starts from rest.) rev/s

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A constant torque of 24.5 N · m is applied to a grindstone whose moment of inertia is 0.130 kg · m2. Using energy principles, and neglecting friction, find the angular speed after the grindstone has made 14.0 revolutions. Hint: the angular equivalent of

Wnet = FΔx = 
1
2
mvf2 − 
1
2
mvi2

is

Wnet = τΔ? = 
1
2
f2 − 
1
2
i2.

You should convince yourself that this last relationship is correct. (Assume the grindstone starts from rest.)
rev/s

A constant torque of 24.5 N • m is applied to a grindstone whose moment of inertia is 0.130 kg · m². Using energy principles, and neglecting friction, find the angular speed after the grindstone has made 14.0 revolutions. Hint: the
= FAx = mv? -mv,? is W,
2
angular equivalent of W,
= tA0 =
2
You should convince yourself that this last relationship is correct. (Assume the grindstone starts from rest.)
net
net
2
2
2
rev/s
еВook
Transcribed Image Text:A constant torque of 24.5 N • m is applied to a grindstone whose moment of inertia is 0.130 kg · m². Using energy principles, and neglecting friction, find the angular speed after the grindstone has made 14.0 revolutions. Hint: the = FAx = mv? -mv,? is W, 2 angular equivalent of W, = tA0 = 2 You should convince yourself that this last relationship is correct. (Assume the grindstone starts from rest.) net net 2 2 2 rev/s еВook
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