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- - **Determining torque required**
(*http://www.physicsforums.com/showthread.php?t=143427*)

Determining torque requiredProblem 2) What is the angular momentum of a 2.8 kg uniform cylindrical grinding wheel of radius 18 cm when rotating at 1500 rpm?
b) How much torque is required to stop it in 7.0 s? 1500 rpm * 2pi rad = 9424.78 rad/m (9424.78 rad/m) /60 s = 157.08 rad/s I = 1/2mr2 I = 1/2(2.8 kg)(.18 m)2 = 0.04536 kgm2 L = Iw = (0.04536 kgm2)(157.08 rad/s) = 7.13 kgm2/s I have determined the angular momentum, but I am unsure of how to determine the torque required to stop it in 7 s How do you determine this with the torque formula and time involved? Any help? |

The definition of torque is
[tex] \mathbf\tau = \frac{d\mathbf{\mathrm L}}{dt} [/tex] If you want to change the angular momentum from your value to 0, then the average torque over 7 s most then be? |

t=(7.13 kgm^2/s)/7.0 s=1.02 kgm^2
Better? |

hmm....well you know the initial angular velocity ''157.08 rad/s''. I would say you need to calculate the angular deceleration of the spinning wheel from 157.08 r/s to 0 r/s in 7 secs using W(final)=W(initial)+a^t. I make the angular deceleration 22.44 rad s-2.
Then using the formula T=I x a......cha ching ! |

w=w0+at
0=157.08+a(7) -157.08=7a a=-22.44 rad/s^2 since stopping T=Ia T=0.04536 kgm^2*-22.44 T=-1.02 m*N |

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