Torque required to stop wheel in 41.2s?

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To determine the torque required to stop a ten-pound car wheel with a moment of inertia of 0.35 kgm², the wheel's angular velocity is first calculated at 101.56 rad/s. The necessary angular deceleration to halt the wheel in 41.2 seconds is found to be 2.465 rad/s². Using the formula for torque, the calculation results in a torque of approximately -0.8675 Nm. The discussion highlights the importance of proper unit notation, correcting "m x n" to "Nm." Overall, the participants engage in clarifying the calculations and ensuring accuracy in the solution process.
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Homework Statement


A ten pound car wheel has a moment of inertia of .35 kgm^2. While rotating at a constant speed the wheel makes 87.61 revolutions in 5.42s. How much torque is required to stop the wheel in 41.2 s?


Homework Equations





The Attempt at a Solution


87.61rev/5.42sec = 16.164 rev/sec x 2pi = 101.56 rad/s / 41.2 sec = 2.465 rad/s^2
t= .35 kgm^2 x -.2.465 = -.8675 m x n?

not sure how relevant all of the steps i did are but that is where i stopped... any help?
 
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welcome to pf!

hi 03sto! welcome to pf! :wink:

yes, that looks ok :smile: (except, "m x n" should be "Nm" or "N.m")

what is worrying you about it? :confused:​
 
Okay thanks for the feedback, i just figured i did it completely wrong haha. Thanks and yes i will switch the mn to nm. Thanks!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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