Calculating Torque on a Potter's Wheel

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To calculate the angular acceleration of a potter's wheel with a radius of 0.50 m and mass of 100.0 kg, the wheel's initial speed of 50.0 rev/min converts to approximately 5.24 rad/sec, leading to an angular acceleration of 0.87 rad/sec² when stopped in 6.0 seconds. For torque, the correct formula is torque = I * alpha, where I is the moment of inertia calculated as 0.5 * m * r², resulting in 12.5 kg·m² for this wheel. By multiplying the moment of inertia by the angular acceleration, the torque applied is found to be approximately 11 Nm. The confusion arose from initially using angular velocity instead of angular acceleration in the torque calculation. The final torque value aligns with the expected answer of 11 Nm.
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Homework Statement


A potter's wheel of radius 0.50 m and mass 100.0 kg is freely rotating at 50.0 rev/min. The potter can stop the wheel in 6.0 seconds by pressing a wet rag against the rim.
a. What is the angular acceleration of the wheel?
b. How much torque does the potter apply to the wheel?

Homework Equations


t=Ia
a=w/t

The Attempt at a Solution


a. (50/60*2*pi)=5.235988 rad/sec
5.235988/6=.87 rad/sec^2
b. I didnt understand this portion of the question. I tried .5^2*100*5.235988 but the answer was not 11 Nm which I know is the answer. Please help! Any help is greatly appreciated. Thank you.
 
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You want

torque = I * alpha

not I * omega, that is angular momentum. Check your calculation.

Also, if you are assuming the wheel is a solid cylindrical disk, then I = 0.5MR^2, so your above I value is not correct if that's the case.
 
Last edited:
So...in this case, i should go .5*100*.5^2=12.5?
But the answer given is 11 Nm. I don't understand. :(
 
chamonix said:
So...in this case, i should go .5*100*.5^2=12.5?
But the answer given is 11 Nm. I don't understand. :(

You didn't multiply by alpha! torque = I*alpha. If you do that you will get the right answer. You found the right value of alpha in part (a).
 
Oh, ok. My fault. Ok. Alpha.
so... .5*100*.5^2*.87=10.9=11Nm! ok. Thank you!
Sorry, I must have overlooked that alpha part. Thank you.
 
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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