Time it takes for the wheel to stop?

  • Thread starter sona1177
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  • #1
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A 34-cm-diameter potter's wheel with a mass of 20 kg is spinning at 180 rpm. Using her hands, a potter forms a pot, centered on the wheel, with a 14 cm diameter. Her hands apply a net friction force of 1.3 N to the edge of the pot. If the power goes out, so that the wheel's motor no longer provides any torque, how long will it take for the wheel to come to a stop in her hands?

Angular acceleration= net Torque/moment of inertia so

(1.3)(.07)/ (20 *.07^2)= .93 radians/second^2

So since angular acceleration=change in angular Velocity/time then:

-.98 = (0-19.8)/change in time so t=19.2 seconds.

Where the initial angular velocity was 19.8 by converting the 180 rev/min to 19.8 radians/sec.

So is my answer of 19.2 seconds correct?

Thank you kindly for your help.
 

Answers and Replies

  • #2
rl.bhat
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Check the moment of inertia of the wheel whose diameter is 34 cm.
 
  • #3
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So that's MR^2 which is 20 * .17^2= .578 but how does this help me?
 
  • #4
rl.bhat
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So that's MR^2 which is 20 * .17^2= .578 but how does this help me?

angular acceleration α = (1.3)(.07)/ (20 *.17^2)= ..... radians/second^2

ω = ωο - α*t

In the problem ω = 0 and ωο = 6*π radians/second.

Now find t.
 
  • #5
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Thank you but she applies the force on the pot not on the wheel so this is why I am confused.
 
  • #6
rl.bhat
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Thank you but she applies the force on the pot not on the wheel so this is why I am confused.
Torque depends on the point of application of the force. Therefore the torque is F*r, where r is the radius of the pot. This torque rotates the wheel with an angular acceleration such that F*r = I*α.
 
  • #7
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Thank you so much for taking the time to help me! :)
 
  • #8
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My answer is 118 seconds. Is that correct? Again, thank you so much for taking the time to help me! :)
 
  • #9
rl.bhat
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My answer is 118 seconds. Is that correct? Again, thank you so much for taking the time to help me! :)
I am getting 119.7 seconds.
 

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