Torque and Net Work

  • Thread starter Mowgli
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  • #1
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Homework Statement



An old 33 1/3 rpm turntable may be approximated by a cylindrical disk of mass 0.250kg and a diameter of 31.0cm.

a.) What average torque will bring it up to operating speed in 1.50 seconds starting from rest?

b.) what net work is done on the disk in bringing it up to operating speed?

Homework Equations





The Attempt at a Solution



Can anyone help me get started on this? We never went over torque!
 

Answers and Replies

  • #2
collinsmark
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Can anyone help me get started on this? We never went over torque!
Really? :confused: Is this for a homework problem?

Well I guess a first step would be to read up on torque. You'll also want to acquaint yourself with the concept of "moment of inertia." As an interim step to this problem, you will need to find the moment of inertia of a 0.250kg cylindrical disk. Before you're finished, you'll also need to research the concept of rotational kinetic energy. :wink:
 
  • #3
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Is this correct?

mass m = 0.25 Kg
diameter d = 31 cm
radius r = 15.5 cm = 0.155 m
moment of inertia I = (1/2)mr^2 = 0.003 Kg m^2
(a)
ω1 = 0
n = 33 1/3 = 100/3 = 33.33 rpm = 0.5555 rps
ω2 = 2πn = 2*3.14*0.5555 = 3.48854 rad/s
t = 1.5 s
we know that
torque τ = I α = I (ω2 - ω1)/t = 0.00697708 N m
(b)
work done w = change in rotational kinetic energy
= (1/2)*I*(ω2)^2
= 0.5*0.003*3.48854*3.48854
= 0.0182 J
 
  • #4
collinsmark
Homework Helper
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Is this correct?

mass m = 0.25 Kg
diameter d = 31 cm
radius r = 15.5 cm = 0.155 m
moment of inertia I = (1/2)mr^2 = 0.003 Kg m^2
(a)
ω1 = 0
n = 33 1/3 = 100/3 = 33.33 rpm = 0.5555 rps
ω2 = 2πn = 2*3.14*0.5555 = 3.48854 rad/s
t = 1.5 s
we know that
torque τ = I α = I (ω2 - ω1)/t = 0.00697708 N m
(b)
work done w = change in rotational kinetic energy
= (1/2)*I*(ω2)^2
= 0.5*0.003*3.48854*3.48854
= 0.0182 J

'Looks okay to me. :approve:
 

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