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A cylinder, on a level surface angularly accelerates as a string wrapped around is put under tension. Because this string causes the cylinder to roll without slipping, the tension force cannot to any work because it is applied to a point on the surface for a zero distance. How then, can the rotational kinetic energy increase if there are no sources of work.

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  • $\begingroup$ This question needs a diagram $\endgroup$ – Aaron Stevens Apr 18 '19 at 14:25
  • $\begingroup$ i cant post a picture $\endgroup$ – TheImproperIntegral Apr 18 '19 at 14:26
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    $\begingroup$ If you post a link to a picture, someone can edit the question to embed it. $\endgroup$ – ja72 Apr 18 '19 at 14:30
  • $\begingroup$ The torque of tension does work so it will change kinetic energy of the system. $\endgroup$ – user213933 Apr 18 '19 at 14:40
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In your question I think that you are asking that the point of application of tension doesn't slips on ground so work done by tension is zero(but it is not zero!) then why rotational K.E is increasing?

Take a look at the situation for a moment carefully then read the next paragraph.

See the tension produces a torque around the axis passing through cylinder's centre.The torque does work which can be calculated by the integral of dot product of torque and infinitesimal angular displacement.

By work energy theorem if torque does work then it must lead to change in kinetic energy of the cylinder.Hope this helps!

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  • $\begingroup$ The torque does work then it would change kinetic energy of the system. $\endgroup$ – user213933 Apr 18 '19 at 14:39
  • $\begingroup$ But the work energy theorem only includes forces- If the point that the force is applied doesn't displace, it can't do work by its definition. $\endgroup$ – TheImproperIntegral Apr 18 '19 at 14:40
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    $\begingroup$ it is the same logic behind friction not doing work such as in this question: physics.stackexchange.com/questions/158878/… $\endgroup$ – TheImproperIntegral Apr 18 '19 at 14:42
  • $\begingroup$ The force doesn't does work on that specific point but its torque produces angular acceleration which can change the kinetic energy of the system. It is important to be acquainted with the fact that torques doesn't does work only on the point of application but its torque does. If you will write the K.E then you have to write change in kinetic energy of that particular point only but when you will look closely its torque will change K.E of whole cylinder. $\endgroup$ – user213933 Apr 18 '19 at 14:44
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    $\begingroup$ @RoshanKlein-Seetharaman The point of contact is instantaneously at rest with respect to the ground, but the cylinder is still rotating as well as moving $\endgroup$ – Aaron Stevens Apr 18 '19 at 14:45

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