enter image description here

Based on intuition, the lesser the velocity the bob is initially given, the sooner will the string slack. I don’t understand how the velocity the bob is given initially is related to tension being created in the string and the bob’s circular path.


closed as off-topic by Norbert Schuch, tpg2114 May 14 at 21:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Norbert Schuch, tpg2114
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Even if the bob is not given an initial velocity, there'd be tension in the string due to the mass of the bob that acts downwards and an equal force, i.e., the tension on the string acts upwards due to the bob being suspended from a suspension point. $\endgroup$ – Tapi May 14 at 17:25
  • $\begingroup$ @Brenda Do you understand about centripetal force? $\endgroup$ – Shivansh J May 14 at 17:25
  • $\begingroup$ @Brenda I suppose the ball is hinged (fixed) about the dotted point? $\endgroup$ – Shivansh J May 14 at 17:27
  • $\begingroup$ @ShivanshJ It is fixed. I understand that it is what enables circular motion. $\endgroup$ – Brenda May 14 at 17:31

One end of the string is affixed to the bob, and the other is affixed to a point. Lets first experiment by giving the bob an instantaneous velocity directly downward, not being able to go downward on account of the string, bob decelerates to 0, the former kinetic energy of the bob is stored as potential energy expressed as tension on the string, and subsequently imparted back into the bob as upward velocity. Presuming a "Perfectly elastic string" (one which cannot be stretched, and whose length is constant) none of the energy would be lost.

Now, in the example we give the bob instantaneous velocity perpendicular to the string. As the bob travels forward the string would be stretched if allowed, hence, the tension on the string. The energy from the forward deceleration of the bob is stored as potential energy expressed as tension on the string, and then released as acceleration of the bob along the string toward the point. In a continuous fashion the bob's trajectory constantly changes such that the bob traverses the arc defined by the point and the length of string, with velocity tangent to that arc, and at every instant where the trajectory changes, the kinetic energy on the perpendicular axises are exchanged through the tension of the string, as a potential energy conduit.

  • 1
    $\begingroup$ The velocity the kinetic energy of the ball is stored as tension on the string [...] That sentence really doesn't rock. Edit? $\endgroup$ – Gert May 14 at 19:50
  • $\begingroup$ I've edited it a bit more than suggested, I hope it's more comprehensible. I am not a blessed technical writer. $\endgroup$ – SteamyThePunk May 15 at 4:07
  • $\begingroup$ @PM2Ring I disagree, an inelastic string does NOT stretch, thus does not deform. The phrasing in the example is careful to talk about situations where the string "would be stretched if allowed" but as it is inelastic, it CANNOT be stretched. This was merely to say that where you might imagine the string might stretch if allowed, is when there is tension on the string. $\endgroup$ – SteamyThePunk May 15 at 4:36
  • $\begingroup$ @PM2Ring I will edit it. $\endgroup$ – SteamyThePunk May 15 at 4:44

Not the answer you're looking for? Browse other questions tagged or ask your own question.