- #1

- 322

- 45

- Homework Statement
- 56-kg student runs at grabs a hanging rope,

and swings out over a lake (Fig. 45). He releases the

rope when his velocity

is zero. (a) What is the angle when he releases the rope? (b) What is the tension in the rope just before he releases it? (c) What is the maximum tension in the rope?

- Relevant Equations
- ##a_n = \frac v^2R

Please tell me if I need to post my solution for this.., but I just have a question more or less 'conceptual' question about (c).

so I know that from Newton's 2nd law for centripetal acceleration --> ##F_{rope} - mgcos(\theta) = ma_n## where ##a_n = \frac {v^2}{R}## such that where the normal acceleration is maximal (i.e. also maximum velocity) is where there's max tension in the rope. But.. how do I find where the velocity is going to be maximum?

I would reason about it in terms of energy that in the beginning of the jump he only has kinetic energy after which ##KE## is changed in ##U_{grav}##. From this I know that the velocity is going to be maximum at the beginning of the jump and thus the tension in the rope as well.

Is this correct reasoning about this or can I look at it in another way?

Thanks in advance.

so I know that from Newton's 2nd law for centripetal acceleration --> ##F_{rope} - mgcos(\theta) = ma_n## where ##a_n = \frac {v^2}{R}## such that where the normal acceleration is maximal (i.e. also maximum velocity) is where there's max tension in the rope. But.. how do I find where the velocity is going to be maximum?

I would reason about it in terms of energy that in the beginning of the jump he only has kinetic energy after which ##KE## is changed in ##U_{grav}##. From this I know that the velocity is going to be maximum at the beginning of the jump and thus the tension in the rope as well.

Is this correct reasoning about this or can I look at it in another way?

Thanks in advance.

**[Mentor Note -- Update requested by OP later in the thread:]**v_{initial}= ##95km/h## and v_{final}= ##35km/h##

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