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I'm doing some revision using a study guide before the semester starts up again, and have gotten stuck on part of a question. The question I’m having trouble with is as follows:

A rock of mass $m = 1.27kg$ is tied to a string and spun in a circle as it slides on a frictionless horizontal surface. The radius of the circle the rock follows is $r = 1.04m$. At a given moment, the string lies along the direction of the arrow in the diagram when the rock is in the position shown. The magnitude of the tension in the string is $T = 18.1N$. What are the speed and rate of change of the speed of the rock at that moment?

enter image description here

Important equations:

  • $\text{Tangential force} = T*\sin(\theta)$
  • $\text{Radial force} = T*\cos(\theta)$
  • $\text{Tangential acceleration} = dv/dt$
  • $\text{Radial acceleration} = v^2/r$

Using these equations you can find the velocity and acceleration, but the question doesn’t supply the angle needed to use these. Working backwards from the known solutions given in the example, $v = 3.82\text{m/s}$ and $a = 2.47\text{m/s}^2$, gives the angle to be about 10 degrees, but there has to be a way to find this using only the given values of $T$, $m$ and $r$.

  • Can anyone suggest ways to get the angle? Every way I think of either leads to incorrect answers or requires already knowing the answer to the question.
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  • $\begingroup$ you should provide the full question so that we can see all the information provided. As it stands, its unclear what the question actually is about. $\endgroup$ – Walter Jan 11 '16 at 8:30
  • $\begingroup$ Ah ok, edited to add full question $\endgroup$ – Tycho 9000 Jan 11 '16 at 12:56
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(1) $v = a_T t = \dfrac{T\sin\theta}{m} t$

(2) $a_R = \dfrac{v^2}{r}$

(3) $a_R = \dfrac{T\cos\theta}{m}$

From these you can conclude that $\dfrac{T^2\sin^2\theta}{m^2r} t^2 = a_R = \dfrac{T\cos\theta}{m}$

Solving for $\theta$ yields a mess you can find using WolframAlpha.

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  • $\begingroup$ What is t in your solution? $\endgroup$ – M. Enns Oct 31 '18 at 3:22
  • $\begingroup$ @M.Enns It's time $\endgroup$ – PiKindOfGuy Oct 31 '18 at 3:38
  • $\begingroup$ Time since when? The only way your first equation makes sense is if the tangential acceleration is constant and the rock starts at rest at *t*=0. I don't think either of those are valid assumptions for this problem. $\endgroup$ – M. Enns Oct 31 '18 at 12:39
  • $\begingroup$ @M.Enns I may have misunderstood the question, but I double checked my answer and I still think it's good. $\endgroup$ – PiKindOfGuy Nov 1 '18 at 3:56
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To me it really looks like this problem is underspecified.

By considering the centripetal acceleration we can get an expression for the speed as a function of $\theta$ (in addition to mass, radius and tension) and we can get an expression of the rate of change of speed as a function of the speed and the $\theta$. (Details below)

So there are different solutions for speed and rate of change of speed depending on the angle. The first one that comes to mind is that you could specify the angle to be zero and the problem simplifies to a standard uniform circular motion situation.

Starting with the radial component of the tension

$$F_r=F\cos \theta$$

Which gives us the centripetal acceleration

$$a_c= \frac {F_r}{m}=\frac{mv^2}{r}$$

So,

$$F \cos \theta = \frac {mv^2}{r}$$ $$ v=\sqrt {\frac{rF \cos \theta}{m}}$$

We can also find a relationship between the speed and the tangential component of the acceleration (which I'll represent as $v\prime$).

The tangential component of the tension is $$F_t=F\sin \theta$$

Dividing this by the mass gives the tangential component of the acceleration (which the problem refers to as the rate of change of speed). $$F \sin \theta =m v\prime$$

So, $$\frac{F \sin \theta}{F \cos \theta}=\frac{v \prime r}{v^2}$$ $$\tan \theta = \frac {v \prime r}{v^2}$$

or $$v\prime = \frac{v^2\tan \theta}{r}$$

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