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Tangential and Centripetal Acceleration

  • Thread starter Ataman
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  • #1
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Again, I was having trouble with something that I think I resolved, but I still lack the confidence in these problems, so I am asking people to check my solution.

Homework Statement


A pendulum of mass M with a massless string length L is released from the horizontal. Theta initial is 90 degrees, and the bob swings to the left. Find the acceleration of the bob, as a function of the angle the string makes with the vertical. You will have to take into account both the tangential and centripetal accelerations.

Homework Equations


[tex]\vec{F}_{_{NET}}=m\vec{a}[/tex]
[tex]\vec{a}=\vec{a}_{_{C}}+\vec{a}_{_{T}}[/tex]
[tex]W = \Delta E[/tex]

The Attempt at a Solution


I set up the coordinate system in a non-rotated frame, simply to keep track of the angle of the acceleration at any point in time. The tension force is unkown, but we always know the direction of the force, and that the centripetal acceleration is always in the direction of the tension. The tangential acceleration is always perpendicular to the centripetal acceleration, so its only the magnitude we need to find out.

Let +i point to the left (the direction in which the bob is swinging), and let +j point up for the sake of simplicity.

Using energy to find out the speed of the bob as a function of the angle:

System: bob, earth

[tex]W = \Delta E[/tex]

[tex]W_{_{T}} = \Delta KE_{_{bob}}+\Delta U_{_{G}}[/tex]

[tex]0 = KE_{_{f}}-KE_{_{i}}-mg\Delta h[/tex]

[tex]0 = KE_{_{f}}-0-mg\Delta h[/tex]

[tex]mg\Delta h = KE_{_{f}}[/tex]

Using simple trigonometry, finding h isn't a hassle:
[tex]mgL\cos\theta = \frac{1}{2}mv_{_{f}}^2[/tex]
[tex]2gL\cos\theta = v_{_{f}}^2[/tex]

Using Newton's Second Law to come up with the accelerations. We always know the direction of the tangential and centripital acceleration, but solving for the magnitude is tricky.

[tex]\vec{F}_{_{NET}}=m\vec{a}[/tex]
[tex]\vec{F}_{_{G}}+\vec{F}_{_{T}}=m\left(\vec{a}_{_{C}}+\vec{a}_{_{T}}\right)[/tex]

[tex]\hat{i}:F_{_{GX}}+F_{_{TX}}=m\left(a_{_{CX}}+a_{_{TX}}\right)[/tex]
[tex]\hat{i}:F_{_{T}}\sin\theta=m\left(a_{_{C}}\sin\theta+a_{_{T}}\cos\theta\right)[/tex]
[tex]\hat{i}:F_{_{T}}=\frac{m\left(a_{_{C}}\sin\theta+a_{_{T}}\cos\theta\right)}{\sin\theta}[/tex]

[tex]\hat{j}:F_{_{GY}}+F_{_{TY}}=m\left(a_{_{CY}}-a_{_{TY}}\right)[/tex]
[tex]\hat{j}:-F_{_{G}}+F_{_{T}}\cos\theta=m\left(a_{_{C}}\cos\theta-a_{_{T}}\sin\theta\right)[/tex]
[tex]\hat{j}:-mg+F_{_{T}}\cos\theta=m\left(a_{_{C}}\cos\theta-a_{_{T}}\sin\theta\right)[/tex]
[tex]\hat{j}:F_{_{T}}=\frac{m\left(a_{_{C}}\cos\theta-a_{_{T}}\sin\theta\right)+mg}{\cos\theta}[/tex]

We can combine the results from the i and j components with the magnitude of the velocity that we solved using energy. We continue by setting the (unknown) tension forces equal.

[tex]F_{_{T}}=F_{_{T}}[/tex]
[tex]\frac{m\left(a_{_{C}}\sin\theta+a_{_{T}}\cos\theta\right)}{\sin\theta}=\frac{m\left(a_{_{C}}\cos\theta-a_{_{T}}\sin\theta\right)+mg}{\cos\theta}[/tex]

Cross multiplying, and canceling the m's:

[tex]a_{_{C}}\sin\theta\cos\theta+a_{_{T}}\cos^2\theta=a_{_{C}}\cos\theta\sin\theta-a_{_{T}}\sin^2\theta+g\sin\theta[/tex]
[tex]a_{_{T}}\cos^2\theta=-a_{_{T}}\sin^2\theta+g\sin\theta[/tex]
[tex]a_{_{T}}\cos^2\theta+a_{_{T}}\sin^2\theta=g\sin\theta[/tex]
[tex]a_{_{T}}=g\sin\theta[/tex]

This goes to show the magnitude of the tangential acceleration is only dependant on the angle, and is independant of the centripetal acceleration.

So
[tex]\vec{a}_{_{T}}=a_{_{TX}}\hat{i}+a_{_{TY}}\hat{j}[/tex]
[tex]\vec{a}_{_{T}}=a_{_{T}}\cos\theta\hat{i}+a_{_{T}}\sin\theta\hat{j}[/tex]
[tex]\vec{a}_{_{T}}=g\sin\theta\cos\theta\hat{i}+g\sin^2\theta\hat{j}[/tex]

Speaking of centripetal acceleration, we solved for the speed of the bob as a function of the angle using energy, and we know the direction is always towards the center.

[tex]2gL\cos\theta = v_{_{f}}^2[/tex]
[tex]a_{_{C}} = \frac{v^2}{r}[/tex]
[tex]a_{_{C}} = 2gL\cos\theta[/tex]

So
[tex]\vec{a}_{_{C}}=a_{_{CX}}\hat{i}+a_{_{CY}}\hat{j}[/tex]
[tex]\vec{a}_{_{C}}=a_{_{C}}\sin\theta\hat{i}+a_{_{C}}\cos\theta\hat{j}[/tex]
[tex]\vec{a}_{_{C}}=2gL\cos\theta\sin\theta\hat{i}+2gL\cos^2\theta\hat{j}[/tex]

Now we can find the resultant acceleration through vector addition of the centripetal and tangential accelerations:

[tex]\vec{a}=\vec{a}_{_{C}}+\vec{a}_{_{T}}[/tex]
[tex]\vec{a}=2gL\cos\theta\sin\theta\hat{i}+2gL\cos^2\theta\hat{j}+g\sin\theta\cos\theta\hat{i}+g\sin^2\theta\hat{j}[/tex]

Cleaning it up:

[tex]\vec{a}=\left(2gL+g\right)\cos\theta\sin\theta\hat{i}+\left(2gL\cos^2\theta+g\sin^2\theta\right)\hat{j}[/tex]

-Ataman
 

Answers and Replies

  • #2
Shooting Star
Homework Helper
1,977
4
The first check for such equations should be the consistency of the dimensions of each term. On the RHS, 2gL does not have the dimension of accn. I think the mistake starts from this step:

[tex]a_{_{C}} = 2gL\cos\theta[/tex]
 

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