# How do I find the total acceleration in a pendulum?

#### Chemist116

The problem is as follows:

The diagram from below shows a pendulum. It is known that the magnitude of the tension of the wire is five times of that the weight of the sphere which is attached to. Acording to this information indicate True or False to the following statements:

I. The magnitude of the centripetal force which acts on the instant shown is $5mg$
II. The magnitude of the tangential force which acts on the instant shown is $0.6mg$
III. The magnitude of the total acceleration in the instant shown is $3\sqrt{2}g$.

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&FFF\\ 2.&FTT\\ 3.&FTF\\ 4.&TTF\\ 5.&TFT\\ \end{array}$

From my attempt I could spot is the centripetal force would be:

$F_{c}=\frac{mv^2}{R}$

The problem mentions:

$T=5mg$

From the sketch I made (and if my interpretation is correct)

Then:

$T=5mg=f_{c}$

Therefore the first statement is True.

But I'm stuck there. How can I find the other two?. Can somebody help me here?. :help:

#### DarnItJimImAnEngineer

Hint: A better coordinate system would be radial and tangential, not horizontal and vertical.

Also, don't forget that the weight has a component in both directions.

#### Chemist116

Hint: A better coordinate system would be radial and tangential, not horizontal and vertical.

Also, don't forget that the weight has a component in both directions.
Because I'm in a rush, I cannot use my drawing software at the moment.

However, following your suggestion, I'm getting in the tangential would be:

Now I'm assuming that $\sin 37^{\circ}=\frac{3}{5}$

$F_{t}=mg\sin 37^{\circ}= mg\frac{3}{5}$

Therefore the tangential yes it is $0.6mg$ so that must be true.

But how about the total acceleration?

The only thing that I could deduce is:

If the tangential force is $0.6 mg$ or $\frac{3}{5}mg$

Then:

$ma_{t}=\frac{3}{5}mg$

$a_{t}=\frac{3}{5}g$

and the centripetal acceleration would be:

For this part I'm assuming:

$T-mg\cos 37^{\circ}=F_{c}$

$5mg-\frac{4}{5}mg=F_{c}$

$F_{c}=\frac{21}{5}mg$

$F_{c}=ma_{c}=\frac{21}{5}mg$

$a_{c}= \frac{21}{5}g$

Therefore the total acceleration would be:

$a_{total}=\sqrt{a_{c}^2+a_{t}^2}$

$a_{total}=\sqrt{\left(\frac{3}{5}g\right)^2+\left(\frac{21}{5}g\right)^2}$

$a_{total}=3\sqrt{2}$

Therefore the third option is also true.

I hope I'm doing what you mentioned. Can you verify whether what I'm doing is right?

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