contestada

A simple pendulum has a bob of mass M. The bob is on a light string of length L. The string is fixed at C. At position A, the string is horizontal and the bob is at rest. The bob is released from A and swings to B, where the string is vertical. The tension in the string when the bob first reaches B is...?

The answer is 3 mg, but I don't know how that was arrived at. Can someone explain this to me?

Respuesta :

whitneytr12

Answer:

[tex]T=3Mg[/tex]

Explanation:

Let's apply the conservation of energy between points A and B.

We just have gravitational potential energy at point A and just kinetic energy at point B.

[tex]MgL=\frac{1}{2}Mv^{2}[/tex] (1)

Now, let's apply force free-body diagram at point B, we have the tension upwards and weight downwards. Now using Newton's second law we have:

[tex]\sum F=Ma_{c}[/tex]

As we have circular motion, we use centripetal acceleration here (a(c)=v²/R). The forces acting are:

[tex]T-W_{bob}=Ma_{c}=M\frac{v^{2}}{L}[/tex], because the radius R is L.

So let's solve this equation for T.

[tex]T=W_{bob}+M\frac{v^{2}}{L}=Mg+M\frac{v^{2}}{L}[/tex] (2)

Using equation (1) we can find v².

[tex]v^{2}=2gL[/tex]

Finally let's put this value in the equation (2).

[tex]T=Mg+M\frac{2gL}{L}[/tex]

[tex]T=3Mg[/tex]

I hope it helps you!

Otras preguntas