Forces at play in a pendulum (conceptual doubt)

[tellmesomething]

Homework Statement

In a pendulum which is not at its mean position why is there a centripetal acceleration? Considering there are only two forces acting on the bob (Tension and weight) wouldn't the component of weight which is anti parallel to the direction of tension be responsible for creating the tension in the string, so wouldn't that cancel out. And the only acceleration left should be tangential acceleration?

Relevant Equations

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[Orodruin]

Homework Statement: In a pendulum which is not at its mean position why is there a centripetal acceleration? Considering there are only two forces acting on the bob (Tension and weight) wouldn't the component of weight which is anti parallel to the direction of tension be responsible for creating the tension in the string, so wouldn't that cancel out. And the only acceleration left should be tangential acceleration?
Relevant Equations: .

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If there was only tangential acceleration the pendulum would move in a straight line.

Tension is created both from the weight component and the need to keep the pendulum moving in a circle.

 

[tellmesomething]

If there was only tangential acceleration the pendulum would move in a straight line.

Tension is created both from the weight component and the need to keep the pendulum moving in a circle

I see so the tension would be greater in magnitude than the component of weight ?

 

[Orodruin]

I see so the tension would be greater in magnitude than the component of weight ?

Yes.

Well, as long as $v \neq 0$.

 

[kuruman]

I see so the tension would be greater in magnitude than the component of weight ?

If you draw a free body diagram and use energy conservation, you can show that when a pendulum is released from rest at angle $\theta_0$ relative to the vertical, the tension as a function of $\theta$ as it swings is given by $$T=mg(3\cos\theta-2\cos\theta_0).$$The tension is equal to the weight in the trivial case $\theta =\theta_0=0$, i.e. when the pendulum is just hanging without moving . When the pendulum is moving, the tension is equal to the weight when $$\cos\theta=\frac{1+2\cos\theta_0}{3}.$$For example, when the pendulum is released at $\theta_0=60^{\circ}$, the tension is equal to the weight at $\theta=\arccos(2/3)=48.2^{\circ}$ and greater than the weight at smaller angles than that.

 

[DrBanana]

If you draw a free body diagram and use energy conservation, you can show that when a pendulum is released from rest at angle $\theta_0$ relative to the vertical, the tension as a function of $\theta$ as it swings is given by $$T=mg(3\cos\theta-2\cos\theta_0).$$The tension is equal to the weight in the trivial case $\theta =\theta_0=0$, i.e. when the pendulum is just hanging without moving . When the pendulum is moving, the tension is equal to the weight when $$\cos\theta=\frac{1+2\cos\theta_0}{3}.$$For example, when the pendulum is released at $\theta_0=60^{\circ}$, the tension is equal to the weight at $\theta=\arccos(2/3)=48.2^{\circ}$ and greater than the weight at smaller angles than that.

Isn't your first equation derived under the assumption that the centripetal acceleration is $\frac{v^2}{r}$? But that's only true when there's no tangential acceleration, which is not the case here.

Edit: Ok I suppose that the total acceleration vector can be resolved into two components: the centripetal and the tangential, in which case the magnitude of the centripetal one will just be $\frac{v^2}{r}$. Am I correct?

 

[jbriggs444]

Edit: Ok I suppose that the total acceleration vector can be resolved into two components: the centripetal and the tangential, in which case the magnitude of the centripetal one will just be $\frac{v^2}{r}$. Am I correct?

Yes. This is the explanation. The tangential acceleration component does not [immediately] affect the centripetal acceleration component.

 

[PeroK]

And the only acceleration left should be tangential acceleration?

The pendulum bob is constrained to move in a circular arc. The purpose of the string or cable holding the bob is to provide the centripetal force to make that happen. There must be tension in the string. Otherwise, the bob would simply fall under gravity.