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If we are in the frame of a pendulum (accelerating frame) it will have a centrifugal force and it's weight acting vertically downwards and the tension of the string acting vertically upwards. Now the force equation will be $$T=mg+\frac{mv^2}{R}$$(T=tension | m=mass of pendullum | v=velocity at its lowest point on the circle | R is the length of string(radius of circular motion)...

Now these forces are balanced in magnitude and opposite in direction so wouldn't the net force at the bottom be$=0$ and the net acceleration$=0$ at the bottom of the circular motion as a result? (There are no others forces acting to give it an acceleration hence why the net acceleration not$=0$?)

The net force along the vertical is zero hence the equation

this is the problem where in the solution it is considered that the bob has acceleration of $\frac{v^2}{R}$ at the bottom most point and not zero.How can this be explained??(https://i.stack.imgur.com/CNAXQ.jpg)

Solution to the above problem (https://i.stack.imgur.com/ajj5O.jpg)

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  • $\begingroup$ In some of the problems where they say the acceleration at the highest and lowest point are equal for some reason they consider g sin(theta)=v^2/r $\endgroup$
    – Rishab
    Feb 13, 2020 at 16:04
  • $\begingroup$ Forces in the perpendicular direction will be equal but you should consider the energy conservation law since you drop the body at some height and it will have a velocity at the bottom. $\endgroup$
    – Monopole
    Feb 13, 2020 at 16:08
  • $\begingroup$ The net acceleration at the lowest point is equal to zero. $\endgroup$
    – Sam
    Feb 13, 2020 at 16:12
  • $\begingroup$ At the bottom your acceleration will be radial but tangential component of your acceleration will be zero. $\endgroup$
    – Monopole
    Feb 13, 2020 at 16:25
  • $\begingroup$ @ monopole if there is velocity doesn't mean it has an acceleration.....and if there is no net force there is no acceleration so your point is invalid $\endgroup$
    – Rishab
    Feb 13, 2020 at 16:36

3 Answers 3

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The drawing is incorrect. If the forces on the pendulum bob were balanced, there would be no net force on the pendulum bob, and its direction of motion at the bottom of the arc would be tangential to the circle (e.g., horizontal).

"Centripetal force" is a catchall term for some force that is causing circular motion. In this case, centripetal force is being caused by tension in the string. This means that "T" should be shown on the drawing to represent tension in the string, but centripetal force should NOT be shown on the drawing. Naturally, this leads to the following equation when the pendulum bob is at the bottom of the arc:

$T = \frac{mv^2}{r} + mg$

which means that the maximum force on the string occurs when the pendulum bob is at the bottom of the arc, and this force is pointing towards the center of the circle that the pendulum bob is swinging through.

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  • $\begingroup$ Yes I have the same equation....I have considered it in an accelerating frame and I have considered centrifugal force not centripetal $\endgroup$
    – Rishab
    Feb 13, 2020 at 16:39
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    $\begingroup$ Centrifugal force is a pseudo force that exists in an accelerating frame of reference. I find it less confusing to draw free body diagrams and work problems in inertial frames. $\endgroup$ Feb 13, 2020 at 16:42
  • $\begingroup$ And yes the velocity at the bottom point is tangential until it slightly crosses that point and acceleration takes over but at the bottom point net force should be zero and hence the acceleration right?? $\endgroup$
    – Rishab
    Feb 13, 2020 at 16:42
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    $\begingroup$ If you were riding on the bob, you would feel a net force at the bottom of the arc. When you quickly turn through a circular corner in your car, you feel a net force on you in your frame of reference, which you call centrifugal force. And note - this isn't "my opinion" ... I taught high school physics for 13 years, including 10 years of AP Physics C. The AP Physics C graders are looking for the concepts that I posted in my answer, and will not award points for showing centripetal force on a free body diagram. $\endgroup$ Feb 13, 2020 at 16:57
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    $\begingroup$ The "force" shown in the diagram is the centrifugal (pseudo) force in the frame of the bob which is undergoing a centripetal acceleration produced by the tension in the string. Sitting on the bob (frame of reference of the bob) makes no difference to the value of the tension in the string which has a value in excess of $mg$ the excess force being the force causing the centripetal acceleration of the bob as observed in the lab frame. In the bob's frame the bob is not acceleration so to make Newton's laws be valid in that frame the centrifugal force is added resulting on net force on the bob. $\endgroup$
    – Farcher
    Feb 13, 2020 at 17:52
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In the frame of the bob ,the acceleration of the bob at its lower point is zero.

So in the frame of the bob there is a force acting on it due to the tension in the string which has a value in excess of $mg$ which in the laboratory frame explains why the bob undergoes a centripetal acceleration. $T (= mg + F') -mg = \frac{mv^2}{r} \Rightarrow F' = \frac{mv^2}{r}$.

In the frame of the bob ,the bob is not accelerating so if Newton's laws are to be used there must be no net force on the bob.

To make the net force zero so that Newton's laws of motion work in the accelerated frame of the bob, an additional (pseudo/fictitious) force is added which in this sort of example is called the centrifugal force which has the same magnitude as the force causing the centripetal acceleration in the laboratory frame but is opposite in direction. Rather than label that fictitious (cenrifugal) force $F'$ in the diagram it is labelled $ \frac{mv^2}{r}$.

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  • $\begingroup$ I did add a pseudo force ,so is the net force zero at the lowest point Including the pseudo force?? Sir u didn't get my question...I asked if the net acceleration is zero at the lowest point $\endgroup$
    – Rishab
    Feb 14, 2020 at 0:59
  • $\begingroup$ In the third paragraph I stated that the acceleration in the frame of the bob is zero. $\endgroup$
    – Farcher
    Feb 14, 2020 at 8:17
  • $\begingroup$ Who is they? Is this from a book? $\endgroup$
    – Farcher
    Feb 14, 2020 at 8:54
  • $\begingroup$ could you please include the solution of the problem I have posted ,in you answer please $\endgroup$
    – Rishab
    Feb 14, 2020 at 10:45
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With respect to the ground (which is an idealised inertial frame of reference), the pendulum is accelerated towards the center (the hinge). Whenever a particle is in circular motion (may or may not be uniform), it is said to be accelerated towards the center of motion (along with some tangential acceleration in non-uniform circular motion). This is because at every successive instant (an infinitesimally small period of time), the particle is changing direction. Otherwise, it won’t be in circular motion. And since, the motion of a pendulum is non-uniform circular motion, it is accelerated at every point of its path.

This acceleration is provided by Tension in the string, i.e. it doesn’t only balance the bob’s weight, but also provides the centripetal acceleration. This extra tension arises as the Bob is trying to stretch the string but in ideal cases we assume the string to be inextensible. Even in a practical situation, even a small extension may cause great tension in many strings due to their elastic properties.

However, you can say that the bob’s tangential acceleration is zero at the bottom most point, since there are no forces on it in the direction tangential to motion at that point.

But, if you see the bob with respect to itself, it would be at rest at every point of its motion. Just try running around while watching your hand (keep it stationary with respect to yourself), you will feel that it is not moving while it is in fact moving with respect to your floor or ground (depending on where you are). This is because your hand is moving with you.

In the same way, if you see the bob with respect to itself, it would seem to be at rest (it is at rest with respect to itself).

NOTE: I believe that bob is the technical term for the mass suspended in a simple pendulum (one that you have described). That is why I have used it in my answer.

I hope this answers your question.

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