# Circular Motion

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Hi,

I cant understand the following concept regarding circular motion. My understanding tells me that centripetal acceleration must be present at every point of the circle for circular motion to take place. For example, if we have two points which are close, lets suppose P and Q. I have attached a diagram as well.

At P (the bottom most point)

u = initial velocity (Initial Velocity)

K.E = Max (Kinetic Energy)

P.E = 0 (Potential Energy)

ac= r*omega^2 (centripetal acceleration)

at = r*d^2theta/dt^2 (tangential acceleration)

At Q ,

v = 0 (Final Velocity)

K.E = 0 (Kinetic Energy)

P.E = Max (Potential Energy)

ac= 0 (centripetal acceleration)

at = r*d^2theta/dt^2 (tangential acceleration)

Furthermore, for Point Q, as " v = 0 ", centripetal acceleration is zero (i.e. ac = 0 because v = r*omega & v = 0), although tangential acceleration still exists. Wouldn't we require some centripetal acceleration for the particle to move from P to Q? If yes, why is centripetal acceleration zero?. Can someone please elaborate the situation? Thanks for helping.

I cant understand the following concept regarding circular motion. My understanding tells me that centripetal acceleration must be present at every point of the circle for circular motion to take place. For example, if we have two points which are close, lets suppose P and Q. I have attached a diagram as well.

At P (the bottom most point)

u = initial velocity (Initial Velocity)

K.E = Max (Kinetic Energy)

P.E = 0 (Potential Energy)

ac= r*omega^2 (centripetal acceleration)

at = r*d^2theta/dt^2 (tangential acceleration)

At Q ,

v = 0 (Final Velocity)

K.E = 0 (Kinetic Energy)

P.E = Max (Potential Energy)

ac= 0 (centripetal acceleration)

at = r*d^2theta/dt^2 (tangential acceleration)

Furthermore, for Point Q, as " v = 0 ", centripetal acceleration is zero (i.e. ac = 0 because v = r*omega & v = 0), although tangential acceleration still exists. Wouldn't we require some centripetal acceleration for the particle to move from P to Q? If yes, why is centripetal acceleration zero?. Can someone please elaborate the situation? Thanks for helping.

Last edited by Tesla3; 7 months ago

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#2

https://www.physicsclassroom.com/cla...endulum-Motion

Is pretty good, in particular the velocity - acceleration animation 1/2 way down.

Is pretty good, in particular the velocity - acceleration animation 1/2 way down.

Last edited by mqb2766; 7 months ago

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(Original post by

https://www.physicsclassroom.com/cla...endulum-Motion

Is pretty good, in particular the velocity - acceleration animation 1/2 way down.

**mqb2766**)https://www.physicsclassroom.com/cla...endulum-Motion

Is pretty good, in particular the velocity - acceleration animation 1/2 way down.

Last edited by Tesla3; 6 months ago

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#4

(Original post by

ummmmmm, still confused. The animation is showing that there is no centripetal acceleration at the point where velocity is zero although there is some tangential acceleration. So as there is no centripetal acceleration at the point where " v = 0 ", wouldn't it contradict with our statement that there must always be some centripetal acceleration?

**Tesla3**)ummmmmm, still confused. The animation is showing that there is no centripetal acceleration at the point where velocity is zero although there is some tangential acceleration. So as there is no centripetal acceleration at the point where " v = 0 ", wouldn't it contradict with our statement that there must always be some centripetal acceleration?

Im not sure what (our) statement you're referring to?

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(Original post by

The linear and angular velocities are instantaneously zero, so the centripetal acceleration is instantaneously zero.

Im not sure what (our) statement you're referring to?

**mqb2766**)The linear and angular velocities are instantaneously zero, so the centripetal acceleration is instantaneously zero.

Im not sure what (our) statement you're referring to?

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#6

From about 1/4 of the way down (with diagram)

First, observe the diagram for when the bob is displaced to its maximum displacement to the right of the equilibrium position. This is the position in which the pendulum bob momentarily has a velocity of 0 m/s and is changing its direction. The tension force (Ftens) and the perpendicular component of gravity (Fgrav-perp) balance each other. At this instant in time, there is no net force directed along the axis that is perpendicular to the motion. Since the motion of the object is momentarily paused, there is no need for a centripetal force.

The two forces add to give a tangential acceleration only.

First, observe the diagram for when the bob is displaced to its maximum displacement to the right of the equilibrium position. This is the position in which the pendulum bob momentarily has a velocity of 0 m/s and is changing its direction. The tension force (Ftens) and the perpendicular component of gravity (Fgrav-perp) balance each other. At this instant in time, there is no net force directed along the axis that is perpendicular to the motion. Since the motion of the object is momentarily paused, there is no need for a centripetal force.

The two forces add to give a tangential acceleration only.

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(Original post by

From about 1/4 of the way down (with diagram)

First, observe the diagram for when the bob is displaced to its maximum displacement to the right of the equilibrium position. This is the position in which the pendulum bob momentarily has a velocity of 0 m/s and is changing its direction. The tension force (Ftens) and the perpendicular component of gravity (Fgrav-perp) balance each other. At this instant in time, there is no net force directed along the axis that is perpendicular to the motion. Since the motion of the object is momentarily paused, there is no need for a centripetal force.

The two forces add to give a tangential acceleration only.

**mqb2766**)From about 1/4 of the way down (with diagram)

First, observe the diagram for when the bob is displaced to its maximum displacement to the right of the equilibrium position. This is the position in which the pendulum bob momentarily has a velocity of 0 m/s and is changing its direction. The tension force (Ftens) and the perpendicular component of gravity (Fgrav-perp) balance each other. At this instant in time, there is no net force directed along the axis that is perpendicular to the motion. Since the motion of the object is momentarily paused, there is no need for a centripetal force.

The two forces add to give a tangential acceleration only.

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#8

(Original post by

Sorry to bother again, I want to ask that whether we would need centripetal acceleration to convert a velocity vector " v = 0.0001 " to a null vector " v = 0 ". Clearly the direction is getting changed, so I think that we would need centripetal acceleration for that. Is it ok?

**Tesla3**)Sorry to bother again, I want to ask that whether we would need centripetal acceleration to convert a velocity vector " v = 0.0001 " to a null vector " v = 0 ". Clearly the direction is getting changed, so I think that we would need centripetal acceleration for that. Is it ok?

You seem to be unsure about things that happen instantaneously. The centripetal acceleration is instantaneously zero at that point. Its not zero before or after. The tangential acceleration is not zero at that point.

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(Original post by

Have you read the previous linked document properly?

You seem to be unsure about things that happen instantaneously. The centripetal acceleration is instantaneously zero at that point. Its not zero before or after. The tangential acceleration is not zero at that point.

**mqb2766**)Have you read the previous linked document properly?

You seem to be unsure about things that happen instantaneously. The centripetal acceleration is instantaneously zero at that point. Its not zero before or after. The tangential acceleration is not zero at that point.

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#10

(Original post by

wait, just let me think again..........

**Tesla3**)wait, just let me think again..........

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(Original post by

The instantaneous net force is tangential. The centripetal acceleration must be instantaneously zero.

**mqb2766**)The instantaneous net force is tangential. The centripetal acceleration must be instantaneously zero.

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#12

(Original post by

I get that instantaneous net force is tangential. ....k so that means it just becomes zero the moment v = 0, there is no question at all of before or after perhaps....

**Tesla3**)I get that instantaneous net force is tangential. ....k so that means it just becomes zero the moment v = 0, there is no question at all of before or after perhaps....

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