Circular motion-centripetal acceleration

In summary, the conversation is about a problem with using a formula to calculate the change of velocity in circular motion. The formula in question is delta(v) = v * delta(T). However, it is argued that this formula may not be accurate when delta(T) is not approaching zero. The correct formula to use in this case is v = r * delta(theta), where r is the circle radius and theta is the angle. There is confusion about why the original formula is not accurate and how it relates to the correct formula.
  • #1
allok
16
0
hi

This latex code is giving me some problems. I write one thing, it displays something completely different

In circular motion velocity only changes direction but not size

change of velocity - [tex]\Delta v[/tex]
Change of angle - [tex]\Delta T[/tex]
Velocity - V
Centripetal acceleration - a

delta(V) = V * delta(T)

When delta(T) approaches its limit (goes to zero), change of velocity has same direction as acceleration vector?

We compute the magnitude of velocity change with :

Delta(v) = v * Delta(T)

I see this being true when change of angle approaches its limit ( goes to zero ), since then length of circular arc ( with radius begin velocity vector ) equals [tex]\Delta v[/tex]. But that is not true if delta(T) is not approaching limit. So how can we use formula

delta(v) = V * delta(T)

in cases were delta(T) is not approaching zero, since I assume length of circle arc is quite different than delta(T) if delta(T) doesn't go to zero?

cheers
 
Last edited:
Physics news on Phys.org
  • #2
I think your original assumption may be incorrect. If you want to relate the linear velocity of the object on the circle circumference with the change in angle, it would be:

[tex]s = r\theta[/tex]
[tex]\frac{ds}{dt} = r\frac{d\theta}{dt}[/tex]
[tex]v = r\frac{d\theta}{dt}[/tex]

where s is the circle arclength, r is the circle radius, and theta is the angle. The [tex]\frac{d\theta}{dt}[/tex] would be equal to your [tex]\Delta T[/tex].
 
  • #3
mezarashi said:
I think your original assumption may be incorrect. If you want to relate the linear velocity of the object on the circle circumference with the change in angle, it would be:

[tex]s = r\theta[/tex]
[tex]\frac{ds}{dt} = r\frac{d\theta}{dt}[/tex]
[tex]v = r\frac{d\theta}{dt}[/tex]

where s is the circle arclength, r is the circle radius, and theta is the angle. The [tex]\frac{d\theta}{dt}[/tex] would be equal to your [tex]\Delta T[/tex].

I don't get it. First of all, [tex]v = r\frac{d\theta}{dt}[/tex] is not same formula as delta(v)=v*delta(theta)

I still don't understand why delta(v) = v * delta(theta) would give us correct result when delta(theta) is anything but [tex]d\theta[/tex] ?
 

1. What is circular motion-centripetal acceleration?

Circular motion-centripetal acceleration is the acceleration experienced by an object moving in a circular path. It is always directed towards the center of the circle and is caused by the object's change in direction.

2. How is circular motion-centripetal acceleration calculated?

Circular motion-centripetal acceleration can be calculated using the formula a = v²/r, where a is the acceleration, v is the velocity, and r is the radius of the circle.

3. What is the difference between centripetal acceleration and centrifugal force?

Centripetal acceleration is the inward acceleration experienced by an object in circular motion, while centrifugal force is the outward force experienced by the object. Centrifugal force is not a real force, but rather an apparent force caused by inertia.

4. How does centripetal acceleration affect the speed of an object?

Centripetal acceleration does not affect the speed of an object, but rather the direction of its velocity. It causes the object to continuously change direction, but not speed, as it moves in a circular path.

5. What is the role of centripetal acceleration in everyday life?

Centripetal acceleration is responsible for many everyday phenomena, such as the motion of planets around the sun, the rotation of a ceiling fan, and the movement of a car around a curve. It is also important in sports, such as when a baseball player throws a curveball or a figure skater spins on the ice.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
890
  • Introductory Physics Homework Help
Replies
6
Views
909
  • Introductory Physics Homework Help
2
Replies
55
Views
512
  • Introductory Physics Homework Help
Replies
2
Views
805
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
924
  • Introductory Physics Homework Help
Replies
1
Views
267
  • Mechanics
Replies
30
Views
761
  • Introductory Physics Homework Help
Replies
16
Views
10K
  • Introductory Physics Homework Help
2
Replies
38
Views
1K
Back
Top