Tangential and radial acceleration

In summary: We assume that the string cannot change its length. That means that any force applied to the end of the string is effectively canceled along the string; but its transverse component acts unopposed.So basically, the transverse component of gravity (ie, the force perpendicular to the string) is the only force acting on the ball at that instant.
  • #1
babysnatcher
91
0
A ball tied to the end of a string 0.50 m in length swings in a vertical circle under the influence of gravity. When the string makes an angle x= 20 degrees with the vertical, the ball has a speed of 1.5 m/s. Find the magnitude of the radial component of acceleration at this instant.

So i have a concept question. How do i know that tangential acceleration = gsin(20 degrees).

Basically, prove the general formula tangential acceleration= gsin(x) for this problem.
 
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  • #2
Draw a free body diagram on the ball at the instant specified in the problem (when it it makes an angle of 20 degrees with the vertical). Then decompose the forces into the radial and tangential directions. That's a good rule of thumb for when you are solving problems using Newton's 2nd law.
 
  • #3
WannabeNewton said:
Draw a free body diagram on the ball at the instant specified in the problem (when it it makes an angle of 20 degrees with the vertical). Then decompose the forces into the radial and tangential directions. That's a good rule of thumb for when you are solving problems using Newton's 2nd law.

we haven't read about Newton's 2nd law yet -- that's the next chapter.

there is already a drawning. i don't see why tangential acceleration=gsin(x)
 
  • #4
Well you don't really need it here, I was just saying it's a good rule of thumb in general. But do what I said: draw a free body diagram on the ball; what are the forces acting on the ball at that instant? How can you decompose these forces into radial and tangential directions?

If there's already a diagram, which part of it is troubling you?
 
  • #5
WannabeNewton said:
Well you don't really need it here, I was just saying it's a good rule of thumb in general. But do what I said: draw a free body diagram on the ball; what are the forces acting on the ball at that instant? How can you decompose these forces into radial and tangential directions?

If there's already a diagram, which part of it is troubling you?

im not there yet. I am on chapter 4, motions in two dimensions. The only force we think of is gravity.

nvm, a teacher helped me with it already. he just moved the radial acceleration vector all the way to the top so that they tip of that vector touches the top of the string. i see the right triangle now.
 
  • #6
We assume that the string cannot change its length. That means that any force applied to the end of the string is effectively canceled along the string; but its transverse component acts unopposed.
 
  • #7
voko said:
We assume that the string cannot change its length. That means that any force applied to the end of the string is effectively canceled along the string; but its transverse component acts unopposed.

I am not thinking about forces in this chapter yet, except for gravity.
hmmm, in g2=at2+y2, what is y2? and what is the physical meaning of this? y doesn't seem to be ar.

I mean pertaining to this triangle at=gsin(x).
 
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  • #8
babysnatcher said:
I am not thinking about forces in this chapter yet, except for gravity.

What I wrote fully applies to gravity.

hmmm, in g2=at2+y2, what is y2? and what is the physical meaning of this? y doesn't seem to be ar.

'y' could be the "radial" acceleration if the ball were allowed to fly freely. But it is not, so 'y', and the entire equation for that matter, does not make a lot of sense physically.

What does make sense is ## (mg)^2 = (ma_t)^2 + P_r^2 ##, where ##P_r## is the radial component of the weight.
 
  • #9
voko said:
What I wrote fully applies to gravity.
'y' could be the "radial" acceleration if the ball were allowed to fly freely. But it is not, so 'y', and the entire equation for that matter, does not make a lot of sense physically.

What does make sense is ## (mg)^2 = (ma_t)^2 + P_r^2 ##, where ##P_r## is the radial component of the weight.

If it does not make sense, then i am still confused about why at=gsin(x).

I haven't read about Mass/weight/Newtons laws yet so can you explain it with just what I know.
 
  • #10
babysnatcher said:
If it does not make sense, then i am still confused about why at=gsin(x).

I haven't read about Mass/weight/Newtons laws yet so can you explain it with just what I know.

I did in #6. Just replace "force" with "gravity" there.
 
  • #11
voko said:
We assume that the string cannot change its length. That means that any force applied to the end of the string is effectively canceled along the string; but its transverse component acts unopposed.

I still don't get how this makes sense of at=gsin(x). The transverse component of gravity?
 
  • #12
babysnatcher said:
I still don't get how this makes sense of at=gsin(x). The transverse component of gravity?
Gravity exerts a force of mg straight down. What is the component of gravity in the direction the ball is traveling when the angle of the string X is 20° from vertical?
 
  • #13
rcgldr said:
Gravity exerts a force of mg straight down. What is the component of gravity in the direction the ball is traveling when the angle of the string X is 20° from vertical?

9.80 m/s2 South. I see a triangle but what is the 3rd side?
 
  • #14
babysnatcher said:
I still don't get how this makes sense of at=gsin(x). The transverse component of gravity?

Gravity acting on the ball on a string has the radial and tangential (transverse) components. Find them, that is simple trigonometry.
 
  • #15
voko said:
Gravity acting on the ball on a string has the radial and tangential (transverse) components. Find them, that is simple trigonometry.

Yes, I get hypotenose g and leg at. I want to know what this other leg is -- I don't understand that.
 
  • #16
babysnatcher said:
Yes, I get hypotenose g and leg at. I want to know what this other leg is -- I don't understand that.

You were given an explanation earlier. You rejected it because you did not know some concepts. I am afraid you will have to wait till you know those concepts.
 
  • #17
voko said:
What I wrote fully applies to gravity.



'y' could be the "radial" acceleration if the ball were allowed to fly freely. But it is not, so 'y', and the entire equation for that matter, does not make a lot of sense physically.

What does make sense is ## (mg)^2 = (ma_t)^2 + P_r^2 ##, where ##P_r## is the radial component of the weight.

Why does Pr not include mass?
 
  • #18
The full equation should be ##(mg)^2 = P^2 = P_t^2 + P_r^2 ##. Tangentially, there is no other force except ##P_t##, so we can write ##P_t = ma_t## and then have ##(mg)^2 = (ma_t)^2 + P_r^2 ##. Radially, however, other forces may be present, ##ma_r = P_r + F##, so we cannot replace ##P_r## with ##ma_r##.
 
  • #19
babysnatcher said:
A ball tied to the end of a string 0.50 m in length swings in a vertical circle under the influence of gravity. When the string makes an angle x= 20 degrees with the vertical, the ball has a speed of 1.5 m/s. Find the magnitude of the radial component of acceleration at this instant.

So i have a concept question. How do i know that tangential acceleration = gsin(20 degrees).

Basically, prove the general formula tangential acceleration= gsin(x) for this problem.

Were you taught the formula for radial acceleration? V2/R. Note that, in this question, it does not depend on g, since, as stated in the problem description, the ball is traveling in a circle.
 
  • #20
Chestermiller said:
Were you taught the formula for radial acceleration? V2/R. Note that, in this question, it does not depend on g, since, as stated in the problem description, the ball is traveling in a circle.

Im only talking about the tangential acceleration.

voko said:
The full equation should be ##(mg)^2 = P^2 = P_t^2 + P_r^2 ##. Tangentially, there is no other force except ##P_t##, so we can write ##P_t = ma_t## and then have ##(mg)^2 = (ma_t)^2 + P_r^2 ##. Radially, however, other forces may be present, ##ma_r = P_r + F##, so we cannot replace ##P_r## with ##ma_r##.

I think Pr=mar+F makes more sense because Pr is mar plus other possible forces. Why is it the other way?
 
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  • #21
whoops xD
 
  • #22
babysnatcher said:
I think Pr=mar+F makes more sense because Pr is mar plus other possible forces. Why is it the other way?

No, ##P_r## is not ##ma_r## plus other possible forces. The basic equation is ## mg = P = P_r + P_t ##, where ##P_r## and ##P_t## are the radial and tangential components of weight; they are uniquely defined by the orientation of the radius and do not depend on any other force.

In addition to that, Newton's second law requires that ##ma_r## be equal to the sum of all the forces, so ##P_r = ma_r + F ## would violate that law.
 
  • #23
rcgldr said:
Gravity exerts a force of mg straight down. What is the component of gravity in the direction the ball is traveling when the angle of the string X is 20° from vertical?

babysnatcher said:
9.80 m/s2 South. I see a triangle but what is the 3rd side?
This is similar to a wedge or inclined plane where the hypotenuse is the direction the ball is moving, which would be perpendicular to the string which is 20° from vertical. The other two sides would be vertical and horizontal. For triangle based on the force from gravity, the hypotenuse is a vertical line (vector), the radial component of gravity would be a line (vector) 20° from vertical, and the tangental component would be a line (vector) perpendicular to the radial component.
 
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  • #24
babysnatcher said:
Im only talking about the tangential acceleration.

Oh, excuse me. My fault. You wanted to focus on the tangential acceleration. You would do well to follow rcgldr and wannabeNewton's advice. When the mass is 20 degrees from the vertical, the gravitational force on the mass can be resolved into components radial and tangential to the circular path. The radial component of the gravitational force is mg cos 20, and the tangential component of the gravitational force is mg sin 20. This is the only force acting on the mass in the tangential direction (in the radial direction, there is also a tension force from the string). Since, in the tangential direction, the only force acting on the mass is mg sin 20, this must be equal to its mass times its acceleration in the tangential direction:

mat = mg sin 20

If you cancel the m's from the above equation, you get

at = g sin 20
 

1. What is the difference between tangential and radial acceleration?

Tangential acceleration is the rate of change of an object's tangential velocity, while radial acceleration is the rate of change of an object's radial distance from a fixed point. In simpler terms, tangential acceleration refers to how fast an object is speeding up or slowing down in a straight line, while radial acceleration refers to how fast an object is changing direction.

2. How are tangential and radial acceleration related?

Tangential and radial acceleration are both components of an object's total acceleration. In circular motion, tangential acceleration is always perpendicular to radial acceleration, creating a right angle. This means that the total acceleration can be calculated using the Pythagorean theorem.

3. What is the formula for calculating tangential and radial acceleration?

The formula for tangential acceleration is a = (vf - vi) / t, where vf is the final tangential velocity, vi is the initial tangential velocity, and t is the time. The formula for radial acceleration is ar = v2 / r, where v is the tangential velocity and r is the radial distance from a fixed point.

4. Can tangential and radial acceleration be negative?

Yes, both tangential and radial acceleration can be negative. A negative tangential acceleration means that the object is slowing down, while a negative radial acceleration means that the object is moving towards the fixed point. However, it is important to note that tangential and radial acceleration can never be negative at the same time, as they are always perpendicular to each other.

5. How is tangential and radial acceleration used in real-world applications?

Tangential and radial acceleration are important concepts in understanding circular motion, which is present in many real-world applications. For example, roller coasters use tangential and radial acceleration to create thrilling and safe rides. In space travel, understanding tangential and radial acceleration is crucial for the trajectory of spacecrafts. Additionally, car manufacturers use these concepts to design vehicles with optimal handling and performance.

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