Normal Acceleration of a Thrown Ball: What Is It?

In summary: N## is then ##angle_N=gtcosx#OK. Here's what I would do, for any time ##t##:##V_x## is constant##V_y = gt## (downward)Then use a bit of trig to find the angle from the vertical. ##a_N## is then ##angle_N=gtcosx#
  • #1
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Homework Statement


a ball is thrown from some height horizontally with some velocity. After 1 second what is its normal acceleration?
Ignore air friction

Homework Equations

The Attempt at a Solution


I was absolutely certain that it is 10m/s2 couse that doesn't change and it has no acceleration in horizontal direction. But somehow the answer is ##5\sqrt{2}## what does this mean?
 
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  • #2
They probably want the component of the ball's acceleration normal to its path at that point. (What was the horizontal velocity?)
 
  • #3
Doc Al said:
They probably want the component of the ball's acceleration normal to its path at that point. (What was the horizontal velocity?)
It was 10m/s so how is that done?
 
  • #4
Start by drawing a free body diagram after 1 second. Mark the acceleration and its tangential and normal components. See if you can find a way to calculate ##a_N## using geometry.
You will have to find a way to ascertain the angles between various acceleration components. What you know about velocity should help.
 
  • #5
diredragon said:
It was 10m/s so how is that done?
Figure out the direction of the velocity vector at that point.
 
  • #6
Doc Al said:
Figure out the direction of the velocity vector at that point.
Is this the picture of the problem?( uploaded )
Well in our case, there isn't any tangential acceleration right?
I can't seem to find a way to calculate the normal acceleration using only the 10m/s couse that doesn't change...what can that tell me?
 

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  • #7
diredragon said:
there isn't any tangential acceleration right?
There is. It's the component of gravitational acceleration along the velocity vector. It's what makes the object accelerate as it follows the curved path. Normal acceleration curves the path.
 
  • #8
diredragon said:
Well in our case, there isn't any tangential acceleration right?
"Tangential" means tangential to the path: Parallel to the velocity vector at that point.

diredragon said:
I can't seem to find a way to calculate the normal acceleration using only the 10m/s couse that doesn't change...what can that tell me?
What direction is the normal to the velocity vector at that point? What angle does that normal make with the direction of gravity?
 
  • #9
Bandersnatch said:
There is. It's the component of gravitational acceleration along the velocity vector. It's what makes the object accelerate as it follows the curved path. Normal acceleration curves the path.

Doc Al said:
"Tangential" means tangential to the path: Parallel to the velocity vector at that point.What direction is the normal to the velocity vector at that point? What angle does that normal make with the direction of gravity?

Isnt the gravitational acceleration the one that curves the path.
If i find the angle between ##a_n## and ##g## its simply ##gcosx=a_n## right?
 
  • #10
diredragon said:
Isnt the gravitational acceleration the one that curves the path.
Yes, it's the gravitational force acting normal to the velocity that curves the trajectory.

diredragon said:
If i find the angle between ##a_n## and ##g## its simply ##gcosx=a_n## right?
Right!
 
  • #11
Doc Al said:
Yes, it's the gravitational force acting normal to the velocity that curves the trajectory.Right!
So it should be 45 degrees as the velocity in the x and y are 10m/s. But what if we are taken 1 s more? How does the angle depend on the time?
 
  • #12
diredragon said:
So it should be 45 degrees as the velocity in the x and y are 10m/s.
Right.

diredragon said:
But what if we are taken 1 s more? How does the angle depend on the time?
You tell me. Only one of the velocity components will change.
 
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  • #13
Doc Al said:
Right.You tell me. Only one of the velocity components will change.
I see, ##V_y=V_tcosx##
 
  • #14
diredragon said:
I see, ##V_y=V_tcosx##
OK. Here's what I would do, for any time ##t##:
##V_x## is constant
##V_y = gt## (downward)

Then use a bit of trig to find the angle from the vertical.
 
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1. What is normal acceleration of a thrown ball?

The normal acceleration of a thrown ball is the rate at which the ball's velocity changes per unit of time in the direction perpendicular to its motion. It is also known as centripetal acceleration.

2. What causes the normal acceleration of a thrown ball?

The normal acceleration of a thrown ball is caused by the force of gravity pulling the ball towards the center of the Earth. This force creates a circular motion and therefore, a centripetal acceleration.

3. How is normal acceleration of a thrown ball calculated?

The normal acceleration of a thrown ball can be calculated using the formula a = v^2/r, where a is the acceleration, v is the velocity, and r is the radius of the circular motion. Alternatively, it can also be calculated using the formula a = ω^2r, where ω is the angular velocity.

4. How does the mass of the ball affect its normal acceleration?

The mass of the ball does not affect its normal acceleration. According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass. As the mass of the ball does not change, its normal acceleration remains constant.

5. Can the normal acceleration of a thrown ball be negative?

Yes, the normal acceleration of a thrown ball can be negative. This occurs when the ball is decelerating or changing direction, causing its velocity to decrease. However, the magnitude of the acceleration will still be positive as it is always measured in the direction of the center of the circular motion.

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