'They' haven't considered centripetal acceleration

  • #1
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Homework Statement


A train is moving counter-clockwise with a constant speed of 10 m/s in a circle of radius ##\frac {16} π## m. The plane of the circle lies in the x-y plane. At time t = 0, the train is at P, when a stone is thrown from it with a speed of 10 m/s relative to the train towards the negative ##x## axis at an angle of 37° with the vertical ##z## axis (assume g = 10 m/s2 and sin 37° = ##\frac 3 5##). Find:
(a) the velocity of the stone relative to the train at the highest point of its trajectory.
(b) the co-ordinates of the point where it finally falls.
(c) the co-ordinates of the point at the highest point of its trajectory.
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Homework Equations


Let vST be the velocity vector of the stone relative to the train.
vST = uST - gt##\hat k##
Also, position vector r = uSt - gt2##\hat k##

The Attempt at a Solution


I know how to solve this problem, but there's just one hitch. The ##x## components of the velocity and co-ordinates they've given as the answer isn't matching with mine. Their answers are:
(a) (-6##\hat i## + 10##\hat j##) m/s
(b) (-4.5 m, 16 m, 0)
(c) (0.3 m, 8 m, 3.2 m)
Later, I found out that they haven't considered the centripetal acceleration (along the negative ##x## direction) that would be imparted to the stone at the instant of projection; due to the circular motion of the train. Why?
 

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  • #2
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Later, I found out that they haven't considered the centripetal acceleration (along the negative xxx direction) that would be imparted to the stone at the instant of projection; due to the circular motion of the train. Why?
There are a few things that are wrong here and it will take a couple different discussions to unravel, but let’s start with the fact that the centripetal acceleration does not matter. You are told the relative velocity a projectile was launched from a moving train. You are not asked how much force it took to launch the projectile. It is true that whoever threw the stone would have to throw with more force because the train, the stone, and the thrower are moving along a curved path. However that is irrelevant. The force to throw the stone isn’t part of the question. Whatever force was necessary was applied, and the stone was thrown with the stated velocity. At that point it is a free flying projectile and the only force on it is gravity (ignoring air resistance). The centripetal acceleration of the train is no longer relevant to the stones motion.
 
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  • #3
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I see several other issues, but I will get in trouble if I say anymore before you show us your attempts at solving the problem.
 
  • #4
Orodruin
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I know how to solve this problem, but there's just one hitch. The xxx components of the velocity and co-ordinates they've given as the answer isn't matching with mine.
Forum rules require you to actually show your attempt. There is no way for us to see if or how you went wrong unless you actually show us what you did.

Later, I found out that they haven't considered the centripetal acceleration (along the negative xxx direction) that would be imparted to the stone at the instant of projection; due to the circular motion of the train. Why?
You are given the velocity relative to the train, that is sufficient. Even if you considered centripetal acceleration for a single instant, it will not change that velocity. The impulse imparted by any finite force in zero time is zero.

However, the solution key's (a) is still wrong for reasons we can get back to when you have posted your actual attempt.
 
  • #5
haruspex
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the centripetal acceleration (along the negative x direction) that would be imparted
You do not impart acceleration, centripetal or otherwise, in the same sense that you impart momentum or energy. The acceleration is not a property that is retained by the thrown object. The acceleration is merely the means by which those properties are imparted.
whoever threw the stone would have to throw with more force because the train, the stone, and the thrower are moving along a curved path.
I don't think that is true.
We tend to think of throwing a stone, in such questions, as an instantaneous event. To decide whether it would take more force in this context we would have to treat the throw as taking some time.
Since merely maintaining the stone in circular path was accelerating it in the right direction, we just have to provide extra force for the extra acceleration. That is, the centripetal acceleration the stone had enjoyed contributes to the total acceleration necessary for the throw.
 
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  • #6
Delta2
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Interesting , a 3D kinematics problems involving relative velocities. The whole catch of this exercise lies at this point
with a speed of 10 m/s relative to the train towards the negative x axis at an angle of 37° with the vertical z axis
If I interpret this correctly I get that the initial velocity of the stone in the absolute reference frame has all three components ##(v_x,v_y,v_z)=(-6,10,8)##. And then the answer key for (a) seems wrong to me cause if we do the math correctly I think the train will be at the highest y point and having only x-axis velocity, at the time the stone reaches the highest point.

But in kinematics we don't care about the forces that cause the acceleration and velocity. We deal only with equations involving acceleration, velocity and position vectors. The centripetal acceleration and force at point P would be relevant only if we were asked to find the force that was applied to the stone to cause that relative velocity.

I think for this problem is better to work in the absolute reference frame, and when to find the relative velocities , we must find the velocity of the train at the requested time in the absolute reference frame.
 
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  • #7
CWatters
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Perhaps it's me but i interpreted "towards the negative X axis" as being to the left. The diagram appears to show it being thrown toward the positive y axis?
 
  • #8
Orodruin
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Perhaps it's me but i interpreted "towards the negative X axis" as being to the left. The diagram appears to show it being thrown toward the positive y axis?
That's not the stone, that's the train velocity.
 
  • #9
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I see several other issues, but I will get in trouble if I say anymore before you show us your attempts at solving the problem.
Forum rules require you to actually show your attempt. There is no way for us to see if or how you went wrong unless you actually show us what you did.
Yes, yes. My original attempt (when I felt that centripetal acceleration should be considered):
The initial velocity of the stone is uST = -6##\hat i## + 10##\hat j## + 8##\hat k##
Acp = vT2/ro → ro is the radius.
Acp = ##\frac {100} {16/π}## = ##\frac {25π} 4## m.
a = ##{\frac {-25π} 4}{\hat i} - 10\hat k##
(a) vST = uST + at
The particle will be at its highest point of trajectory when the ##z## component of velocity becomes 0 (the ##x## component won't become 0 because ax ∥ uST-x)
Hence, 8 -10t = 0 ⇒ t = ##\frac 4 5## s. Therefore, the total time of flight T = ##\frac 8 5## s.
vST = (-6 - 5π)##\hat i## + 10##\hat j##
(b) r - ro = uSTt + ##\frac 1 2##at2
Substituting with T = ##\frac 8 5## s,
r - ##\frac {16} π \hat i## = ##(\frac {-48} 5 - 8π)\hat i + 16\hat j##
r = ##(\frac {16} π - \frac {48} 5 - 8π)\hat i + 16\hat j##
(c) Substituting with t = ##\frac 4 5## s,
r = ##(\frac {16} π - \frac {24} 5 - 2π)\hat i + 8\hat j + \frac {16} 5 \hat k##
 
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  • #10
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There are a few things that are wrong here and it will take a couple different discussions to unravel, but let’s start with the fact that the centripetal acceleration does not matter. You are told the relative velocity a projectile was launched from a moving train. You are not asked how much force it took to launch the projectile.
But in kinematics we don't care about the forces that cause the acceleration and velocity. We deal only with equations involving acceleration, velocity and position vectors. The centripetal acceleration and force at point P would be relevant only if we were asked to find the force that was applied to the stone to cause that relative velocity.
But after all, an acceleration is an acceleration. So should it not be considered even if there's no force or momentum involved?
 
  • #11
Delta2
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But after all, an acceleration is an acceleration. So should it not be considered even if there's no force or momentum involved?
@haruspex already pointed this out at post #5 but I ll try to repeat it here. You cant impart acceleration to an object the way you impart velocity. The stone has centripetal acceleration and what other acceleration is imparted on it for the short duration during the process of throwing but it loses all the acceleration after the process of throwing has ended and it has gain the velocity ##v_0=(-6,10,8)##. The velocity "remains" but acceleration is "lost" at least this is what happens in this case of throwing the stone. The only acceleration that remains after the process of throwing has ended, is the gravitational acceleration.
 
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  • #12
Orodruin
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But after all, an acceleration is an acceleration. So should it not be considered even if there's no force or momentum involved?
##F = ma##. There is no acceleration without a force.
 
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  • #13
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The velocity "remains" but acceleration is "lost" at least this is what happens in this case of throwing the stone. The only acceleration that remains after the process of throwing has en
##F = ma##. There is no acceleration without a force.
Ok, got it.
However, the solution key's (a) is still wrong for reasons we can get back to when
So, you were saying?
 
  • #14
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##F = ma##. There is no acceleration without a force.
To be clear, there is the force of gravity and the acceleration from it. However there is no force or acceleration that needs to be considered in the launching of the stone. Those forces and accelerations have already happened. You are given the velocity of the stone AFTER it is launched. Gravity is the only force acting on the stone for the entire problem. Initial velocity and gravity. That is all you need.
 
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  • #15
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The given solution to a is not relative to the train, but is correct relative to the stationary coordinates. Either the didn’t mean to say “relative to the train” or the answer is wrong.
 
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  • #16
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vST = (-6 - 5π)^ii^\hat i + 10^jj^\hat j
You were doing well to here. Where did the 5 pi come from?
 
  • #17
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And then the answer key for (a) seems wrong to me cause if we do the math correctly I think the train will be at the highest y point and having only x-axis velocity, at the time the stone reaches the highest point
The given solution to a is not relative to the train, but is correct relative to the stationary coordinates. Either the didn’t mean to say “relative to the train” or the answer is wrong.
So should it be -6##\hat i## + 10##\hat j## -(-10##\hat i##) ,i.e., (4##\hat i## + 10##\hat j##) m/s?
Since the velocity of the train would have become -10##\hat i## m/s.
 
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  • #18
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You were doing well to here. Where did the 5 pi come from?
As I said, my original attempt taking the Acp into account.
 
  • #19
Orodruin
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So should it be -6##\hat i## + 10##\hat j## -(-10##\hat i##) ,i.e., (4##\hat i## + 10##\hat j##) m/s?
Since the velocity of the train would have become -10##\hat i## m/s.
Yes.
 
  • #21
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So should it be -6##\hat i## + 10##\hat j## -(-10##\hat i##) ,i.e., (4##\hat i## + 10##\hat j##) m/s?
Since the velocity of the train would have become -10##\hat i## m/s.
It would be if the train was still at the same place it was when the stone launched. (But it isn’t)
 
  • #22
Orodruin
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It would be if the train was still at the same place it was when the stone launched. (But it isn’t)
No, it is correct. The train has travelled to the y axis and is moving in the negative y-direction, which he correctly used.
 
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  • #23
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No, it is correct. The train has travelled to the y axis and is moving in the negative y-direction, which he correctly used.
I stand corrected. I glanced and ass-u-me-d. Should have looked closer.
 
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  • #24
Delta2
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No, it is correct. The train has travelled to the y axis and is moving in the negative y-direction, which he correctly used.
You mean in the negative x-direction (that's where the instantaneous velocity of the trains lies , when the train is at the top point of the circular orbit).
 
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  • #25
Orodruin
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You mean in the negative x-direction (that's where the instantaneous velocity of the trains lies , when the train is at the top point of the circular orbit).
Yes, early morning.
 
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