What am I Missing? Solving Conservation of Energy

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SUMMARY

The discussion centers on solving the conservation of energy for a spherical ball rolling down a loop track, specifically calculating the tangential acceleration at point A. The derived linear velocity is expressed as [10g(H-R-Rsin(theta))/7]^(1/2), leading to a tangential acceleration of -(5gcos(theta))/7, which aligns with the textbook solution. However, confusion arises regarding the forces acting on the ball, particularly the role of static friction and the transition between linear and circular motion, which affects the acceleration continuity at the point of transition.

PREREQUISITES
  • Understanding of conservation of energy principles in physics
  • Familiarity with free body diagrams and force resolution
  • Knowledge of tangential and radial acceleration concepts
  • Basic understanding of rolling motion and static friction effects
NEXT STEPS
  • Study the effects of static friction on rolling motion in detail
  • Learn about the continuity of acceleration and its implications in physics
  • Explore the mathematical derivation of tangential acceleration in rolling objects
  • Investigate the concept of 'jerk' and its relevance in motion analysis
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Students studying classical mechanics, physics educators, and anyone interested in the dynamics of rolling motion and energy conservation principles.

Vivek98phyboy
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Homework Statement
Below figure shows a small spherical ball of mass m rolling down the loop track. The ball is released on the linear portion at a vertical height H from the lowest point. The circular part shown has a radius R.

Find the tangential acceleration of the centre of mass of the ball when it is at A.
Relevant Equations
Started solving with Conservation of energy

mgH= mgR(1+sin(theta)) + ((Iw²)/2) + mv²/2
IMG_20200725_103208.jpg

By solving conservation of energy, I was able to find the linear velocity which is
[10g(H-R-Rsin(theta))/7]^½ and by differentiating this with respect to "t", I arrived at the tangential acceleration value of -(5gcos(theta))/7 and found it to be in agreement with the solution provided in the book.

But when I draw a free body diagram of the ball by resolving it into radial and tangential component, If I had resolved it properly, i found that the Normal force and mg.sin(theta) points along the radius inwards and mgcos(theta) points along the tangential direction.

Given the scenario that mgcos(theta) is the only force along tangential direction, shouldn't the tangential acceleration be -gcos(theta)?

What am I missing?
 
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Vivek98phyboy said:
Homework Statement:: Below figure shows a small spherical ball of mass m rolling down the loop track. The ball is released on the linear portion at a vertical height H from the lowest point. The circular part shown has a radius R.

Find the tangential acceleration of the centre of mass of the ball when it is at A.
Relevant Equations:: Started solving with Conservation of energy

mgH= mgR(1+sin(theta)) + ((Iw²)/2) + mv²/2

View attachment 266809
By solving conservation of energy, I was able to find the linear velocity which is
[10g(H-R-Rsin(theta))/7]^½ and by differentiating this with respect to "t", I arrived at the tangential acceleration value of -(5gcos(theta))/7 and found it to be in agreement with the solution provided in the book.

But when I draw a free body diagram of the ball by resolving it into radial and tangential component, If I had resolved it properly, i found that the Normal force and mg.sin(theta) points along the radius inwards and mgcos(theta) points along the tangential direction.

Given the scenario that mgcos(theta) is the only force along tangential direction, shouldn't the tangential acceleration be -gcos(theta)?

What am I missing?
You took rolling into account when calculating the tangential acceleration from the energy, but ignored rolling in the other case. it is also the static friction causing rolling that acts on the ball.
 
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An interesting thing appears at the point where the straight line conjugates with the circle: the acceleration jumps
 
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wrobel said:
An interesting thing appears at the point where the straight line conjugates with the circle: the acceleration jumps
You mean because centripetal acceleration "suddenly" appears there? As far as i know acceleration need not be continuous, its velocity that is usually continuous.
 
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Delta2 said:
You mean because centripetal acceleration "suddenly" appears there?
Delta2 said:
As far as i know acceleration need not be continuous, its velocity that is usually continuous.
yes, but I believe that such things should be stressed
 
ehild said:
You took rolling into account when calculating the tangential acceleration from the energy, but ignored rolling in the other case. it is also the static friction causing rolling that acts on the ball.
You are right. When I took friction into account, I got it right.

Thank you
 
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wrobel said:
An interesting thing appears at the point where the straight line conjugates with the circle: the acceleration jumps
I didn't notice this. Thank you for pointing out this fact
 
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Vivek98phyboy said:
I didn't notice this. Thank you for pointing out this fact
Rate of change of acceleration is sometimes called 'jerk'. It is important in ergonomics. A standing passenger on a slowing bus has to brace herself against the inertial force; at the instant the bus stops the inertial force vanishes, causing her to lurch backwards.
 
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