Distance traveled by a Ball affected by friction after t seconds

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The discussion focuses on calculating the distance traveled by a ball affected by friction and rolling resistance on a flat surface. Participants suggest that the ball experiences a constant rate of deceleration due to rolling resistance, complicating the calculation of distance. There is also consideration of whether air resistance should be factored in, particularly for a golf ball rolling on carpet at an initial velocity of 6.5 m/s. Accurate estimates of deceleration may require direct measurements to account for both rolling and air resistance. Further testing is recommended to refine the calculations and obtain precise results.
Volt582
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Thread moved from the technical math forums to the schoolwork forums
Hi everyone,
i have been trying to find an answer to this problem I have but couldnt find any good answers...
(I dont know much about this stuff, but need a formula for a Project I am currently working on).

So The problem goes as follows:
Assuming we have a ball with a mass of m which is rolling on a flat surface with an initial velocity of v. Lets call the friction coefficiant between the surface and the Ball p. How much distance will the ball have traveled after t seconds when it is constantly affected by friction?
 
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Volt582 said:
Hi everyone,
i have been trying to find an answer to this problem I have but couldnt find any good answers...
(I dont know much about this stuff, but need a formula for a Project I am currently working on).

So The problem goes as follows:
Assuming we have a ball with a mass of m which is rolling on a flat surface with an initial velocity of v. Lets call the friction coefficiant between the surface and the Ball p. How much distance will the ball have traveled after t seconds when it is constantly affected by friction?
A rolling ball is slowly by rolling resistance, rather than friction. In any case, we can assume a constant rate of deceleration. What do you know about motion with constant acceleration?
 
PeroK said:
A rolling ball is slowly by rolling resistance, rather than friction. In any case, we can assume a constant rate of deceleration. What do you know about motion with constant acceleration?
When I have a constant acceleration I should be able to calculate the distance... But how could you calculate the deceleration?

And maybe another question: Do you think that air resistance should be accounted for as well when we are working with a golf ball on a carpet which is approximately traveling with an initial velocity of 6.5 m/s?
 
Volt582 said:
When I have a constant acceleration I should be able to calculate the distance... But how could you calculate the deceleration?
Rolling resistance is quite complicated. In any case, you are probably going to have to measure things if you want an accurate estimate.
Volt582 said:
And maybe another question: Do you think that air resistance should be accounted for as well when we are working with a golf ball on a carpet which is approximately traveling with an initial velocity of 6.5 m/s?
Probably. But, it would take a bit of work to distinguish the effects of air resistance from rolling resistance. What you will measure directly is a deceleration that involves them both.
 
PeroK said:
Rolling resistance is quite complicated. In any case, you are probably going to have to measure things if you want an accurate estimate.

Probably. But, it would take a bit of work to distinguish the effects of air resistance from rolling resistance. What you will measure directly is a deceleration that involves them both.
mhh ok... Then we will have to perform some more tests. Thanks anyway!
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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