What is the angle of a ramp given a marble's velocity and time of descent?

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The discussion revolves around calculating the angle of a ramp based on the velocity and time of a marble's descent. The marble rolls down a 2.26m slope in 3.12 seconds, leading to an average velocity of 0.724 m/s and an acceleration of 0.464 m/s². Despite these calculations suggesting a slope angle of approximately 3 degrees, participants note that the actual slope appears steeper, likely between 10 to 20 degrees. The discrepancy is attributed to the assumption of no friction, which does not account for the marble's rotational motion and energy loss. Ultimately, the calculations are deemed correct, but real-world factors such as friction and measurement errors may have influenced the results.
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Homework Statement


This was actually a lab for class today. Basically, a marble rolls down a 2.26m slope, in 3.12s. Find the angle of the slope. Assume there is no friction.

Homework Equations


v=d/t
a=v t
a=g sin (theta)

The Attempt at a Solution



v=2.26/3.12 = 0.724m/s
since that is Avg. V, multiply by 2 to get Total V. V=1.45m/s
a=1.45/3.12=0.464m/s2

so with that, i change the equation with the angle to solve for theta, getting
Invsin (a/g) = theta
Invsin (0.464/9.81) = theta
theta = 2.71

This seems good on paper, but in the lab, this is to find the angle of a slope outside. The actual slope is steeper than 3 degrees, it's probably about 10-20 degrees. So is there a problem with my calculations? Or are there sources of error that cause the angle to come out as so small?
 
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Hi, welcome to PF.

(What do you mean by the total velocity?)

Use the equation linking distance with time and acceleration. The initial speed I think you know. Do you know that acceleration of the body along the slope in terms of g and theta? Give it a try.
 
By total velocity, I mean the final velocity of the marble in both the x and the y when it reaches the bottom of the ramp.

The equation you are referring to would be x=0.5at2, correct?
So 2.26=(0.5)a(3.12)2
2.26=4.87a
a=0.464 m/s2

So I still end up with the same acceleration.
The second equation you are referring to would be a=g sin (theta) correct?
The equation I used, Invsin (a/g) = theta, is just derived from that, so I would end up with the same answer.
But again, I would get about 3 degrees, which seems unreasonable for the angle of the slope in this lab, as it looks at least about 10 degrees. Is there something wrong with my math or are there sources of error?
 
Based on the numbers you have presented, the calculations are correct. The assumption of no friction introduces some error because it models the marble as sliding down the ramp as opposed to rolling down. However, compensating for that gives a calculated incline of about 4 degrees, not enough.
 
KDawgAtsu said:
But again, I would get about 3 degrees, which seems unreasonable for the angle of the slope in this lab, as it looks at least about 10 degrees. Is there something wrong with my math or are there sources of error?

We are considering the marble as a point mass sliding down the ramp with no friction, whereas in reality it is a solid sphere with some radius rolling down the ramp with friction. Part of the potential energy of the marble goes into its rotational motion and another part goes toward overcoming that little bit of rolling fiction. This means that under real conditions, a higher theta for the ramp will show less acceleration than the calculation, and vice versa. This is possibly the explanation you are looking for.
 
I agree with Kuruman. Even modeling as rolling motion of a rigid object the angle is only about 4 degrees. Either friction cannot be neglected or there is some experimental error. Perhaps in measuring the time it takes the object to roll down the ramp.
 
Thank you all for the help, I think I understand this now.
 
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