3 balls rolling down different-shaped inclined platforms

In summary: I think I have some numerical answers. I will provide the method, but not the numbers to comply with the question's rules for not giving a complete solution.In summary, the question asks which of three identical balls will reach the end of the board first, reach the ground first, and be the fastest at both points. The heights are the same and there is no friction. The solution involves using the equation for potential energy, mgh=1/2mv^2, to show that all three balls have the same speed when they reach the ground and the end of the board. However, the ball on track B may appear to be the fastest as it loses contact with the track and is in free fall, but this does not necessarily
  • #1
jaguar___
6
0

Homework Statement


which ball reaches at the end of the board first, which ball reaches ground first, which ball would be fastest when reached ground and at the end of the board?
heights are same and balls are identical

Homework Equations

The Attempt at a Solution


i think since ball rolls there's no friction and mgh=1/2mv^2 => so the all have same speed when they reached ground and at end of board
and since in condition b ball reaches the max speed fastest it reaches first at end of the board
 

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  • #2
jaguar___ said:

Homework Statement


which ball reaches at the end of the board first, which ball reaches ground first, which ball would be fastest when reached ground and at the end of the board?
heights are same and balls are identical

Homework Equations

The Attempt at a Solution


i think since ball rolls there's no friction and mgh=1/2mv^2 => so the all have same speed when they reached ground and at end of board
and since in condition b ball reaches the max speed fastest it reaches first at end of the board
That sounds about right.
 
  • #3
jaguar___ said:
i think since ball rolls there's no friction
No, if there were no friction the balls would slide. It is static friction, so does not reduce the total KE, but it diverts some of the KE into rotation.
jaguar___ said:
all have same speed when they reached ground and at end of board
That is true because for each ball the same fraction of energy goes into rotation. The speed will be less than √(2gh), but to the same extent for each.
jaguar___ said:
since in condition b ball reaches the max speed fastest
That does not follow. B starts with a gentler gradient. It could take longer to get past that than C takes for the whole descent. As against that, C has a greater horizontal distance to cover.
I cannot see that it is possible to decide merely from the diagram. The exact shape of B needs to be specified.
 
  • #4
haruspex said:
That does not follow. B starts with a gentler gradient. It could take longer to get past that than C takes for the whole descent.
It could but then the question would be difficult to answer even when the details of track B are given. This is a conceptual question after all. I interpreted the intent of the question to be the realization that in B the ball loses contact with the track and is in free fall thereafter (after acquiring sufficient horizontal velocity to clear the bottom of the track). Therefore, for a good portion of the trip the vertical acceleration of B is g while for A and C the vertical acceleration is less than g during the entire trip.
 
  • #5
kuruman said:
This is a conceptual question after all.
Sure, but which concept?!
It is quite possible that if the profiles were exactly as shown B would be still be approaching the drop when C reached the bottom.
I tried setting this up using a web digitising tool (https://apps.automeris.io/wpd/), but it's too sensitive to the exact slope at the start to produce a reliable result.
 
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  • #6
haruspex said:
Sure, but which concept?!
It is quite possible that if the profiles were exactly as shown B would be still be approaching the drop when C reached the bottom.
I tried setting this up using a web digitising tool (https://apps.automeris.io/wpd/), but it's too sensitive to the exact slope at the start to produce a reliable result.
Assuming that there is identical conversion of potential to rotational energy, I think that the main idea of the question is that mechanical energy conservation can be used to predict that the speed of an object dropping by height ##h## will be the same regardless of the path followed; additionally the question addresses the misconception that equal final speeds do not necessarily imply equal travel times.

I admire your effort to quantify the problem. My idea to this effect is to approximate the track as the superposition of two quarter-circles and a straight piece. (See yellow line below). I suspect that's how the creator of the question generated the picture. The dimensions are radius of 1/4 circle, R = 0.8 u, radius of sphere r = 0.15 u, length of straight piece h = 0.7 u. I don't have the time right now to play with it, but it's reasonable to assume that the ball starts moving from rest because it's off the vertical by angle θ = r/(R+r).

Track.png
 

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  • #7
kuruman said:
approximate the track as the superposition of two quarter-circle
Good idea. The key then is the ratio between the radius of the curve and the radius of the ball. The greater the ratio the shallower the initial gradient and the longer the ball will take ... without limit.
I'll try using the measurements you have.
 
  • #8
I think I have some numerical answers. I will provide the method, but not the numbers to comply with PF rules. I assumed frictionless sliding with no rolling.
For b
1. Found the critical angle ##\theta_{crit}## at which contact is lost.
2. Found that while contact is maintained ##\frac{d\theta}{dt}=\sqrt{\frac{2g}{R+r}(\cos\theta_0-\cos\theta)}##.
3. Integrated numerically $$I_{crit}=\int_{\theta_0}^{\theta_{crit}}\frac{d\theta}{\sqrt{\cos\theta_0-\cos\theta}}$$
to obtain the value of the time required for the ball to lose contact, ##t_{crit}=\sqrt{\frac{R+r}{2g}}I_{crit}##.
4. Found the y-component of the velocity at ##t=t_c## and then used projectile kinematics to find the time of flight, ##t_f##.
5. Found the time of transit using ##T_b=t_{crit}+t_f##.

Also, just for fun I calculated that the horizontal displacement of ball b is very close to ##2R##, clearly a coincidence because the height of the vertical straight piece ##h## affects the time of flight.
Note: I assigned meters to the units previously posted so that g = 9.8 m/s2 could be used without rescaling.

For c
1. Found the upper corner of the incline to be at ##(x_0,~y_0)= (2.1, 2.3)## m.
2. Calculated the down-the-incline acceleration ##a = g \sin\beta = g \frac{y_0}{\sqrt{x_0^2+y_0^2}}## and ...
3. Used it to find the time of transit ##T_c## from the kinematic equation ##\sqrt{x_0^2+y_0^2}-r=\frac{1}{2}aT_c^2.##
I found the value of ##T_b## to be about 33% higher than the value of ##T_c##. Frankly, I did not expect that.
 
  • #9
kuruman said:
I assumed frictionless sliding with no rolling.
That does give C an advantage. If rolling is assumed, when the ball loses contact in B it will no longer need to acquire rotational KE, so all goes into linear KE.
On the other hand, I would say the problem setter did not intend loss of contact to be considered.
 

Related to 3 balls rolling down different-shaped inclined platforms

1. How does the shape of the inclined platform affect the speed of the rolling balls?

The shape of the inclined platform can significantly impact the speed of the rolling balls. Generally, a steeper incline will result in a faster speed, while a gentler incline will result in a slower speed. Additionally, the shape of the platform can also affect the amount of friction, which can further impact the speed of the balls.

2. What factors influence the acceleration of the rolling balls?

The acceleration of the rolling balls is influenced by several factors, including the angle of the incline, the shape and material of the balls, and the amount of friction between the balls and the platform. The force of gravity also plays a significant role in the acceleration of the balls.

3. How does the mass of the balls affect their movement on the inclined platforms?

The mass of the balls can affect their movement on the inclined platforms in a few ways. A heavier ball will have more inertia and may require more force to get it moving, resulting in a slower speed. However, a heavier ball will also have more gravitational potential energy, which can result in a faster speed once it starts rolling down the incline.

4. Can the shape of the inclined platform affect the direction of the rolling balls?

Yes, the shape of the inclined platform can affect the direction of the rolling balls. For example, a curved or concave platform may cause the balls to roll towards the center, while a convex platform may cause the balls to roll away from the center. This is due to the change in the direction of the force of gravity on the balls.

5. How does the coefficient of friction impact the movement of the rolling balls?

The coefficient of friction between the balls and the inclined platform can greatly impact their movement. A higher coefficient of friction will result in a slower speed and shorter distance traveled by the balls. On the other hand, a lower coefficient of friction will result in a faster speed and longer distance traveled by the balls.

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