Which Body Reaches the Bottom First?

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In summary, a sphere is fastest to the bottom of an incline, while a cylinder and hoop are the same speed at the top. However, the hoop has a greater V for a given drop in height.
  • #1
brainracked
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Homework Statement


Okay..here's d question:
A sphere, a cylinder and a hoop start from rest and roll down the same incline.
Determine which body reaches the bottom first.

Homework Equations


Sphere: I=2/5 mr2
Cylinder: I=1/2 mr2
Hoop: I= mr2
F=ma

The Attempt at a Solution


To find out which of the three reach first I suspect that their acceleration should b found 1st..but..how do I go about it? I also feel as if there is not much data to work on..is tis question complete?
If it is solvable..a lil clue or idea would b much appreciated... =)
 
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  • #2
brainracked said:

Homework Statement


Okay..here's d question:
A sphere, a cylinder and a hoop start from rest and roll down the same incline.
Determine which body reaches the bottom first.

Hi brainracked! :smile:

Hint: "roll" means without slipping …

so the instantaneous velocity of the point of contact is zero …

so the speed of the centre of mass must always be canceled by the rotational speed of the rim (relative to the centre of mass).

That gives you a relationship between velocity and angular velocity :smile:
 
  • #3


Hiyya tiny-tim..we meet again..haha..
So sorry..but..i don't quite understand..the instantaneous velocity of point of contact?
Relationship?
w=v^2/r?

So sorry for the trouble and thank you so much for ur time and help.. :smile:
 
  • #4
rolling

brainracked said:
So sorry..but..i don't quite understand..the instantaneous velocity of point of contact?
Relationship?

Hi brainracked! :smile:

The point of contact of a wheel with the ground is always stationary, just for an instant.

If it wasn't, the wheel would skid.

So if the centre is moving horizontally at speed v, and the wheel is rotating with angular speed ω, the point at the top of the rim has speed v + rω, and the point at the bottom of the rim (the point of contact) has speed v - rω.

So rolling requires v - rω = 0. (that is the "relationship")

This is geometry, not physics. :wink:

Now that the geometry has told you the relationship between v and ω, plug that into KE + PE = constant to get the acceleration. :smile:
 
  • #5


Haha..okay..i get it..thanks!
 
  • #6


Wait wait wait..I ended having two variables..h and v

PE + KE
= mgh +1/2mv^2 + 1/5 mv^2
=gh + 7/10v^2

I don't see how tis can solve the answer..
 
  • #7


*solve the question
 
  • #8
brainracked said:
Wait wait wait..I ended having two variables..h and v

v = (dh/dt)/sinθ :smile:
 
  • #9


and to get the acceleration?
I'm lost..
 
  • #10


brainracked said:
Wait wait wait..I ended having two variables..h and v

PE + KE
= mgh +1/2mv^2 + 1/5 mv^2
=gh + 7/10v^2

I don't see how tis can solve the answer..

[tex]\Delta PE = \Delta KE = mgh = \frac{mv^2}{2} + \frac{I\omega^2}{2} = \frac{mv^2}{2} + \frac{I}{2}*\frac{v^2}{r^2}[/tex]

Determine this equation for each of the objects and compare them by inspection. All you are asked is which is fastest.

You can always solve V for each.

As to acceleration you were asking about, lest you forget dv/dt is acceleration.
 
  • #11


the h is still the problem though..i still don't understand what tiny tim means by v=(dh/dt)/sin [tex]\theta[/tex]
I can only manage to simplify the equations down to:
Sphere:
mgh=[tex]\frac{7}{10}[/tex]mv2

Cylinder:
mgh=[tex]\frac{3}{4}[/tex]mv2

Hoop:
mgh=mv2
 
  • #12


brainracked said:
the h is still the problem though..i still don't understand what tiny tim means by v=(dh/dt)/sin [tex]\theta[/tex]
I can only manage to simplify the equations down to:
Sphere:
mgh=[tex]\frac{7}{10}[/tex]mv2

Cylinder:
mgh=[tex]\frac{3}{4}[/tex]mv2

Hoop:
mgh=mv2

For a given drop in height which has the greater V?

Won't the fastest object at the bottom necessarily be the one there first, if acceleration is uniform?
 
  • #13


LowlyPion said:
For a given drop in height which has the greater V?

Won't the fastest object at the bottom necessarily be the one there first, if acceleration is uniform?

For a given drop in height, the hoop has a greater V? It is possible to find the acceleration of each of the objects right?
 
  • #14


brainracked said:
For a given drop in height, the hoop has a greater V? It is possible to find the acceleration of each of the objects right?

You might want to check again which has the greater speed for a given h.

Of course you can figure acceleration if you wish. But if you know that one object is faster at one point of h below the top in a uniformly accelerated field then you know from induction that it is faster at all points on the incline don't you?
 
  • #15


Forgot to add that if it is faster at every point on the incline then it would be fastest to the bottom wouldn't it?
 
  • #16
Hi brainracked! :smile:
brainracked said:
the h is still the problem though..i still don't understand what tiny tim means by v=(dh/dt)/sin [tex]\theta[/tex]

v = dx/dt.

It's a slope, with angle θ, say …

so x (along the slope) = h/sinθ.

This isn't physics … this is geometry. :smile:
 
  • #17


LowlyPion said:
You might want to check again which has the greater speed for a given h.

Of course you can figure acceleration if you wish. But if you know that one object is faster at one point of h below the top in a uniformly accelerated field then you know from induction that it is faster at all points on the incline don't you?

The equations mentioned at top, are they right?
I assume the sphere has a greater speed for a given h. Followed by the cylinder and the hoop. This is known by looking at the equations? Is that right?

LowlyPion said:
Forgot to add that if it is faster at every point on the incline then it would be fastest to the bottom wouldn't it?

Yea..it would.. :smile:


tiny-tim said:
Hi brainracked! :smile:


v = dx/dt.

It's a slope, with angle θ, say …

so x (along the slope) = h/sinθ.

This isn't physics … this is geometry. :smile:


I understand that..it's just that I don't understand how there could be values for x and h..
 
  • #18


brainracked said:
The equations mentioned at top, are they right?
I assume the sphere has a greater speed for a given h. Followed by the cylinder and the hoop. This is known by looking at the equations? Is that right?

Yes, that is correct.

Here is a demonstration you might enjoy:

http://hyperphysics.phy-astr.gsu.edu/hbase/hoocyl.html#hc1
 

1. What factors affect the speed at which bodies reach the bottom?

The speed at which a body reaches the bottom is affected by several factors, including its mass, shape, and the force of gravity acting on it. Objects with greater mass will typically fall faster, while objects with a more streamlined shape will experience less air resistance and fall faster as well. Additionally, the force of gravity varies depending on the location and can influence the speed at which objects fall.

2. How does air resistance affect the speed of a falling object?

Air resistance, also known as drag, can slow down the speed of a falling object. As an object falls, it pushes against the air particles in its path, which creates a force in the opposite direction of its motion. This force increases as the object's speed increases, but eventually, the force of gravity will become greater and the object will reach a constant speed known as terminal velocity.

3. Does the shape of an object affect its acceleration?

Yes, the shape of an object can affect its acceleration. Objects with a greater surface area, such as a parachute, will experience more air resistance and have a slower acceleration compared to objects with a smaller surface area. As mentioned before, air resistance can slow down the speed of a falling object, thus affecting its acceleration.

4. How does the force of gravity differ on different planets?

The force of gravity is dependent on the mass and distance between two objects. Therefore, the force of gravity will differ on different planets due to variations in their mass and size. For example, the force of gravity on Earth is stronger than the force of gravity on the moon because the Earth has a greater mass.

5. Is there a limit to how fast an object can fall?

Yes, there is a limit to how fast an object can fall, known as terminal velocity. This is the point at which the force of air resistance becomes equal to the force of gravity, causing the object to reach a constant speed. The exact terminal velocity may vary depending on the object's shape, size, and the density of the surrounding air.

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