Two ramps, one friction, one frictionless

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Homework Statement

A ball rolls down a frictionless ramp that has an inclination angle of 25 degrees to the horizontal. The ball then slides across a horizontal frictionless floor that is 20 m to the next ramp. The next ramp has a coefficient of friction of 0.10 and has an inclination angle of 10 degrees. How far up the second ramp will the ball go?

Homework Equations

• Fnet = m*a
• Fnet = F1 + F2 + F3 etc.
• Fg = m*g (g = 9.8 m/s^2)
• given that there's distance involved, probably a few kinematics equations also

The Attempt at a Solution

okay well first things first, free body diagram. keep in mind this is my interpretation from the question, it wasn't given so tell me if there's anything wrong with it:

since the force of gravity is at an angle, i'll have a rotated coordinate system to accommodate that:

on ramp #1, there's no movement in the y-direction but there is in the x-direction. so fnetx = fgx, or max = mgsin25, cancel out the masses and you have ax = gsin25. okay, i have acceleration, i know vi = 0, and uh, that's pretty much where I'm at.

it asks for the distance up the second incline, I'm guessing i'll need to find the speeds along each "section" and then use vf for the frictionless floor as vi for the second ramp, i can get acceleration in the same way except i'll have to involve friction, but still that's only vi and a again.

how do i do this, guys?

 

[spikethekitty]

The only thing I see that you need to solve the problem is the initial height or distance up the first ramp of the ball. Follow through on your plan and see what you get.

 

[page123]

The only thing I see that you need to solve the problem is the initial height or distance up the first ramp of the ball. Follow through on your plan and see what you get.

how do i follow through though? i'll only get the acceleration basically for the first ramp, and I'm guessing vi for the first ramp is 0.

 

[spikethekitty]

The general kinematics equation can be manipulated to find what you are looking for.

x_f = x_i + v_i t + \frac{1}{2} a t^2

Since the problem doesn't state that there is an initial velocity it is safe to assume that it is 0.
Things to keep in mind:
the ball doesn't lose any velocity on the frictionless floor
you are trying to find where the ball stops (v final)
the process of the ball rolling up the hill is the same physically forward as it is backward. (This applies to initial and final velocities.)
Also think about the relationship between velocity, acceleration, and distance in terms of time.

Is the initial position given? If not there is a very different way this must be attacked.

 

[SammyS]

How far up the frictionless ramp is the ball at the start?

The ball will not rotate until it reaches the second ramp. The friction will cause the ball to begin to rotate. What effect if any do you think the rotation and the friction will have?

 

[page123]

The general kinematics equation can be manipulated to find what you are looking for.

x_f = x_i + v_i t + \frac{1}{2} a t^2

Since the problem doesn't state that there is an initial velocity it is safe to assume that it is 0.
Things to keep in mind:
the ball doesn't lose any velocity on the frictionless floor
you are trying to find where the ball stops (v final)
the process of the ball rolling up the hill is the same physically forward as it is backward. (This applies to initial and final velocities.)
Also think about the relationship between velocity, acceleration, and distance in terms of time.

Is the initial position given? If not there is a very different way this must be attacked.

The initial position is not given. I do think the ball rolling up is the same as rolling down, but not when there's one that's frictionless and one that's friction. In fact, on a frictionless surface the ball should continue moving forever, right? So would that mean the acceleration would be the same during the horizontal surface phase? If so, I would have distance, and acceleration in my assets to get the vi of the second incline.

I think I still need help.

 

[RANDHIR SINGH]

Find the solution on attached doc file

 

[SammyS]

The initial position is not given. I do think the ball rolling up is the same as rolling down, but not when there's one that's frictionless and one that's friction. In fact, on a frictionless surface the ball should continue moving forever, right? So would that mean the acceleration would be the same during the horizontal surface phase? If so, I would have distance, and acceleration in my assets to get the vi of the second incline.

I think I still need help.

The velocity constant on the horizontal floor, so the acceleration is zero there.

Regarding RANDHIR SINGH's solution: Not all of the work done by friction is lost (to heat). Some of it goes into angular acceleration of the ball. Once the ball's rotation matches the ball's speed, friction will do no more work, but merely keep the contact point from sliding.

 

[page123]

Find the solution on attached doc file

Uh wow, that's more of an answer than I was expecting. Still, I think you may condensed it a bit too much, and one of the symbols is just a rectangle. I think it might be better if you referred to (or editted) my own one and just post it here - sorry, it's just a little confusing. But more importantly, what equations were you using? vi^2 = a + ... ? And also, the answer is 65% of the original height but I'm pretty sure the answer should be in m (without percentages).

 

[SammyS]

...
And also, the answer is 65% of the original height but I'm pretty sure the answer should be in m (without percentages).

Expressing the final height as a % of the initial height makes perfect sense!

How can you express the final height in meters, when the initial height is not given?

 

[page123]

Expressing the final height as a % of the initial height makes perfect sense!

How can you express the final height in meters, when the initial height is not given?

But doesn't Randhir use 'h' (the initial height) in his solution? Can't he solve for it? The solution is kind of confusing it's like

([] sin25 degrees = h/S)​

 

[SammyS]

But doesn't Randhir use 'h' (the initial height) in his solution? Can't he solve for it? The solution is kind of confusing it's like

([] sin25 degrees = h/S)​

So, do you know S in meters? That's equivalent to knowing h, or knowing the starting distance up the ramp.

 

[RANDHIR SINGH]

Some Explannation regarding my given answer:
In the figure, h is the initial height from which the body starts rolling down from smooth ramp & S is the distance it travels to come down to the first ramp.
and h', S' are the height distance on the second ramp respectively.
The equation used by me is basic equation of uniformly accelerated motion namely V²-u²=2aS.
Regarding confusing symbol, it is degree symbol over the angle.
I ignored rotational motion of the ball because the mass and radius of the ball are not given in the statement, which means its moment if inertia {I=(2/5)mr²} is negligibel (i.e. radius of ball is negligible) so that rotational Kinetic energy is also ignored.
At last to get the answer in metres, one can put the value of h to find answer of h' in m. (e.g. if h=100m, then h'=0.64x100=64m)
Thanks

 

[SammyS]

...

I ignored rotational motion of the ball because the mass and radius of the ball are not given in the statement, which means its moment if inertia {I=(2/5)mr²} is negligible (i.e. radius of ball is negligible) so that rotational Kinetic energy is also ignored.
...
Thanks

If this is the case, then friction up the ramp also is ignorable, unless we model the ball as a flat object which slides and doesn't roll.

 

[spikethekitty]

While you can't ignore friction on the ramp, you could consider the ball with coefficient of rolling friction of .1 to be a block sliding with coefficient of sliding friction of .1 also. That doesn't change the physics or calculations involved unless you get very specific. But I am sure that is a lot more in depth than you are expected to go.

 

[page123]

Some Explannation regarding my given answer:
In the figure, h is the initial height from which the body starts rolling down from smooth ramp & S is the distance it travels to come down to the first ramp.
and h', S' are the height distance on the second ramp respectively.
The equation used by me is basic equation of uniformly accelerated motion namely V²-u²=2aS.
Regarding confusing symbol, it is degree symbol over the angle.
I ignored rotational motion of the ball because the mass and radius of the ball are not given in the statement, which means its moment if inertia {I=(2/5)mr²} is negligibel (i.e. radius of ball is negligible) so that rotational Kinetic energy is also ignored.
At last to get the answer in metres, one can put the value of h to find answer of h' in m. (e.g. if h=100m, then h'=0.64x100=64m)
Thanks

What equation is V²-u²=2aS in basic form using vi, vf, a, d, and t?

I'm still not able to fully to understand your solution, the rectangle is a degree symbol?

So it's '(degree symbol sin 10 degree symbol)' I think it would be better if it was posted on the forum?

 

[page123]

What equation is V²-u²=2aS in basic form using vi, vf, a, d, and t?

I'm still not able to fully to understand your solution, the rectangle is a degree symbol?

So it's '(degree symbol sin 10 degree symbol)' I think it would be better if it was posted on the forum?

Just wondering if someone could reply to this.

 

[SammyS]

What equation is V²-u²=2aS in basic form using vi, vf, a, d, and t?

I'm still not able to fully to understand your solution, the rectangle is a degree symbol?

So it's '(degree symbol sin 10 degree symbol)' I think it would be better if it was posted on the forum?

That's v_f² - v_i² = 2ad .

You may have seen it as: v_f² = v_i² + 2ad .

... maybe even: v² = v₀² + 2ad