Motion with frictional force but without driving force

In summary, The engine of mass m moves without driving force but under the influence of the frictional force f(v)=(alpha)+(Beta)v2 on horizontal rails. The initial velocity is v0. The problem involves determining the time it takes for the engine to come to rest and the distance it covers. To solve this, the frictional force function is used to derive an expression for the acceleration, which is then integrated to obtain a function for the velocity with respect to time. The acceleration is not constant, as the velocity and force both change, and therefore the velocity cannot be linear with time.
  • #36
so.. for the part A 2nd part.. is ask for the maximum deceleration as initial velocity approaching infinity.. Does that means my final velocity still remain the same? which is 0?
 
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  • #37
LockeZz said:
so.. for the part A 2nd part.. is ask for the maximum deceleration as initial velocity approaching infinity.. Does that means my final velocity still remain the same? which is 0?
Yes, the final velocity is still zero, but you are asked to find the maximum deceleration time as the initial velocity v0 becomes larger and larger.
 
  • #38
alright.. i have substitute the final velocity to be 0 and initial velocity which is very large within the arctangent will result in 900 which is half of the pi.

so the time for maximum deceleration in which v0 approaching 0 will be:

[tex]
t=(\frac{\pi}{2})\sqrt{\frac{1}{\alpha\beta}}
[/tex]
 
  • #39
Almost correct. You forgot to multiply by the mass.
 
  • #40
oops another careless mistake... so the time for maximum deceleration is this:

[tex]

t=m(\frac{\pi}{2})\sqrt{\frac{1}{\alpha\beta}}

[/tex]

Do i need to find the acceleration function?
 
  • #41
The problem is not asking for the acceleration. You need to move to the next question. It is asking to find the distance traveled when v0 is very large. To do this

1. Go back to

[tex]
t=m\sqrt{\frac{1}{\alpha\beta}}(tan^-1(v_0\sqrt{\frac{\beta}{\alpha}})-tan^-1(v\sqrt{\frac{\beta}{\alpha}}))
[/tex]

and find what it looks like when v0 is very large.
2. Find an expression for v(t) by inverting the equation.
3. Integrate to find x(t).
4. Evaluate x(t) at the limiting time [tex]

t=(\frac{\pi}{2})\sqrt{\frac{1}{\alpha\beta}}

[/tex]
 
  • #42
i had try to substitute the v0 as infinity and then invert the tangent to get v(t) which i gained :

[tex]
v=\sqrt{\frac{\alpha}{\beta}}(tan(\frac{\pi}{2}-\frac{t}{m}(\sqrt{\alpha\beta})))
[/tex]

then i integrate it by using the substitution method ( taking x=distance traveled and x0 as starting point which is 0 and t0= 0 and
[tex] t_1=(\frac{\pi}{2})\sqrt{\frac{1}{\alpha\beta}} [/tex]

i get :

[tex]
x-x_0=\frac{m}{\beta}(\ln|\frac{\sqrt{\alpha\beta}}{m}t_1|-\ln|\frac{\sqrt{\alpha\beta}}{m}t_0|
[/tex]

then after substitute t0 and x0 with zero together with the t1 .. i get the distance as :

[tex]
x=\frac{m}{\beta}\ln|\frac{\pi}{2m}|
[/tex]
 
  • #43
I am sorry, but I misled you earlier. First you need to invert the equation to find v(t), then you integrate to find x(t), then you take v0 to its limiting case and evaluate at the calculated time. The way I suggested at first gives an infinity (you have to take log[0]) which is wrong. Also, check your algebra as you go along and do dimensional analysis to make sure you didn't miss anything.
 
  • #44
i was stucked at here... how to invert the subtraction of two arctangent?

[tex]
\frac{\sqrt{\alpha\beta}}{m}t=\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\arctan(v\sqrt{\frac{\beta}{\alpha}})
[/tex]
 
  • #45
Don't forget that

[tex]arctan(v_0 \sqrt{ \frac{\beta}{\alpha}})[/tex]

is a constant. Move the two terms that don't have v in them over to the other side of the equation, and take the tangent on both sides.
 
Last edited:
  • #46
im sorry.. i don't quite understand what u mean.. can explain in another word?
 
  • #47
I mean simple algebra like

[tex]

\arctan(v\sqrt{\frac{\beta}{\alpha}})=\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t

[/tex]

Now take the tangent on both sides and see what you get.
 
Last edited:
  • #48
i give it a try.. and i get this :

[tex]
v=\sqrt{\frac{\alpha}{\beta}}tan(\arctan(v_0\ sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t)
[/tex]
 
  • #49
Good. Now multiply both sides by dt and integrate to find x(t).
 
  • #50
By integration with substitution in which i assign the limit for dx(x,x0)and limt for dt(t1,t0) :

[tex]
x-x_0=-\frac{m}{\beta}(\ln(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1)-\ln(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_0))
[/tex]
 
  • #51
Your expression does not look right. The integral is of the form

[tex] \int Tan(\delta - \theta)d \theta[/tex]

Can you find what the indefinite integral is?
 
  • #52
hm.. i think i know where did i went wrong, i miss out the secant:

[tex]

x-x_0=-\frac{m}{\beta}(\ln(\sec(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1))-\ln(\sec(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_0)))
[/tex]
 
  • #53
Incorrect. I suggest you look it up on the web. Just google "Integral Tables" and take your pick.
 
  • #54
i have check on the table.. is cosine..
[tex] x-x_0=-\frac{m}{\beta}(\ln(\cos(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1))-\ln(\cos(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_0)))

[/tex]
 
  • #55
Can you make the expression more compact? What is the difference between two logarithms? Also set t0=0 and see what you get.
 
  • #56
ok.. the logarithm still can be simplify :

[tex]
x-x_0=-\frac{m}{\beta}(\ln(\frac{\cos(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1)}{\cos(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_0)})


[/tex]
so.. for t0=0.. i will get :

[tex]
x-x_0=-\frac{m}{\beta}(\ln(\frac{\cos(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1)}{\cos(\arctan(v_0\sqrt{\frac{\beta}{\alpha}}))})


[/tex]
 
  • #57
Good. I can see the light at the end of the tunnel. Can you? What do you think you should do next?
 
  • #58
get rid of the x0 ad substitute the t1?
 
  • #59
LockeZz said:
get rid of the x0 ad substitute the t1?
Yes, you may assume that x0=0. Substitute t1 and also the time has come to let v0 become very large.
 
  • #60
after the substitution, i get :
[tex]
x=-\frac{m}{\beta}(\ln(\frac{\cos(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\pi}{2})}{\cos(\arctan(v_0 \sqrt{\frac{\beta}{\alpha}}))})
[/tex]

since
[tex]


t_1=m(\frac{\pi}{2})\sqrt{\frac{1}{\alpha\beta}}


[/tex]
should i bother about v0?
 
  • #61
LockeZz said:
should i bother about v0?
Not yet. Do you recognize the ratio in the argument of the log for what it is? Specifically, can you simplify the numerator some?
 
  • #62
i think i can convert to tangent:

[tex]
x=-\frac{m}{\beta}(\ln(\tan(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})))
[/tex]

then simplify again:

[tex]
x=-\frac{m}{\beta}(\ln(v_0\sqrt{\frac{\beta}{\alpha}}))
[/tex]
 
  • #63
Note that if you let v0 become very large, you get an infinity. I hadn't appreciated that in the beginning. One thing that bothers me is the negative sign up front. Certainly x must be positive and the argument of the log is greater than 1 especially if v0 becomes very large. Check your integral tables and don't forget that you are integrating

[tex]
\int Tan(\delta - t)dt
[/tex]

There is a negative sign in front of t.

Check that and you are done. :smile:
 
  • #64
hm.. does that mean that the negative sign represent a mistake in the expression?
 
  • #65
All I am saying is that

[tex]

\int Tan(\delta - t)dt=Log[Cos(t-\delta)]

[/tex]
 
  • #66
I shall check back from the beginning the see if any mistake again.. thanks for the guidance by the way.. i really appreciate on your help.. Thank you very much :)
 
<h2>1. What is frictional force?</h2><p>Frictional force is a resistance force that occurs when two surfaces come into contact and slide against each other. It acts in the opposite direction of motion and is caused by the microscopic irregularities on the surfaces.</p><h2>2. How does friction affect motion without a driving force?</h2><p>Frictional force can slow down or stop the motion of an object without a driving force. This is because the force of friction acts in the opposite direction of motion, causing the object to lose its momentum and eventually come to a stop.</p><h2>3. Can frictional force be beneficial in motion without a driving force?</h2><p>Yes, frictional force can be beneficial in certain situations. For example, it allows us to walk without slipping and helps vehicles come to a stop when the brakes are applied. However, in most cases, frictional force is considered a hindrance to motion.</p><h2>4. How is frictional force calculated in motion without a driving force?</h2><p>The magnitude of frictional force can be calculated using the formula F = μN, where F is the frictional force, μ is the coefficient of friction, and N is the normal force between the two surfaces in contact. The direction of frictional force is always opposite to the direction of motion.</p><h2>5. How can frictional force be reduced in motion without a driving force?</h2><p>Frictional force can be reduced by using lubricants, such as oil or grease, between the two surfaces in contact. This creates a thin layer that reduces the surface roughness and allows for smoother motion. Additionally, using smoother surfaces or reducing the weight of the object can also help reduce frictional force.</p>

1. What is frictional force?

Frictional force is a resistance force that occurs when two surfaces come into contact and slide against each other. It acts in the opposite direction of motion and is caused by the microscopic irregularities on the surfaces.

2. How does friction affect motion without a driving force?

Frictional force can slow down or stop the motion of an object without a driving force. This is because the force of friction acts in the opposite direction of motion, causing the object to lose its momentum and eventually come to a stop.

3. Can frictional force be beneficial in motion without a driving force?

Yes, frictional force can be beneficial in certain situations. For example, it allows us to walk without slipping and helps vehicles come to a stop when the brakes are applied. However, in most cases, frictional force is considered a hindrance to motion.

4. How is frictional force calculated in motion without a driving force?

The magnitude of frictional force can be calculated using the formula F = μN, where F is the frictional force, μ is the coefficient of friction, and N is the normal force between the two surfaces in contact. The direction of frictional force is always opposite to the direction of motion.

5. How can frictional force be reduced in motion without a driving force?

Frictional force can be reduced by using lubricants, such as oil or grease, between the two surfaces in contact. This creates a thin layer that reduces the surface roughness and allows for smoother motion. Additionally, using smoother surfaces or reducing the weight of the object can also help reduce frictional force.

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