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Frisbee physics problem

by shiri
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shiri
#1
Oct10-07, 02:25 PM
P: 85
One side of the roof of a building slopes up at 39.0. A student throws a Frisbee onto the roof. It strikes with a speed of 15.0 m/s and does not bounce, but slides straight up the incline. The coefficient of kinetic friction between the plastic and the roof is 0.380. The Frisbee slides 10.0 m up the roof to its peak, where it goes into free-fall, following a parabolic trajectory with negligible air resistance. Determine the maximum height the Frisbee reaches above the point where it struck the roof.


I think this one is the hardest question I have seen yet. In this question it says
the frisbee slides 10.0m up the roof to its peek, does it mean frisbee goes up diagonally
not vertically. If it goes diagonally I got the wrong answer for it. So how I can solve
this problem? Also, if a frisbee strikes the roof, does a frisbee stays 15.0m/s or it was
instantaneously slows down after it hits the roof?
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Doc Al
#2
Oct10-07, 02:44 PM
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Quote Quote by shiri View Post
I think this one is the hardest question I have seen yet. In this question it says
the frisbee slides 10.0m up the roof to its peek, does it mean frisbee goes up diagonally
not vertically.
Yes. It slides up the roof, so it moves along the incline of the roof.
If it goes diagonally I got the wrong answer for it. So how I can solve
this problem?
Find the net force on the frisbee as it slides up. Then you can find the acceleration and its speed as it flies off.
Also, if a frisbee strikes the roof, does a frisbee stays 15.0m/s or it was
instantaneously slows down after it hits the roof?
I would assume that the frisbee loses no speed when it hits the roof--its speed when it begins sliding up the roof is 15 m/s.
shiri
#3
Oct10-07, 04:14 PM
P: 85
Quote Quote by Doc Al View Post
Yes. It slides up the roof, so it moves along the incline of the roof.

Find the net force on the frisbee as it slides up. Then you can find the acceleration and its speed as it flies off.

I would assume that the frisbee loses no speed when it hits the roof--its speed when it begins sliding up the roof is 15 m/s.
Find the net force on the frisbee as it slides up. Then you can find the acceleration and its speed as it flies off.

I would assume that the frisbee loses no speed when it hits the roof--its speed when it begins sliding up the roof is 15 m/s.[/QUOTE]



Here are the calculations for this question

[x-component for F] = 0 - [Kinetic friction of force] - [mg*sin39] = ma (1)
[y-component for F] = [Normal force] - [mg*cos39] = 0 (2)


[Kinetic friction of force] = [Coefficient of kinetic friction]*[Normal force] (3)


then find "normal force" of (2)

[Normal force] = [mg*cos39] (4)


then substitute (4) into (3)

[Kinetic friction of force] = [Coefficient of kinetic friction]*[mg*cos39] (5)


then substitute (5) into (1)

-[Coefficient of kinetic friction]*[mg*cos39] - [mg*sin39] = ma


then find acceleration

a = [-([Coefficient of kinetic friction]*[mg*cos39])-(mg*sin39)] / m

= (-[Coefficient of kinetic friction]*[g*cos39])-(g*sin39)

= -g([(Coefficient of kinetic friction)*cos39]+[sin39])

= -9.0707m/s^2



Now I have to find time

v = at + [Initial velovity] *Now do the anti-derivative
d = [(at^2)/2] + [(Initial velocity)*t] + [Initial displacement]
d = [(at^2)/2] + [(Initial velocity)*t] *Initial displacement is gone because its zero


then substitute a into this equation which becomes

10 = [(-4.5353m/s^2)*(t^2)] + [15.0m/s*t]
0 = [(-4.5353m/s^2)*(t^2)] + [15.0m/s*t] - 10


then use quadratic equation

(-b[+-]sqrt[(b^2)-(4ac)])/(2a)
(-15[+-]sqrt[(15^2)-4(-4.5353)(-10)])/(2*[-4.5353])

t = 0.9258s, 2.3815s I choose 0.9258s for this calculation


substitute t into the velocity equation

v = at + [Initial velovity]
= (-9.0707m/s^2)*(0.9258s) + 15.0m/s
= 6.6020m/s


then substitute v into this formula

[Final velocity] = [Initial velocity] + at
0 = [Initial velocity]*sin39 - gt


find t

t = ([Initial velocity]*sin39)/g


then subsitute t into this formula

d = ([Initial velocity]*t) + (0.5*a*t^2)
h = (Initial velocity*sin39)*[([Initial velocity]*sin39)/g] - [(0.5*g)(([Initial velocity]*sin39)/g)]^2
h = [([Initial velocity]^2)*((sin 39)^2)]/(2g)
h = [((6.6020m/s)^2)*((sin39)^2)]/(2*9.81m/s^2)
h = 0.8798m


then find a height of the roof

h = (sin 39)*10
h = 6.2932m

then add it together

[Maximum height] = 0.8798m + 6.2932m
[Maximum height] = 7.1730m = 7.17m

Doc Al
#4
Oct10-07, 04:21 PM
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Frisbee physics problem

Quote Quote by shiri View Post
Here are the calculations for this question

[x-component for F] = 0 - [Kinetic friction of force] = ma (1)
[y-component for F] = [Normal force] - [mg*cos39] = 0 (2)
Don't forget the x-component of gravity.
shiri
#5
Oct10-07, 04:22 PM
P: 85
Quote Quote by Doc Al View Post
Don't forget the x-component of gravity.
What do you mean x-component of gravity?

I don't understand what you mean?
shiri
#6
Oct10-07, 05:00 PM
P: 85
So it becomes like this:


[x-component for F] = 0 - [Kinetic friction of force] - [mg*sin39] = ma (1)
[y-component for F] = [Normal force] - [mg*cos39] = 0 (2)
Doc Al
#7
Oct10-07, 05:08 PM
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Yes. Looks good.
shiri
#8
Oct10-07, 05:46 PM
P: 85
Quote Quote by Doc Al View Post
Yes. Looks good.
So I recalculate the question (look above) do I do all the calculations right?
Doc Al
#9
Oct10-07, 07:38 PM
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Quote Quote by shiri View Post
So I recalculate the question (look above) do I do all the calculations right?
Yes, looks OK.

But you could save yourself some effort if you learned another kinematic formula:

[tex]v^2 = v_0^2 + 2 a \Delta x[/tex]


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