An 8-inch dinner knife is sitting on a circular plate so that its ends are on the edge of the plate. if the minor arc that is in
1 answer:
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The magnitude of the air drag when the object is traveling at terminal velocity is C. 850 Newtons
Explanation:
There are only two forces acting on the object here:
- The force of gravity, of magnitude , acting downward
- The air drag, , acting upward
Therefore, the equation of motion for the object is
where m is the mass of the object and a its acceleration.
The object in this problem is traveling at terminal velocity: this means that the acceleration is zero, so
a = 0
Therefore the equation becomes
which means that the magnitude of the air resistance is equal to the magnitude of the force of gravity:
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a) See free-body diagram in attachment
b) Net force in the y-direction: [/tex]
c) The velocity at which the roller coaster will fall is [/tex]
d) The speed of the roller coaster must be 17.1 m/s
e) The roller coaster should start from a height of 90 m
f) The roller coaster should start from a height of 100 m
Explanation:
a)
See the free-body diagram in attachment. There are only two forces acting on the roller coaster at the top of the loop:
- The weight of the roller coaster, acting downward, indicated by
(where m is the mass of the roller coaster and g is the acceleration of gravity)
- The normal reaction exerted by the track on the roller coaster, acting downward, and indicated with N
The two forces are represented in the diagram as two downward arrows (the length is not proportional to their magnitude, in this case)
b)
Since there are only two forces acting on the roller coaster at the top of the loop, and both forces are acting downward, then we can write the vertical net force as follows (we take downward as positive direction):
where
mg is the weight
N is the normal reaction
Since the roller coaster is in circular motion, this net force must be equal to the centripetal force, therefore
where v is the speed of the car at the top of the loop and r is the radius of the loop.
c)
For this part of the problem, we start from the equation written in part b)
where the term on the left represents the centripetal force, and the terms on the right are the weight and the normal reaction.
We now re-arrange the equation making v (the speed) as the subject:
However, the velocity at which the roller coaster will fall is the velocity at which the normal reaction becomes zero (the roller coaster loses contact with the track), so when
N = 0
And as a result, the minimum velocity of the cart is
d)
In this part, we are told that the radius of the loop is
r = 30 m
And the mass of the cart is
m = 50 kg
Moreover, the acceleration of gravity is
We said that the minimum velocity that the cart must have in order not to fall at the top is
And substituting, we find
e)
According to the law of conservation of energy, the initial gravitational energy of the roller coaster at the starting point must be equal to the sum of the kinetic energy + gravitational potential energy at the top of the loop, therefore:
where
h is the initial height at the starting point
(2r) is the height of the roller coaster at the top of the loop
We can re-arrange the equation making h the subject,
And substituting the minimum speed of the cart,
this becomes
And since r = 30 m, we find
f)
In this case, 10% of the initial energy is lost during the motion of the roller coaster. We can rewrite the equation of the previous part as
Because only 90% (0.90) of the initial energy is converted into useful energy (kinetic+potential) when the cart reaches the top of the loop.
Re-arranging the equation, this time we get
Again, by substituting , we get
And therefore, the new initial height must be
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