A space shuttle with a mass of approximately 7.08 E5 kg is sitting on the launch pad. What would be the weight of the space shut
1 answer:
The weight of the shuttle is
Explanation:
The weight of an object on Earth is the gravitational force exerted by the Earth on the object.
The magnitude of the weight of an object is given by:
where
m is the mass of the object
is the acceleration of gravity on Earth's surface
And the direction is downward (towards the Earth's centre).
For the shuttle in this problem, its mass is
So, its weight is
Note that while the mass of an object (m) does not change, its weight (W) changes according to the location, since the value of g can be different at different location (for example, on the Moon, the value of g is about 1/6 the value of g on the Earth).
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The object's speed will not change.
In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:
where the first term is the resultant of the forces acting on the body, m is the mass of the object and a its acceleration, we see that if no forces act on the object, then the acceleration is zero. Therefore, the acceleration of the object is zero, and its velocity remains constant.
In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:
where the first term is the resultant of the forces acting on the body, m is the mass of the object and a its acceleration, we see that if no forces act on the object, then the acceleration is zero. Therefore, the acceleration of the object is zero, and its velocity remains constant.
Answer:
<em>a) 6738.27 J</em>
<em>b) 61.908 J</em>
<em>c) </em>
<em></em>
Explanation:
The complete question is
A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.
Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?
Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?
Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.
moment of inertia is given as
=
where m is the mass of the flywheel,
and r is the radius of the flywheel
for the flywheel with radius 1.1 m
and mass 11 kg
moment of inertia will be
=
= 6.655 kg-m^2
The maximum speed of the flywheel = 35 m/s
we know that v = ωr
where v is the linear speed = 35 m/s
ω = angular speed
r = radius
therefore,
ω = v/r = 35/1.1 = 31.82 rad/s
maximum rotational energy of the flywheel will be
E = = 6.655 x
= <em>6738.27 J</em>
<em></em>
b) second flywheel has
radius = 2.8 m
mass = 16 kg
moment of inertia is
=
=
= 62.72 kg-m^2
According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels
for the first flywheel, rotational momentum = = 6.655 x 31.82 = 211.76 kg-m^2-rad/s
for their combination, the rotational momentum is
where the subscripts 1 and 2 indicates the values first and second flywheels
= (6.655 + 62.72)ω
where ω here is their final angular momentum together
==> 69.375ω
Equating the two rotational momenta, we have
211.76 = 69.375ω
ω = 211.76/69.375 = 3.05 rad/s
Therefore, the energy stored in the first flywheel in this situation is
E = = 6.655 x
= <em>61.908 J</em>
<em></em>
<em></em>
c) one third of the initial energy of the flywheel is
6738.27/3 = 2246.09 J
For the car, the kinetic energy =
where m is the mass of the car
is the velocity of the car
Equating the energy
2246.09 =
making m the subject of the formula
mass of the car m =