Answer:
c. V = 2 m/s
Explanation:
Using the conservation of energy:
so:
Mgh = 
where M is the mass, g the gravity, h the altitude, I the moment of inertia of the pulley, W the angular velocity of the pulley and V the velocity of the mass.
Also we know that:
V = WR
Where R is the radius of the disk, so:
W = V/R
Also, the moment of inertia of the disk is equal to:
I = 
I = 
I = 10 kg*m^2
so, we can write the initial equation as:
Mgh = 
Replacing the data:
(5kg)(9.8)(0.3m) = 
solving for V:
(5kg)(9.8)(0.3m) = 
V = 2 m/s
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
Answer:
605447.7066 kgm²/s
Explanation:
= Mass of sphere = 10000 kg
= Mass of rod = 10 kg
r = Radius of sphere = 2 m
l = Length of antenna = 3 m
Angular speed
Angular momentum is given by
Moment of inertia of the satellite is
Moment of antenna of the satellite is
The angular momentum of the system is
The angular momentum of the satellite is 605447.7066 kgm²/s
Answer: -0.84 rad/sec (clockwise)
Explanation:
Assuming no external torques act on the system (man + turntable), total angular momentum must be conserved:
L1 = L2
L1 = It ω + mm. v . r = 81.0 kg . m2 .21 rad/s – 56.0 kg. 3.1m/s . 3.1 m
L1 = -521.15 kg.m2/sec (1)
(Considering to the man as a particle that is moving opposite to the rotation of the turntable, so the sign is negative).
Once at rest, the runner is only a point mass with a given rotational inertia respect from the axis of rotation, that can be expressed as follows:
Im = m. r2 = 56.0 kg. (3.1m)2 = 538.16 kg.m2
The total angular momentum, once the runner has come to an stop, can be written as follows:
L2= (It + Im) ωf = -521.15 kg.m2/sec
L2= (81.0 kg.m2 + 538.16 kg.m2) ωf = -521.15 kg.m2/sec
Solving for ωf, we get:
ωf = -0.84 rad/sec (clockwise)
