The real world application of the atwood’s machine is to be
able to demonstrate specific principles such as the acceleration and dynamics
in a way to understand and know how it is being demonstrated in the real world
and in different situations.
Explanation:
SI unit of force is <em>Newton</em>
Kepler's three laws of planetary motion can be stated as follows: (1) All planets move about the Sun in elliptical orbits, having the Sun as one of the foci. (2) A radius vector joining any planet to the Sun sweeps out equal areas in equal lengths of time.
Answer:
a) I = 0.0198 kg m²
, b) I = 21.85 kg m²
Explanation:
For this exercise we will use the definition of moment of inertia
I = ∫ r² dm
For body with high symmetry they are tabulated
sphere I = 2/5 m r²
bar with respect to center of mass I = 1/12 m L²
let's calculate the mass of each body
bar
ρ = m / V
m = ρ V
m = ρ l w h
where we are given the density of the bar rho = 32840 kg / m³ and its dimensions 1 m, 0.8 cm and 4 cm
m = 32820 1 0.008 0.04
m = 10.5 kg
Sphere
M = ρ V
V = 4/3 pi r³
M = rgo 4/3 π r³
give us the density 37800 kg / m³ and the radius of 5 cm
M = 37800 4/3 π 0.05³
M = 19.8 kg
a) asks us for the moment of inertia of the sphere with respect to its center of mass
I = 2/5 M r²
I = 2/5 19.8 0.05²
I = 0.0198 kg m²
b) the moment of inertia with respect to the turning point, for this we will use the theorem of parallel axes
I = I_cm + M d2
where d is the distance from the body to the point of interest
I_cm = 0.0198 kg m²
the distance to the pivot point is
l = length of the bar + radius of the sphere
l = 1 + 0.05 = 1.005 m
I = 0.0198 + 19.8 1.05²
I = 21.85 kg m²
Answer:
V = 0.714m/s
Explanation:
Full solution calculation can be found in the attachment below.
From the principle of conservation of linear momentum, the sum of momentum before collision equals the sum of momentum after collision.
Before collision only the train had momentum. After the collision the train and the boxcars stick together and move as one body. The initial momentum of the train is now shared with the boxcars as they move together as one body. The both move with a common velocity v.
See the attachment below for the solution calculation.
