1 T(x, y) = = m (x² + y²) 2

icon
Related questions
Question

Please show how to convert the cartesian coordinates (highlighted in picture)

Carriage on rotating beam
A carriage runs without friction along rails on a beam, as shown in the figure below. The carriage is
attached to a spring, of equilibrium length ro and force constant k. The other end of the spring is fixed.
On the carriage, another set of rails is perpendicular to the first along which a mass m moves, held by a
spring fixed on the beam, of zero equilibrium length and force constant K. The length of this second
spring is at all times considered small compared to ro. Beam, rails, springs, and carriage are massless
and non-bending. The whole system is forced to rotate in the plane of the beams about the point of
attachment of the first spring, with a constant angular speed w.
m
1000
k
(a) Using Cartesian coordinates in the laboratory (non-rotating) frame, construct the energy of the
system.
(b) Is the energy of the system conserved in the laboratory frame? Explain.
(a) The kinetic energy of the m in Cartesian coordinates in the laboratory (non-rotating) frame is
1
T(x, y) == m (x² + j;³²)
2
The potential energy of m is
1
V (11, 12)== (k (l₁ - ro)² + K 12)
2
where (t) are the total lengths of the springs.
The relation between 1;(t) and the Cartesian coordinates is
x = l₁ cos(t w) – ½ sin(t w)
-
y = l₁ sin(t w) + l½₂ cos(t w)
Transcribed Image Text:Carriage on rotating beam A carriage runs without friction along rails on a beam, as shown in the figure below. The carriage is attached to a spring, of equilibrium length ro and force constant k. The other end of the spring is fixed. On the carriage, another set of rails is perpendicular to the first along which a mass m moves, held by a spring fixed on the beam, of zero equilibrium length and force constant K. The length of this second spring is at all times considered small compared to ro. Beam, rails, springs, and carriage are massless and non-bending. The whole system is forced to rotate in the plane of the beams about the point of attachment of the first spring, with a constant angular speed w. m 1000 k (a) Using Cartesian coordinates in the laboratory (non-rotating) frame, construct the energy of the system. (b) Is the energy of the system conserved in the laboratory frame? Explain. (a) The kinetic energy of the m in Cartesian coordinates in the laboratory (non-rotating) frame is 1 T(x, y) == m (x² + j;³²) 2 The potential energy of m is 1 V (11, 12)== (k (l₁ - ro)² + K 12) 2 where (t) are the total lengths of the springs. The relation between 1;(t) and the Cartesian coordinates is x = l₁ cos(t w) – ½ sin(t w) - y = l₁ sin(t w) + l½₂ cos(t w)
Expert Solution
steps

Step by step

Solved in 4 steps with 3 images

Blurred answer