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nexus9112 [7]
4 years ago
7

A carnival merry-go-round has a large disk-shaped platform of mass 120kg that can rotate about a center axle. A 60−kg student st

ands at rest at the edge of the platform 4.0m from its center. The platform is also at rest. The student starts running clockwise around the edge of the platform and attains a speed of 2.0m/s relative to the ground. (a) Determine the rotational velocity of the platform. (b) Determine the change of kinetic energy of the system consisting of the platform and the student.
Physics
1 answer:
amid [387]4 years ago
6 0

Answer:

a) The rotational velocity = 0.25rad/s

b) change in KE= 180JOULES

Explanation:

Applying the principles of rotational momentum conservation.

Let m be merry-go-round

s be student

L be the rotational momentum

Li = Lf

Imwf - IsVs/r

Imwf= IsVs/f

1/2 MmWf = 1/2MsVsr

Wf= MsVs/(Mmr)

Wf= 60×2/120×4

Wf= 0.25 rad/s

b)The change in system's KE= Ki - Kf

Change in KE= 1/2[(Im- Is)Wf^2 + MsVs^2 - 0]

Change in KE= 1/2[ 1/2 Mmr^2 + Msr^2)Wf^2 + MsVs^2]

Change in KE= 1/2[(1/2×120×4^2+60×4^2)× 0.25^2 +60×2^2]

CHANGE IN KE = 180 Joules

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Una atracción de feria consiste en lanzar un trineo de 2 kg por una rampa ascendente que forma un ángulo de 30º con la horizonta
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Answer:

La velocidad con la que se debe lanzar el trineo es aproximadamente 9.96 m/s

Explanation:

La masa del trineo, m = 2 kg

El ángulo de inclinación del trineo, θ = 30 °

El coeficiente de fricción, μ = 0,15

La altura a la que debe ascender el trineo, h = 4 m

La reacción normal del trineo en la rampa, N = m · g · cos (θ)

La fuerza de fricción F_f = N × μ

Dónde;

g = Th aceleración debida a la gravedad ≈ 9.81 m/s²

∴ F_f  = 0.15 × 2 kg × 9.81 m/s² × cos (30°) ≈ 2.55 N

La longitud de la rampa que se mueve el trineo, l = h/(sin(θ))

∴ l = 4/(sin(30°)) = 8

La longitud de la rampa que se mueve el trineo, l = 8 m

El trabajo realizado sobre la fricción, W_f = F_f × l

W_f = 2.55 × 8 ≈ 20.4

El trabajo realizado sobre la fricción, W_f ≈ 20.4 Julios

El trabajo realizado para levantar el trineo a 4 metros de altura, P.E. = m·g·h

∴ P.E. = 2 kg × 9.81 m/s² × 4 m ≈ 78.48 Joules

El trabajo realizado para levantar el trineo a 4 metros de altura, P.E. ≈ 78.48 Joules

El trabajo total requerido para levantar el trineo 4 metros por la rampa, W_t = W_f + P.E.

W_t = 20.4 J + 78.84 J ≈ 99.24 J

Energía cinética, K.E. = 1/2 × m × v²

La energía cinética necesaria para mover el trineo por la rampa, K.E. se da de la siguiente manera;

K.E. = 1/2 × 2 kg × v² ≈ 99.24 J

∴ v² ≈ 99.24 J/(1/2 × 2 kg)

La velocidad con la que se debe lanzar el trineo para que ascienda 4 metros por la rampa, v ≈ 9.96 m/s

4 0
3 years ago
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