contestada

A mass m = 0.6 kg is released from rest at the top edge of a hemispherical bowl with radius = 1 meters. The mass then slides without friction down the inner surface toward the bottom of the bowl. At a certain point of its path the mass achieves a speed v = 4.22 m/s. At this point, what angle \theta\:θ ( in degrees) does the mass make with the top of the bowl?

Respuesta :

lcmendozaf

Answer:

θ = 65.31°

Explanation:

By conservation of energy:

Eo: m*g*R

Ef: [tex]m*g*( R - R*sin\theta ) + 1/2*m*V^2[/tex]

[tex]m*g*R=m*g*( R - R*sin\theta ) + 1/2*m*V^2[/tex]

[tex]m*g*R=m*g*R - m*g*R*sin\theta + 1/2*m*V^2[/tex]

[tex]0=- m*g*R*sin\theta + 1/2*m*V^2[/tex]

[tex]g*R*sin\theta = 1/2*V^2[/tex]

[tex]sin\theta = \frac{V^2}{2*g*R}[/tex]

θ = 65.31°

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