Question

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?

Answer 1

θ = 65.31°

Step-by-step explanation:

By conservation of energy:

Eo: m*g*R

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

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

`m*g*R=m*g*R - m*g*R*sin\theta + 1/2*m*V^2`

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

`g*R*sin\theta = 1/2*V^2`

`sin\theta = (V^2)/(2*g*R)`

θ = 65.31°