Question

A mass m = 0.6 kg is released from rest at the top edge of a hemispherical bowl with radius = 1.1 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 = 3.57 m/s. At this point, what angle \theta\:θ ( in degrees) does the mass make with the top of the bowl?

Answer 1

The angle is
`\theta = 36.24 ^o`

Step-by-step explanation:

From the question we are told that

The mass is
`m = 0.6 \ kg`

The radius is
`r = 1.1 \ m`

The speed is
`v = 3.57 \ m /s`

According to the law of energy conservation

The potential energy of the mass at the top is equal to the kinetic energy at the bottom i.e

`m * g * h = (1)/(2) * m * v^2`

=>
`h = (1)/(2 g ) * v^2`

Here h is the vertical distance traveled by the mass which is also mathematically represented as

`h = r * sin (\theta )`

So

`\theta = sin ^(-1) [ (1)/(2* g* r ) * v^2]`

substituting values

`\theta = sin ^(-1) [ (1)/(2* 9.8* 1.1 ) * (3.57)^2]`

`\theta = 36.24 ^o`