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

Answer:

The  angle is  $\theta = 36.24 ^o$

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

      $m * g * h = \frac{1}{2} * m * v^2$

 =>    $h = \frac{1}{2 g } * v^2$

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

      $h = r * sin (\theta )$

So

     $\theta = sin ^{-1} [ \frac{1}{2* g* r } * v^2]$

substituting values

     $\theta = sin ^{-1} [ \frac{1}{2* 9.8* 1.1 } * (3.57)^2]$

     $\theta = 36.24 ^o$