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?
Answers: 2
![\theta = sin ^{-1} [ \frac{1}{2* g* r } * v^2]](/tpl/images/0721/8913/8c5f8.png)
![\theta = sin ^{-1} [ \frac{1}{2* 9.8* 1.1 } * (3.57)^2]](/tpl/images/0721/8913/0bd82.png)
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A mass m = 0.6 kg is released from rest at the top edge of a hemispherical bowl with radius = 1.1 me...
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