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A 2 gram marble is placed in the bottom of a frictionless, hemispherical bowl with a diameter of 4.9 m and it is set in circular motion around the lowest point in the bowl such that its radius of "orbit" is 16 cm. What is its speed? (If you find this problem confusing, consider a simple pendulum with a string length half the diameter given here and suppose its amplitude is 16 cm.)

Respuesta :

Answer:

6.93 m/s

Explanation:

mass of marble, m = 2 g

Diameter of the bowl = 4.9 m

radius of bowl, r = half of diameter of the bowl = 2.45 m

The length of the string is 2.45 m

Now it is executing simple harmonic motion, thus the potential energy at the bottom is equal to the kinetic energy at the highest point.

[tex]mgr=\frac{1}{2}mv^{2}[/tex]

where, m be the mass of marble, v be the velocity of marble at heighest point, g be the acceleration due to gravity and r be the radius of the path which is equal to the length of the string

By substituting the values

9.8 x 2.45 = 0.5 x v^2

v = 6.93 m/s

Thus, the speed is 6.93 m/s.

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