Please help me out with this question. I\'d much appreciate you showing all the

Question

Please help me out with this question. I'd much appreciate you showing all the work

Thanks :)

Jess

A marble of mass m is held against the side of a hemispherical bowl, as shown in the figure, and then released. It rolls without slipping. The initial position of the marble is such that it make angle of 30 degree with the vertical (sec figure). The marble radius is Rm = 10 mm, and the radius of the bowl is Rb = 100 mm. Find the angular speed of the marble when it reaches the bottom. What is the linear speed of the center-of-mass?

Explanation / Answer

Here ,

radius of marble , Rm = 10 mm

radius of bowl , Rb = 100 mm

theta = 30 degree

let the angular speed is w

speed of the marble is v

A)
as the marble is rolling without slipping ,

v = Rm * w

Using conservation of energy

- change in potential energy = kinetic energy of marbel

m * g * (Rb - Rm) * (1 - cos(30)) = 0.5 * I * w^2 + 0.5 * m * v^2

m * 9.8 * (0.1 - 0.01) * (1 - cos(30)) = 0.5 * 0.4 * m * 0.01^2 * w^2 + 0.5 * m * w^2 * 0.01^2

9.8 * (0.1 - 0.01) * (1 - cos(30)) = 0.5 * 0.4* 0.01^2 * w^2 + 0.5 * w^2 * 0.01^2

solving for w

w = 41.1 rad/s

the angular speed of the marble is 41.1 rad/s

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b)

linear speed , v= Rm * w

v = 0.01 * 41.1 m/s

v = 0.411 m/s

the linear speed of the centre of mas sis 0.411 m/s

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