How Does a Marble's Speed Change as It Rolls Down a Hemispherical Bowl?

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The discussion centers on determining the rotational speed of a marble as it rolls down a hemispherical bowl. The marble starts at an angle of 30 degrees with the vertical and rolls without slipping. Key equations mentioned include the relationship between angular velocity and linear velocity, as well as the potential use of energy conservation principles. Participants express uncertainty about the initial steps in solving the problem, particularly regarding the application of energy conservation. The conversation highlights the importance of understanding both rotational dynamics and energy conservation in solving the problem effectively.
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Homework Statement


A marble of inertia m is held against the side of a hemispherical bowl as shown in (Figure 1) and then released. It rolls without slipping. The initial position of the marble is such that an imaginary line drawn from it to the center of curvature of the bowl makes an angle of 30∘ with the vertical. The marble radius is Rm = 10 mm, and the radius of the bowl is Rb = 200 mm .

Determine the rotational speed of the marble about its center of mass when it reaches the bottom.
Mazur1e.ch12.p69.jpg

Homework Equations


w=v/r
v=R2pi/T
maybe idk w=(delta(theta))/(delta(time))

The Attempt at a Solution


i wasn't really sure where to start but i know that change in theta would be pi/6
 
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L Knope said:

Homework Statement


A marble of inertia m is held against the side of a hemispherical bowl as shown in (Figure 1) and then released. It rolls without slipping. The initial position of the marble is such that an imaginary line drawn from it to the center of curvature of the bowl makes an angle of 30∘ with the vertical. The marble radius is Rm = 10 mm, and the radius of the bowl is Rb = 200 mm .

Determine the rotational speed of the marble about its center of mass when it reaches the bottom.
Mazur1e.ch12.p69.jpg

Homework Equations


w=v/r
v=R2pi/T
maybe idk w=(delta(theta))/(delta(time))

The Attempt at a Solution


i wasn't really sure where to start but i know that change in theta would be pi/6

Is energy conserved? If yes, why not use it?
 
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