210_lab6 - gR(Min value of “w” for which the ball can...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Introduction A small ball rolls around a horizontal circle at height y inside a frictionless hemispherical bowl of radius R. Part a: We will find the expression for the ball’s angular velocity in terms of R, y, g. Part b: We will find the minimum value of “w” for which the ball can move in a circle. Part c: We will determine “w” in rpm if R = .2 m and the ball is halfway up. Data a) Using Newton’s second law along the r and z directions we find: ΣF r =ncosθ=mr w2 ΣF z =nsinθ-F G =0N Using F G and dividing these equations produces: tanθ= grw2 = - R yr The figure displays tan θ = - R yr and: w= - gR y (The ball’s angular velocity) b) w min =
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: gR (Min. value of “w” for which the ball can move in a circle) c) Substitution: w=-gR y = . / .- . 9 8 m s2 2m 1m = 9.9 rads x 60 s1 min x 1 rev2πrad = 95 rpm Conclusion The critical angular velocity occurs when gravity alone is sufficient to cause circular motion at the top of the bowl. Circular motion with an angular velocity less than w min is impossible because there is excessive downward force. The angular velocity depends on the height at which you want to revolve the ball. The height of the angle will change and vertical component of the normal reaction will change accordingly....
View Full Document

Ask a homework question - tutors are online