A hemispherical bowl of radius R is sitting upside down on a table. A small bead

Question

A hemispherical bowl of radius R is sitting upside down on a table. A small bead of mass M is placed at the "top" of the upside down bowl and given a (negligible) push to get it moving. The bead then slides along the surface of the bowl without friction. At some point, h, above the table, the bead will lose contact with the surface. Hint: cos=(2/3)
(a) What is the radial force?
(b) What math statement describes the situation of the bead losing contact with the bowl?
(c) Find the angle, measured from the vertical, at which point the bead loses contact with the bowl. It is possible to get a numerical answer even though no numbers are given!!

Explanation / Answer

a). radial force is the gravitational force exert on the bead on the radius component. b). The Gravitational force exert on the bead (radius component) equal to the centripetal force of the circular motion of the bead. c). assume the point is at angle alpha. Conservation of energy. mv^/2+mg*R*sin(alpha)=mg*R. so v^2=2*gR(1-sin(alpha)) centripetal force mv^2/R radial force. mg*sin(alpha). mv^2/R=mg*sin(alpha) so v^2=gR*sin(alpha). plug this in the first equation. gR*sin(alpha)=2*gR(1-sin(alpha)) so sin(alpha)=2/3. so cos(0)=2/3

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.