A marble is placed at the top of an inverted hemispherical bowl of radius R = 0.

Question

A marble is placed at the top of an inverted hemispherical bowl of radius R = 0.21 m. It starts from rest and slides down the bowl without friction. Draw a free body diagram when the marble reaches an angular position = 30.9°. From your FBD, sketch the approximate direction of the acceleration.

Calculate the radial component of the acceleration (assuming that the radius of the marble is negligible).
(Hint: First find the velocity at = 30.9°.)
What is the tangential component of the acceleration when = 30.9°? (Hint: Use Newton's second law.)
What is the magnitude of the normal force when the marble loses contact with the bowl?
Draw a free body diagram for the angular position where the marble loses contact with the bowl. Draw the direction of the acceleration next to your FBD. What is the angle when the marble loses contact with the bowl (in degrees; don't enter unit)?

Explanation / Answer

The velocity at any given value of can be found using energy equations.

KE = PE
0.5 * m * v² = m * g * h
v² = 2 * g * h

where h is the distance dropped from the apex or

h = R * (1 - cos)

v² = 2 * g * R(1 - cos)

The centripetal acceleration needed to maintain contact at any angle is

ac = v² / r

ac = 2 * g * R(1 - cos) / R
ac = 2 * g(1 - cos)

The component of gravity available to act as centripetal acceleration is given by

ac' = g * cos

When the needed centripetal acceleration equals the available acceleration from gravity, the marble to surface force goes to zero.

ac = ac'
2 * g(1 - cos) = gcos
2 * (1 - cos) = cos
2 - 2cos = cos
2 = 3cos
cos = 2/3
= 48.2º

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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