Particle in hemispherical bowl

[sailsinthesun]

[SOLVED] Particle in hemispherical bowl

Homework Statement

A 195 g particle is released from rest at point A along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 31.0 cm.
http://www.webassign.net/pse/p8-52.gif

(a) Calculate the gravitational potential energy of the particle-Earth system when the particle is at point A relative to point B.
(b) Calculate the kinetic energy of the particle at point B.
(c) Calculate its speed at point B.
(d) Calculate its kinetic energy when the particle is at point C.
(e) Calculate the potential energy when the particle is at point C.

Homework Equations

The Attempt at a Solution

(a)PE=mgh so PE=.195kg*9.8m/s^2*.31m=.59241J

(b)KE=.5mv^2 but I have no idea how to find the velocity at the point. Any help?

(c)Needed to answer B..

(d)Also need to know how to calculate velocity at this point...

(e)PE=mgh=(.195kg)(9.8m/s^2)(.2067m)=.39494

Any help on b, c, and d?

 

[tiny-tim]

Hi sailsinthesun!

(a) and (e) seem ok.

(b)KE=.5mv^2 but I have no idea how to find the velocity at the point. Any help?

(c)Needed to answer B..

KE=.5mv^2 is the answer to (b)! That's all (b) asks you for!

For (c) and for (d), you use "conservation of energy": KE + PE = constant (which, btw, should be in your "Relevant equations").

 

[keldon888]

(b)KE=.5mv^2 but I have no idea how to find the velocity at the point. Any help?

(c)Needed to answer B..

(d)Also need to know how to calculate velocity at this point...

B) At point B the PE is 0 and due to the conservation of energy KE is the same as the answer to A.

C) Now knowing what KE is work backwards with KE=.5mv^2 to find the velocity.

D) Also due to the conservation of energy, in A you found PE and since KE was 0 the entire energy of the system KE + PE = .59241J. So having the PE you found in part E you can substitute and find the KE.