Uniform ball rolling without slipping problem

[pentazoid]

Homework Statement

A uniform ball of mass M and raduis a can roll without slipping on the rough outer surface of a fixed sphere of raduis b and centre O. Initially the ball is at rest at the highest point of the phere when it is slightly disturbed . Find the speed of the center the G of the ball in terms of the variable theta , the angle between the line OG and the upward vertical. [Assume planar motion]. Show that the ball will leave the sphere when cos (theta)=10/17

Homework Equations

linear momentum principle

M*dV/dt=dP/dt=F

The Attempt at a Solution

answer: v^2=(10/7)*g(a+b)(1-cos(theta)

since they give you v^2 and v^2 is associated with the kinetic energy of a particle, should I apply the Energy principle rather than the linear momentum principle.

I probably should break the x and y components of the ball with radius b and the hemisphere with to sin(theta) and cos(theta) components. the y- componet will contained the weight of the balls while the x-component will not.

 

[tiny-tim]

Hi pentazoid!

should I apply the Energy principle rather than the linear momentum principle.

Yes! You must use energy to find the speed …

how could you apply linear momentum when there's no linear motion and there's gravity?

But you will need Newton's second law to find when the normal force is zero.

I probably should break the x and y components of the ball with radius b and the hemisphere with to sin(theta) and cos(theta) components. the y- componet will contained the weight of the balls while the x-component will not.

hmm … probably easier to use radial and tangential components …

the ball will lose contact when the normal (ie radial) force is zero.

 

[pentazoid]

Hi pentazoid!


Yes! You must use energy to find the speed …

how could you apply linear momentum when there's no linear motion and there's gravity?

But you will need Newton's second law to find when the normal force is zero.

V=Mgh = Mg(b+a)cos(theta) (h=0 at theta=90 degrees)

E=T+V

at top , ball is at rest
E(initial)=1/2*M*(0)^2+1/2*I*(0)+Mg(a+b)=Mg(a+b)

E(final)= 1/2*Mv^2+1/2*(2/5*Ma^2)(v/a)^2+ Mg(b+a)cos(theta)

E(i)=E(f) ==> Mg(b+a)=1/2*M*(7/5)v^2+Mg(b+a)cos(theta)

7/10*v^2==g(b+a)(1-cos(theta))

v^2=10/7* g(b+a)(1-cos(theta))

not sure how to find the angle between the line OG and the up ward vertical. How would I show that cos(theta)=10/17 when ball leaves sphere?

to get the normal force would I differentiate v

m*dv/dt=dP/dt=F=0

and I can now find theta?

 

[tiny-tim]

Hi pentazoid!

(have a theta: θ and a squared: ² )

V=Mgh = Mg(b+a)cos(theta) (h=0 at theta=90 degrees)

E=T+V

at top , ball is at rest
E(initial)=1/2*M*(0)^2+1/2*I*(0)+Mg(a+b)=Mg(a+b)

E(final)= 1/2*Mv^2+1/2*(2/5*Ma^2)(v/a)^2+ Mg(b+a)cos(theta)

E(i)=E(f) ==> Mg(b+a)=1/2*M*(7/5)v^2+Mg(b+a)cos(theta)

7/10*v^2==g(b+a)(1-cos(theta))

v^2=10/7* g(b+a)(1-cos(theta))

Very good!

(except most people use U for KE, since V looks too much like v )

not sure how to find the angle between the line OG and the up ward vertical. How would I show that cos(theta)=10/17 when ball leaves sphere?

to get the normal force would I differentiate v

Nooo … as I said, use Newton's second law …

what are the forces on the ball? …

they have to equal the centripetal acceleration …

so work out when the normal force becomes zero.