Work of gravity on an object sliding down a frictionless sphere

[ac7597]

Homework Statement

A perfect hemisphere of frictionless ice has radius R=6.5 meters. Sitting on the top of the ice, motionless, is a box of mass m=6 kg.

The box starts to slide to the right, down the sloping surface of the ice. After it has moved by an angle 20 degrees from the top, how much work has gravity done on the box?

How fast is the box moving?

Relevant Equations

g=9.8m/s^2

Homework Statement: A perfect hemisphere of frictionless ice has radius R=6.5 meters. Sitting on the top of the ice, motionless, is a box of mass m=6 kg.

The box starts to slide to the right, down the sloping surface of the ice. After it has moved by an angle 20 degrees from the top, how much work has gravity done on the box?

How fast is the box moving?
Homework Equations: g=9.8m/s^2

Force	x	y
normal force of sphere on box	0	Fn
gravity	mgsin(theta)	-mgcos(theta)
total	m(ax)	m(ay)=0

From the force diagram, Fn=mgcos(theta) and m(ax)=mgsin(theta). Net force=m(ax). Thus the integral of net force=mgsin(theta)*(distance traveled). Since the circumference of half a sphere = 2(pie)(6.5m)/2 = 20.42m, the distance traveled= (20.42m)*(20degree)/(180degree)= 2.27m. The final equation should be: work=(6kg)(9.8m/s^2)sin(20deg)*(2.27m)=45.65J.

 

[ac7597]

 

[willem2]

You have a little more work to do to compute that integral. theta is not constant. It's certainly possible to do it this way, but there is an easier way to compute the work done by gravity that is valid for all paths, if you only know the starting point and the ending point.

 

[ac7597]

Like: integral of net force= -mgcos(theta) ?

 

[PeroK]

Like: integral of net force= -mgcos(theta) ?

Never integrate unless you have to.

Think energy!

 

[ac7597]

Is it kinetic energy = (1/2)mv^2 ?

 

[PeroK]

Is it kinetic energy = (1/2)mv^2 ?

That is the formula for KE. But the question is where did the KE come from?

 

[ac7597]

work done by gravity=∫ (G*(mass of hemisphere)*(mass of box) )/(x^2) dx. G=6.67*10^(-11) N*(m^2)/kg^2.
This becomes work= -(G*(mass of hemisphere)*(mass of box) )/ (x).
mass of ice hemisphere=(2/3)⋅π⋅(radius^3)⋅(mass density of ice) = (2/3)π (6.5m)^3 (0.92 g/cm^3) = 529159.39 kg.

Thus work1= -6.67*10^(-11) (529159.39kg)(6kg)/(6.5m) = -32.58 *10^(-6) J
work2= -6.67*10^(-11) (529159.39kg)(6kg)/(6.5m-2.27m) = -50.06 *10^(-6) J
work1-work2= -32.58 *10^(-6) J -(-50.06*10^(-6) J )= 17.48*10^(-6) J

 

[haruspex]

G*(mass of hemisphere)*(mass of box)

What force is causing the box to slide down? Indeed, what defines "down"?

 

[PeroK]

work done by gravity=∫ (G*(mass of hemisphere)*(mass of box) )/(x^2) dx. G=6.67*10^(-11) N*(m^2)/kg^2.
This becomes work= -(G*(mass of hemisphere)*(mass of box) )/ (x).
mass of ice hemisphere=(2/3)⋅π⋅(radius^3)⋅(mass density of ice) = (2/3)π (6.5m)^3 (0.92 g/cm^3) = 529159.39 kg.

Thus work1= -6.67*10^(-11) (529159.39kg)(6kg)/(6.5m) = -32.58 *10^(-6) J
work2= -6.67*10^(-11) (529159.39kg)(6kg)/(6.5m-2.27m) = -50.06 *10^(-6) J
work1-work2= -32.58 *10^(-6) J -(-50.06*10^(-6) J )= 17.48*10^(-6) J

Have you ever seen the term $mgh$?

 

[ac7597]

is it work=mg(radius-radius(cos(theta)))? I got work=23.05J

 

[ac7597]

mass(velocity^2)(1/2)=23.05J. Thus velocity=2.77m/s

 

[haruspex]

mass(velocity^2)(1/2)=23.05J. Thus velocity=2.77m/s

Yes.