Work and energy of an object on a ice slope

[ac7597]

Homework Statement

A perfect hemisphere of frictionless ice has radius R=7 meters. Sitting on the top of the ice, motionless, is a box of mass m=7 kg.

The box starts to slide to the right, down the sloping surface of the ice. After it has moved by an angle 11 degrees from the top, how much work has gravity done on the box?

How fast is the box moving?

At some point, as the box slips down the ice and speeds up, it loses contact with the ice and flies off into the air. At what angle from the top does the box leave the ice? I'll provide the units for you.

(Hint: At the critical point, the component of the gravitational force pointing toward the center of the hemisphere is exactly equal to the force required to keep the box moving in a circular path around the center of the hemisphere)

Relevant Equations

KE=(1/2)mv^2

Homework Statement: A perfect hemisphere of frictionless ice has radius R=7 meters. Sitting on the top of the ice, motionless, is a box of mass m=7 kg.

The box starts to slide to the right, down the sloping surface of the ice. After it has moved by an angle 11 degrees from the top, how much work has gravity done on the box?

How fast is the box moving?

At some point, as the box slips down the ice and speeds up, it loses contact with the ice and flies off into the air. At what angle from the top does the box leave the ice? I'll provide the units for you.

(Hint: At the critical point, the component of the gravitational force pointing toward the center of the hemisphere is exactly equal to the force required to keep the box moving in a circular path around the center of the hemisphere)
Homework Equations: KE=(1/2)mv^2

The work of gravity on the box across= mg(distance) = mg(radius-radius(cos(theta)))
Thus work of gravity= (7kg)(9.8m/s^2)(7-7cos(11))= 8.8J

velocity = (2(8.8J)/(7kg))^(1/2)=1.59m/s

I am stuck on how to find the angle when the box slips

 

[Orodruin]

Homework Statement: A perfect hemisphere of frictionless ice has radius R=7 meters. Sitting on the top of the ice, motionless, is a box of mass m=7 kg.

The box starts to slide to the right, down the sloping surface of the ice. After it has moved by an angle 11 degrees from the top, how much work has gravity done on the box?

How fast is the box moving?

At some point, as the box slips down the ice and speeds up, it loses contact with the ice and flies off into the air. At what angle from the top does the box leave the ice? I'll provide the units for you.

(Hint: At the critical point, the component of the gravitational force pointing toward the center of the hemisphere is exactly equal to the force required to keep the box moving in a circular path around the center of the hemisphere)
Homework Equations: KE=(1/2)mv^2

I am stuck on how to find the angle when the box slips

The hint is in the hint.

 

[ac7597]

The component of the gravitational force pointing at the center is mg*cos(theta). The force needed to keep a circular path is m(a)=m(v^2 /radius). We don't know the velocity at the point the ball slips.

 

[Orodruin]

What is the velocity as a function of theta?

 

[ac7597]

sin(theta)= velocity/radius ?

 

[Orodruin]

sin(theta)= velocity/radius ?

That cannot be, it is dimensionally inconsistent. Consider how you solved the first part of the problem.

 

[ac7597]

If final velocity^2 = 2g(R-Rcos(theta))
velocity= 2g(R-Rcos(theta))^(1/2)
thus mgcos(theta)= (velocity)^2/R
finally theta= 48.2 degrees

 

[haruspex]

If final velocity^2 = 2g(R-Rcos(theta))
velocity= 2g(R-Rcos(theta))^(1/2)
thus mgcos(theta)= (velocity)^2/R
finally theta= 48.2 degrees

Some of those equations are still dimensionally inconsistent, but you have the right answer, so I assume they are just typos in posting your work.
You should have $\frac{v^2}{rg}=2(1-\cos(\theta))$ (energy) and $\frac{v^2}{rg}=\cos(\theta)$ (centripetal acceleration), whence $\cos(\theta)=\frac 23$.