What Is the Height at Which the Boy Leaves the Ice?

[Kudo Shinichi]

HELP!physics prove question

Homework Statement

A boy is seated on the top of a hemispherical mound of ice. He is given a very small push and starts sliding down the ice surface, assumed to be fricitionless. show that he leaves the ice at a point who height is 2R/3

The Attempt at a Solution

I don't really know how to do this question. However, I know that the normal force disappears as he leaves the ice.

 

[tiny-tim]

A boy is seated on the top of a hemispherical mound of ice. He is given a very small push and starts sliding down the ice surface, assumed to be fricitionless. show that he leaves the ice at a point who height is 2R/3
…
I know that the normal force disappears as he leaves the ice.

Hi Kudo Shinichi!

Hint: calculate his speed at an angle θ, then caclulate the normal force.

 

[Kudo Shinichi]

Hi Kudo Shinichi!

Hint: calculate his speed at an angle θ, then caclulate the normal force.

x-direction:
v_0=v_0*cosθ
then use V_f=V_0+at
y-direction:
V_0=V_0*sinθ
then use V_f=V_0+at
but I have some many unknown variables...
Normal force=mg
also, How can i relate the speed to the normal force since F_n=mg

 

[tiny-tim]

… then use V_f=V_0+at …

Nooo … that equation (and the other two kinematic equations) only work if a is constant … which will only happen in free-fall, or on a slope.

This is a sphere, not a slope.

Hint: use conservation of energy.

 

[Kudo Shinichi]

Nooo … that equation (and the other two kinematic equations) only work if a is constant … which will only happen in free-fall, or on a slope.

This is a sphere, not a slope.

Hint: use conservation of energy.

Do you mean conservation of momentum and conservation of kinetic energy?
P=P'
m1v1+m2v2=m1v1'+m2v2'

KE=KE'
1/2m1v1^2+1/2m2v2^2=1/2m1(v1')^2+1/2m2(v2')^2
Since the initial velocity is equal to 0
therefore,
m1v1'=-m2v2'
1/2m1(v1')^2=-1/2m2(v2')^2
we don't know the mass of the boy, how can we find out the velocity...also you said find out the velocity by using angles, but i didn't need to use any angle on the above equations. I still don't really understand how to relate the velocity to the normal force, because normal force is equal to mg and both variables are nothing to do with velocity.

 

[tiny-tim]

Do you mean conservation of momentum and conservation of kinetic energy?
P=P'
m1v1+m2v2=m1v1'+m2v2'

conservation of momentum (m1v1+m2v2=m1v1'+m2v2') and conservation of energy are entirely different things

conservation of momentum will not work on a curved surface.

KE=KE'

conservation of kinetic energy does not exist

it's conservation of energy … that's total energy, not just kinetic energy:

KE + PE = constant

m1v1'=-m2v2'
1/2m1(v1')^2=-1/2m2(v2')^2

this is obviously the wrong equation … whatever is m₂?

Try again.

 

[Kudo Shinichi]

conservation of momentum (m1v1+m2v2=m1v1'+m2v2') and conservation of energy are entirely different things

conservation of momentum will not work on a curved surface.


conservation of kinetic energy does not exist

it's conservation of energy … that's total energy, not just kinetic energy:

KE + PE = constant


this is obviously the wrong equation … whatever is m₂?

Try again.

Total mechanical energy=PE+KE
PE+KE=0
PE=-KE
mgh=-1/2mv^2
m(9.8)(2R/3)=-1/2mv^2
m cancels out
v^2=9.8(2R/3)(-2)
v=sqrt(9.8*(2R/3)*-2)

I think this is the velocity that i need to find
and we know that the normal force =0 when the both is at the h=2R/3, but how do we relate the velocity to the normal force and how do we show that the boy leaves at height=2R/3

 

[tiny-tim]

… but how do we relate the velocity to the normal force and how do we show that the boy leaves at height=2R/3

Use Newton's second law in the radial (normal) direction.

 

[Kudo Shinichi]

Use Newton's second law in the radial (normal) direction.

Total mechanical energy=PE+KE
PE+KE=0
PE=-KE
mgh=-1/2mv^2
m(9.8)(2R/3)=-1/2mv^2
m cancels out
v^2=9.8(2R/3)(-2)
v=sqrt(9.8*(2R/3)*-2)

f=m*dv/dt
f=m*d(sqrt(9.8*(2R/3)*-2))/dt

this is the force we got
sorry, but I still don't get how can we prove that the boy starts from 2R/3 by using the force we got from above.

 

[tiny-tim]

f=m*dv/dt

No!

f_net = ma.

(a only = dv/dt if the motion is in a straight line … this is along a curve)

What is f_net?

 

[Kudo Shinichi]

No!

f_net = ma.

(a only = dv/dt if the motion is in a straight line … this is along a curve)

What is f_net?

F_net is the net force, which is also stands for all the forces we applied.
F_net=F_1+F_2+F_3+...
F_net=ma
in this case a is equal to v because
a=v(T2)-v(T1)
and the initial velocity in the problem is zero
therefore a= v-0 a=v a=v=sqrt(9.8*(2R/3)*-2)

F_net=m*v=m*sqrt(9.8*(2R/3)*-2)
= sqrt(-13.1R)*m

 

[tiny-tim]

What is f_net?

F_net=F_1+F_2+F_3+...

So what are F_1 F_2 and F_3?

 

[Kudo Shinichi]

So what are F_1 F_2 and F_3?

F_1=ma_1
F_2=ma_2
F_3=ma_3