Conservation of Energy with Mass on Hemisphere

[ccndy]

Homework Statement

An object of mass m slides without friction on a hemispherical surface of radius, R. For t < 0, the
object is at rest at the very top of the sphere, i.e. θ = 0. At t = 0, it is given an instantaneous impulse, ∆~p = ∆pˆi. Determine the angle at which the mass leaves the hemisphere.

Relevant Equations

E = K + U
∆p = mv_f - mv_0

I tried approaching this question like this:

F_N - mgcos(theta) = -mR(theta_dot)^2

and theta_dot = v/R since R is constant

F_N = m(gcos(theta) - (v - v_0)^2/R) (with v being final velocity and v_0 being the initial velocity from the impulse)

and then using energy conservation:

at t = 0: E = 1/2(mv_0^2) + mgR
at t > 0: E = 1/2(mv^2) + mgRcos(theta)

Equating both equations, I got that:

(v - v_0)^2 = 2gR(1-cos(theta))

which would just yield the same angle as the situation where there is no impulse. Am I approaching the question wrong or am I incorporating impulse incorrectly?

Thanks.

 

[erobz]

Homework Statement:: An object of mass m slides without friction on a hemispherical surface of radius, R. For t < 0, the
object is at rest at the very top of the sphere, i.e. θ = 0. At t = 0, it is given an instantaneous impulse, ∆~p = ∆pˆi. Determine the angle at which the mass leaves the hemisphere.
Relevant Equations:: E = K + U
∆p = mv_f - mv_0

I tried approaching this question like this:

F_N - mgcos(theta) = -mR(theta_dot)^2

and theta_dot = v/R since R is constant

F_N = m(gcos(theta) - (v - v_0)^2/R) (with v being final velocity and v_0 being the initial velocity from the impulse)

and then using energy conservation:

at t = 0: E = 1/2(mv_0^2) + mgR
at t > 0: E = 1/2(mv^2) + mgRcos(theta)

Equating both equations, I got that:

(v - v_0)^2 = 2gR(1-cos(theta))

which would just yield the same angle as the situation where there is no impulse. Am I approaching the question wrong or am I incorporating impulse incorrectly?

Thanks.

Try formatting your equations using Latex. It makes your math easily readable.

For example you have written:

F_N - mgcos(theta) = -mR(theta_dot)^2

In latex that is parsed as:

$$ F_N - mg \cos( \theta) = -mR(\dot \theta)^2$$

 

[kuruman]

Homework Statement:: An object of mass m slides without friction on a hemispherical surface of radius, R. For t < 0, the
object is at rest at the very top of the sphere, i.e. θ = 0. At t = 0, it is given an instantaneous impulse, ∆~p = ∆pˆi. Determine the angle at which the mass leaves the hemisphere.
Relevant Equations:: E = K + U
∆p = mv_f - mv_0

which would just yield the same angle as the situation where there is no impulse. Am I approaching the question wrong or am I incorporating impulse incorrectly?

Why would it yield the same angle? Is the speed the same at a fixed angle with and without impulse at the top?

 

[haruspex]

Equating both equations, I got that:

(v - v_0)^2 = 2gR(1-cos(theta))

You have confused us by leaving out your final step, which would have produced gcos(theta)=2g(1-cos(theta)).
The equation I quote above is wrong. To get it you turned ##v^2-v_0^2## into ##(v-v_0)^2##.

 

[ccndy]

You have confused us by leaving out your final step, which would have produced gcos(theta)=2g(1-cos(theta)).
The equation I quote above is wrong. To get it you turned ##v^2-v_0^2## into ##(v-v_0)^2##.

You're right... that was a stupid algebraic mistake on my part.

However, does the incorporation of ##v_0## from the impulse make sense?

 

[ccndy]

Try formatting your equations using Latex. It makes your math easily readable.

For example you have written:

F_N - mgcos(theta) = -mR(theta_dot)^2

In latex that is parsed as:

$$ F_N - mg \cos( \theta) = -mR(\dot \theta)^2$$

Thank you!

 

[erobz]

You're right... that was a stupid algebraic mistake on my part.

However, does the incorporation of ##v_0## from the impulse make sense?

If it's given an instantaneous impulse such that the angle ##\theta## remains virtually ##0##, I would think yes...you basically start at ##t=0## with ##v_o## ( or ##\dot \theta_o## ).