# Conservation of Energy with Mass on Hemisphere

• ccndy
And then you fall back to your original equation for the energy, ##1/2 m v^2 = mgh##, where ##h=R-Rcos(\theta)##. So, this yields the same answer as before, but with an initial velocity.f

#### 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.

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.

For example you have written:

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

In latex that is parsed as:

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

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?

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##.

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?

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!

• SammyS and 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## ).

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