Conservation of Energy with Mass on Hemisphere

AI Thread Summary
The discussion centers on the conservation of energy for a mass sliding on a hemispherical surface after receiving an instantaneous impulse. The approach involves balancing forces and applying energy conservation principles, leading to the equation (v - v_0)^2 = 2gR(1-cos(theta)). A key point of confusion arises from an algebraic mistake that misrepresents the relationship between velocities and angles. The participants clarify that incorporating the initial velocity from the impulse is valid, but the algebraic error needs correction to accurately determine the angle at which the mass leaves the hemisphere. The conversation emphasizes the importance of precise mathematical representation in solving physics problems.
ccndy
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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.
 
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ccndy said:
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:

$$ F_N - mg \cos( \theta) = -mR(\dot \theta)^2$$
 
ccndy said:
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?
 
ccndy said:
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##.
 
haruspex said:
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?
 
erobz said:
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!
 
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ccndy said:
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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