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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.

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.