a particle of mass m sits directly atop a frictionless ice dome of radius R. by very slightly displacing the particle, the particle begins to slide down the dome. at which angle will the particle fall off the slide?
To determine the angle at which the particle will fall off the slide, we first need to understand the forces acting on the particle as it slides down the dome.
Since the surface is frictionless, the only force acting on the particle is its weight (mg) pointing downward. As the particle moves downward, this weight force will have two components: one in the radial direction (toward the center of the dome) and one in the tangential direction (along the surface of the dome).
Let's consider a small angle θ from the vertical, as shown in the diagram below:
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R
The radial component of the weight force (mgR/R) provides the centripetal force necessary for circular motion. Therefore, we can equate the weight force component to the centripetal force:
mgR/R = mv²/R
where v is the velocity of the particle.
Canceling the R terms and rearranging the equation, we get:
g = v²/R
We can determine the velocity of the particle at any instant by using energy conservation. The initial potential energy when the particle is displaced slightly is mgh, where h is the height above the bottom of the dome. The final energy when the particle falls off the slide is all converted to kinetic energy, so we have:
mgh = (1/2)mv²
Canceling the mass m and rearranging, we get:
v² = 2gh
Substituting this expression for v² back into the equation g = v²/R, we obtain:
g = (2gh)/R
Canceling the h terms, we find:
g = 2g/R
Now, we can solve for the angle at which the particle will fall off the slide using trigonometry. The tangent of the angle θ is given by the opposite side (R) divided by the adjacent side (R/g):
tan(θ) = R / (R/g)
= g / R
Taking the inverse tangent of both sides, we find:
θ = tan^(-1)(g/R)
Therefore, the angle at which the particle will fall off the slide is θ = tan^(-1)(g/R).