Final answer:
The acceleration of a marble rolling inside a conical bowl with an incline angle θ can be shown to be g/tan(θ) by analyzing the gravitational and normal force components, and applying Newton's second law of motion and the constraint of rolling without slipping.
Step-by-step explanation:
The student is asking about the acceleration of a marble that rolls around a circular path on the inner surface of a conical bowl, with the surface of the bowl making an angle θ with the vertical. To demonstrate that the acceleration of the marble is g/tan(θ), we must analyze the forces involved in the marble's motion.
Considering the forces along the direction of the incline, we have the force due to gravity, mg, which can be resolved into two components: one parallel to the incline (mg sin(θ)) and one perpendicular to the incline (mg cos(θ)). The perpendicular component is balanced by the normal force, and hence, does not affect the marble's acceleration. The parallel component causes the marble to accelerate down the incline.
Therefore, the net force causing the marble to accelerate along the incline is mg sin(θ), and according to Newton's second law (F=ma), the acceleration a is given by:
a = F/m = (mg sin(θ)/m) = g sin(θ)
Since the marble is rolling in a circular path, there is a centripetal force that must be considered, which is directed towards the center of the circular path. This force is provided by the horizontal component of the normal force, which is N sin(θ), since θ is the angle from the vertical. Equating this to the required centripetal force (mv²/r, where v is the velocity of the marble and r is the radius of the circular path), we have
N sin(θ) = mv²/r
However, there is a constraint that the marble rolls without slipping, meaning the tangential acceleration must equal the radial acceleration (centripetal). Therefore, setting the radial acceleration ar = v²/r equal to tangential acceleration at = a, we get:
a = v²/r
For equilibrium in the vertical direction, the vertical component of the normal force is N cos(θ) = mg.
Thus, the acceleration of the marble can be obtained by solving for v²/r in terms of g and θ:
a = g/tan(θ)