Question

A puck is sliding around a circular track that is banked at an angle with respect to the horizontal such that no friction between the puck and the track is required to keep the puck from slipping up or down the track. If the puck is moving at a speed 31.9 m/s and the radius of the track is 421 m, determine θ .

Answer 1

Answer:

θ = 14.27°

Explanation:

The only force acting on the puck is the gravitational force. Since the track is banked with an angle θ, we have to separate the components of the weight.

For the sake of simplicity, I will denote the perpendicular direction to the truck as the y-direction, and the direction along the radius as the x-direction.

So, the free-body diagram of the puck is as follows:

1- x-component of the weight of the puck: mgsinθ

2- y-component of the weight of the puck: mgcosθ

3- Normal force in the y-direction perpendicular to the track.

Since there is no motion on the y-direction, normal force is equal to the y-component of the weight of the puck.

The x-component of the weight of the puck is equal to the centripetal force according to Newton's Second Law:

$F = ma\\mg\sin{\theta} = \frac{mv^2}{R}\\\sin{\theta} = \frac{v^2}{gR}$

Substituting the variables given in the question, the angle of the track can be found:

$\sin{\theta} = \frac{(31.9)^2}{(9.8)(421)} = 0.2466\\\theta = 14.27^\circ$