Question

A snowboarder drops from rest into a halfpipe of radius R and slides down its frictionless surface to the bottom

(a) Show that the snowboarder's speed at the bottom of the halfpipe is
v = 2gR (use conservation of energy)
(b) Show that the snowboarder's centripetal acceleration at the bottom is
ac = 2g (Apply the expression for centripetal acceleration)
(c) Show that the normal force on the snowboarder at the bottom of the halfpipe has a magnitude of 3mg (Use Newton's second law of motion)

Answer 1

A snowboarder drops from rest into a halfpipe of radius R and slides down its frictionless surface to the bottom, the snowboarder's speed is $\mathbf{v = \sqrt{2gR}}$

The snowboarder's centripetal acceleration at the bottom is $\mathbf{a_c = 2g}$. The normal force acting on the snowboarder at the bottom of the halfpipe = 3mg

According to the conservation of energy, at any given point of motion, the energy of a system will always be constant.

Using the conservation of energy at rest and also at the bottom of the pipe.

• Total potential energy of the person(since its at rest = Total kinetic energy at the bottom of the pipe.

$\mathbf{mgR = \dfrac{1}{2}mv^2}$

The objective is to make the velocity (v) the subject of the formula:

$\mathbf{gR = \dfrac{1}{2}v^2}$

$\mathbf{2gR = v^2}$

$\mathbf{v = \sqrt{2gR}}$

(b)

Using the expression for centripetal acceleration;

$\mathbf{a_c = \dfrac{v^2}{R}}$

here,

let's replace $\mathbf{v \ \ \ with \ \ \sqrt{2gR}}$;

Then;

$\mathbf{a_c = \dfrac{(\sqrt{2gR})^2}{R}}$

$\mathbf{a_c =2g}$

From the image attached below;

Provided that:

• n = normal force
• w = mg  weight of the person
• $\mathbf{F_c}$ = centripetal force

Then, we can infer that the normal net force acting on the person can be computed as:

• $\mathbf{F_c = n -mg}$

By rearrangement;

• $\mathbf{n =F_c +mg}$

where;

• $\mathbf{F_c = 2 mg}$

∴

• n = 2mg + mg

n = 3mg

Therefore, we can conclude that the snowboarder's speed, centripetal force, and normal force are proved.

Learn more about centripetal force here:

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