# A rollercoaster ride has one giant 12 m radius loop that has the cars upside down on top. The...

## Question:

A rollercoaster ride has one giant 12 m radius loop that has the cars upside down on top. The ride is at ground level at the bottom of the loop and initially starts at a height of 32 m. What will a 75 kg passenger feel their weight is at the very top of the loop?

## Apparent weight Experienced by a Person :

Weight experienced by a person is actually the normal reaction exerted by the surface owing to Newtons third's Law of motion. In almost all practical cases this experienced weight is equal to the actual weight.

But in some cases if a person experiences a centripetal or centrifugal force the normal reaction decreases and the person experiences an apparent weight which is different from the actual one.

Given:

• Radius of the loop : {eq}\displaystyle r = 12 \space m {/eq}
• The initial height of start : {eq}\displaystyle h = 32 \space m {/eq}
• Mass of the passenger : {eq}\displaystyle m = 75 \space kg {/eq}

Let:

• The velocity of roller coaster at the top of the loop be = {eq}\displaystyle v m/s {/eq}
• The normal reaction or the apparent weight experienced by the passenger be = {eq}\displaystyle W \space N {/eq}

Considering conservation of energy at the initial point and at the top of the loop we get:

{eq}\displaystyle mgh = mg(2r) + \frac{1}{2}mv^2 {/eq}

{eq}\displaystyle \implies \frac{mv^2}{r} = \frac{2mgh}{r} - 4mg {/eq}

Now balancing all the forces at the top of the loop we get:

{eq}\displaystyle W + mg = \frac{mv^2}{r} {/eq}

{eq}\displaystyle \implies W = \frac{2mgh}{r} - 5mg {/eq}

Plugging in the known values we get:

{eq}\displaystyle W = \frac{2\times 75\times 9.8 \times 32}{12} - 5\times 75 \times 9.8 {/eq}

{eq}\displaystyle \implies W = 245 \space N {/eq} 