Question

A student of mass M = 58 kg takes a ride in a frictionless loop-the-loop at an amusement park. The radius of the loop-the-loop is R = 12 m. The force due to the seat on the student at the top of the loop-the-loop is FN = 108 N and is vertically down. What is the apparent weight of the student at the bottom of the loop-the-loop?

*The picture are the wrong answers
(not 250.1 neither)

Answer 1

To find the apparent weight of the student at the bottom of the loop-the-loop, we can use the concept of centripetal force and gravitational force.

At the top of the loop-the-loop, the student's apparent weight is equal to the sum of the gravitational force (mg) and the normal force (FN) provided by the seat. Since the normal force is directed downward, we can write this as:

Apparent weight at the top (W_top) = mg + FN

Where:

- m is the mass of the student (58 kg)

- g is the acceleration due to gravity (approximately 9.81 m/s²)

- FN is the normal force (108 N)

Now, let's calculate W_top:

W_top = (58 kg) * (9.81 m/s²) + 108 N

W_top = 568.98 N + 108 N

W_top = 676.98 N

So, at the top of the loop-the-loop, the apparent weight of the student is approximately 676.98 Newtons.

Now, let's find the apparent weight at the bottom of the loop-the-loop. At the bottom, the student is moving in a circular path, and the net force is providing the centripetal acceleration. The apparent weight is the sum of the gravitational force and the net inward force.

Centripetal force (Fc) = m * ac

Where:

- m is the mass of the student (58 kg)

- ac is the centripetal acceleration

The centripetal acceleration is given by:

ac = v² / R

Where:

- v is the velocity of the student at the bottom of the loop

The velocity at the bottom of the loop can be calculated using energy conservation. At the top of the loop, the potential energy is converted into kinetic energy, so:

Potential energy at top = Kinetic energy at bottom

m * g * 2R = (1/2) * m * v²

Now, we can solve for v:

(58 kg) * (9.81 m/s²) * (2 * 12 m) = (1/2) * (58 kg) * v²

Solve for v:

v² = (2 * 9.81 m/s² * 12 m)

v² = 235.44 m²/s²

v = √235.44 m/s

v ≈ 15.34 m/s

Now that we have the velocity, we can find the centripetal force (Fc):

Fc = (58 kg) * (15.34 m/s)² / 12 m

Now, calculate Fc:

Fc ≈ 753.67 N

So, at the bottom of the loop-the-loop, the centripetal force is approximately 753.67 Newtons.

Now, let's find the apparent weight at the bottom:

Apparent weight at the bottom (W_bottom) = mg - Fc

W_bottom = (58 kg) * (9.81 m/s²) - 753.67 N

W_bottom = 568.98 N - 753.67 N

W_bottom ≈ -184.69 N

The negative sign indicates that at the bottom of the loop-the-loop, the student's apparent weight is reduced, and they may feel lighter, but it's not a true weightlessness like in free fall.

So, the apparent weight of the student at the bottom of the loop-the-loop is approximately -184.69 Newtons.