A mass m = 88 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 15.5 m and finally a flat straight section at the same height as the center of the loop (15.5 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.)

3) If the mass has just enough speed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop?

6) It turns out the engineers designing the loop-the-loop didn’t really know physics – when they made the ride, the first drop was only as high as the top of the loop-the-loop. To account for the mistake, they decided to give the mass an initial velocity right at the beginning.
How fast do they need to push the mass at the beginning (now at a height equal to the top of the loop-the-loop) to get the mass around the loop-the-loop without falling off the track?

when it reaches the top of the loop it must have centripetal acceleration = gravitational acceleration of it will drop off the track.

v^2/R = g = 9.8 m/s^2 approx

v^2 / R = g
so
v^2 = g R at top
so how far above the top of the loop must it start?
call that distance h above the top
m g h = (1/2) m v^2
so
g h =(1/2) g R
h = (1/2) R

If it starts at the level of the top of the loop, it better be going at speed v there :)

oh, and at ground level of course

(1/2) m V^2 = m g (h + 2R)

To find the speed of the mass at the bottom of the loop, we can use the principle of conservation of mechanical energy.

3) At the bottom of the loop, the mass will have both kinetic energy and potential energy. Since the track is frictionless, there is no loss of mechanical energy.
The total mechanical energy at the bottom of the loop is the sum of the kinetic energy and potential energy:

E_b = K + U

The kinetic energy (K) can be calculated using the formula:

K = 1/2 * m * v^2

where m is the mass and v is the velocity.
The potential energy (U) can be calculated using the formula:

U = m * g * h

where m is the mass, g is the acceleration due to gravity, and h is the height.
In this case, the height at the bottom of the loop is equal to the radius of the loop (R).

Since the mass just has enough speed to make it around the loop, its kinetic energy at the top of the loop (E_t) will be equal to its total mechanical energy at the bottom (E_b), as there is no loss of energy.

E_t = E_b

Therefore, we can set the kinetic energy at the top of the loop equal to the sum of kinetic and potential energy at the bottom:

1/2 * m * v_t^2 = 1/2 * m * v_b^2 + m * g * R

Simplifying and rearranging the equation, we can solve for the velocity at the bottom of the loop (v_b):

1/2 * m * v_t^2 - 1/2 * m * v_b^2 = m * g * R

v_b^2 = v_t^2 - 2 * g * R

Taking the square root of both sides, we get the velocity at the bottom of the loop:

v_b = sqrt(v_t^2 - 2 * g * R)

where v_t is the speed at the top of the loop, g is the acceleration due to gravity (approximately 9.8 m/s^2), and R is the radius of the loop (15.5 m).

Now, let's move to question 6.

6) To find the initial velocity needed at the beginning of the track to get the mass around the loop-the-loop without falling off the track, we need to consider the conservation of mechanical energy again.

At the top of the loop, the total mechanical energy (E_t) will be the sum of kinetic and potential energy:

E_t = K + U

where K is the kinetic energy and U is the potential energy.

Since the mass is released from a height equal to the top of the loop, its potential energy will be zero.

Therefore, at the top of the loop, the total mechanical energy is equal to the kinetic energy:

E_t = K

Using the formula for kinetic energy:

K = 1/2 * m * v_t^2

where m is the mass and v_t is the velocity at the top of the loop.

Now, at the beginning of the track, the mass will have kinetic energy and potential energy:

E_beginning = K_beginning + U_beginning

The kinetic energy at the beginning of the track can be calculated using the formula:

K_beginning = 1/2 * m * v_beginning^2

where v_beginning is the initial velocity at the beginning of the track.

The potential energy at the beginning of the track is equal to the mass times the height, which is the same as the height at the top of the loop:

U_beginning = m * g * R

where g is the acceleration due to gravity (approximately 9.8 m/s^2) and R is the radius of the loop (15.5 m).

Since the total mechanical energy is conserved throughout the motion, we can set the mechanical energy at the top of the loop equal to the mechanical energy at the beginning:

E_t = E_beginning

1/2 * m * v_t^2 = 1/2 * m * v_beginning^2 + m * g * R

Simplifying and rearranging the equation, we can solve for the initial velocity at the beginning of the track (v_beginning):

v_beginning^2 = v_t^2 - 2 * g * R

Taking the square root of both sides, we get the initial velocity:

v_beginning = sqrt(v_t^2 - 2 * g * R)

where v_t is the speed at the top of the loop, g is the acceleration due to gravity (approximately 9.8 m/s^2), and R is the radius of the loop (15.5 m).

To answer question 3) and question 6) let's analyze the motion of the mass at different points along the track.

When the mass reaches the top of the loop-the-loop, it experiences a downward force from its weight and a normal force from the track. The net force acting on the mass at the top of the loop must be sufficient to provide the necessary centripetal force to keep it moving in a circular path.

1) At the top of the loop-the-loop:
The net force can be calculated using the equations of motion. The force of gravity acting on the mass is given by F_gravity = m * g, where m is the mass (88 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²). The normal force at the top of the loop is equal to the weight of the mass, which is F_normal = m * g. Since the mass is moving in a circular path at the top of the loop, the net force acting on it is the difference between the gravitational force and the normal force, which is given by F_net = F_gravity - F_normal = m * g - m * g = 0. This means that at the top of the loop, there is no net force acting on the mass, and it is momentarily weightless.

2) At the bottom of the loop-the-loop:
At the bottom of the loop, the mass is moving in a circular path and also experiencing a gravitational force and a normal force. The net force acting on the mass at the bottom of the loop must be equal to the centripetal force required to keep it moving in a circular path.

The centripetal force is given by the equation F_centripetal = (m * v^2) / R, where v is the velocity of the mass and R is the radius of the loop (15.5 m). The force of gravity acting on the mass is F_gravity = m * g. The normal force at the bottom of the loop is equal to the weight of the mass, which is F_normal = m * g. The net force acting on the mass at the bottom of the loop is the sum of the gravitational force and the normal force, which is given by F_net = F_gravity + F_normal = m * g + m * g = 2 * m * g. Equating the net force to the centripetal force, we have 2 * m * g = (m * v^2) / R.

Now, let's solve question 3):
Since the mass has just enough speed to make it around the loop without leaving the track, the net force at the bottom of the loop must be equal to the centripetal force. We can rearrange the equation mentioned above to solve for the velocity v at the bottom of the loop.

2 * m * g = (m * v^2) / R [Equation 1]
Simplifying, we get:
2 * g = (v^2) / R
(v^2) = 2 * g * R
v = sqrt(2 * g * R)

Substituting the values of g (approximately 9.8 m/s²) and R (15.5 m) into the equation, we can calculate the velocity v at the bottom of the loop:
v = sqrt(2 * 9.8 * 15.5) = sqrt(303.2) ≈ 17.41 m/s

Therefore, the speed of the mass at the bottom of the loop will be approximately 17.41 m/s.

Now, let's solve question 6):
To account for the mistake made by the engineers and ensure that the mass can make it around the loop-the-loop without falling off, they need to give the mass an initial velocity at the beginning. This initial velocity must be sufficient to provide the necessary centripetal force to keep the mass moving in a circular path.

The net force at the top of the loop should be equal to the centripetal force. Let's use the same equation we derived in Equation 1 to find the minimum initial velocity required.

2 * m * g = (m * v_initial^2) / R
(v_initial^2) = 2 * g * R
v_initial = sqrt(2 * g * R)

Substituting the values of g (approximately 9.8 m/s²) and R (15.5 m) into the equation, we can calculate the minimum initial velocity required to get the mass around the loop-the-loop:
v_initial = sqrt(2 * 9.8 * 15.5) = sqrt(303.2) ≈ 17.41 m/s

Therefore, the engineers need to push the mass at the beginning with a velocity of approximately 17.41 m/s to ensure it can make it around the loop without falling off the track.