A mass m = 76.0 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.7 m and finally a flat straight section at the same height as the center of the loop (19.7 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.) 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?

To find the required initial velocity at the top of the loop, we need to calculate the minimum velocity the mass must have to stay on the loop-the-loop without falling off the track.

At the top of the loop, the net force on the mass has to be equal to the centripetal force required to keep it moving in a circular path. The net force can be given by the difference between the gravitational force and the normal force.

1. Calculate the gravitational force acting on the mass at the top of the loop.
Gravitational force (Fg) = mass (m) * acceleration due to gravity (g)

Here, mass (m) = 76.0 kg and acceleration due to gravity (g) = 9.8 m/s².

Plugging in the values, Fg = 76.0 kg * 9.8 m/s² = 744.8 N

2. Calculate the normal force at the top of the loop.
At the top of the loop, the normal force is directed inward, providing the necessary centripetal force. It can be calculated using the equation:
Normal force (Fn) = mass (m) * (velocity (v))^2 / radius of the loop (R)

Here, mass (m) = 76.0 kg and radius of the loop (R) = 19.7 m.

We need to find the velocity (v) that will allow the normal force (Fn) to equal the gravitational force (Fg).

Fn = Fg
76.0 kg * (v)^2 / 19.7 m = 744.8 N

Solving for (v):

(v)^2 = (744.8 N * 19.7 m) / 76.0 kg
(v)^2 = 193.6426 m²/s²

Taking the square root of both sides:

v = √(193.6426 m²/s²)
v ≈ 13.918 m/s

Therefore, the mass needs to be pushed at an initial velocity of approximately 13.918 m/s at the beginning to successfully make it around the loop without falling off the track.

To determine the required initial velocity for the mass to make it around the loop-the-loop without falling off the track, we can use the principle of conservation of mechanical energy.

At the top of the loop, the mass has gravitational potential energy and kinetic energy. At the bottom of the loop, the mass only has kinetic energy. Assuming no energy losses due to friction or air resistance, the total mechanical energy (E) of the system is conserved.

The gravitational potential energy at the top of the loop is given by:

E_p = mgh_1

where m is the mass, g is the acceleration due to gravity, and h_1 is the height of the top of the loop. The kinetic energy of the mass at the top of the loop is:

E_k = (1/2)mv^2

where v is the velocity. At the bottom of the loop, the kinetic energy is:

E_k' = (1/2)mv'^2

where v' is the velocity.

Since the loop is a complete circle, the centripetal force required to keep the mass moving in a circular path at the top of the loop is provided solely by the gravitational force. This means:

mg = (mv^2) / R

where R is the radius of the loop.

Now, we can set up the conservation of mechanical energy equation:

E_p + E_k = E_k'

or

mgh_1 + (1/2)mv^2 = (1/2)mv'^2

Rearranging the equation, we get:

v'^2 - v^2 = 2gh_1

Substituting the expression for v^2 from the centripetal force equation, we have:

v'^2 - (gR) = 2gh_1

Simplifying further, we get:

v'^2 = 2g(h_1 + R)

Finally, taking the square root of both sides, we can find the required initial velocity (v'):

v' = √(2g(h_1 + R))

Substituting the given values of g = 9.8 m/s^2, h_1 = 19.7 m, and R = 19.7 m into the equation will give the required initial velocity for the mass to make it around the loop-the-loop without falling off the track.

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