A string is wound around a solid cylindrical spool of mass 9.7 kg and radius 0.1 m and is tied to the ceiling. If the spool is released from rest and rolls on the string, and the distance to the floor is 6.14 m, how long will it take the spool to hit the floor?

To find the time it takes for the spool to hit the floor, we can use the principles of rotational motion and energy.

First, let's consider the energy of the system. The spool starts from rest, so its initial kinetic energy is zero. As it falls, the potential energy of the spool is converted into kinetic energy.

The potential energy of the spool can be calculated as the product of its gravitational potential energy and its mass. The gravitational potential energy can be calculated as the product of the height and the mass of the spool, multiplied by the acceleration due to gravity (g = 9.8 m/s^2). So the potential energy is given by:

Potential energy = m * g * h

where m = mass of the spool = 9.7 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height of the spool above the floor = 6.14 m

Next, the kinetic energy of the spool at the bottom is given by:

Kinetic energy = (1/2) * I * ω^2

where I = moment of inertia of the spool
ω = angular velocity of the spool

The moment of inertia of a solid cylinder rotating about its axis is given by:

I = (1/2) * m * r^2

where r = radius of the spool = 0.1 m

We can equate the initial potential energy to the final kinetic energy:

m * g * h = (1/2) * I * ω^2

Substituting the expressions for I and rearranging the equation, we get:

m * g * h = (1/2) * ((1/2) * m * r^2) * ω^2

Cancelling out the mass (m) and rearranging further, we find:

g * h = (1/4) * r^2 * ω^2

Simplifying, we get:

ω^2 = (4 * g * h) / r^2

Finally, we can use the relationship between linear velocity (v) and angular velocity (ω) for a rolling object:

v = ω * r

Substituting the expression for ω, we have:

v = √((4 * g * h) / r^2) * r

Since we know the initial velocity (v0) of the spool is zero, we can use the equation for uniformly accelerated motion to find the time (t) it takes for the spool to hit the floor:

v = v0 + a * t

Solving for time (t), we have:

t = (v - v0) / a

Since the spool is released from rest and rolls without slipping, the linear acceleration (a) is equal to the acceleration due to gravity (g).

Plugging in the values and calculating, we can find the time it takes for the spool to hit the floor.