Question

A 35.30-kg box is attached to a light string that is wrapped around a cylindrical frictionless spool of radius 10.0 cm and moment of inertia 4.00 kg * m^2. The spool is suspended from the ceiling, and the box is then released from rest a distance from rest a distance 3.50 m above the floor. How long does it take for the box to reach the floor?

Answer 1

The box takes approximately 0.324 seconds to reach the floor.

Step-by-step explanation:

To find the time it takes for the box to reach the floor, we can use the principle of conservation of energy.

The potential energy of the box at its initial position can be calculated using the formula PE = mgh, where m is the mass of the box (35.30 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the initial height (3.50 m).

The potential energy is then converted into kinetic energy as the box falls. The kinetic energy can be calculated using the formula KE = (1/2)mv^2, where v is the final velocity of the box.

Equating the potential energy and the kinetic energy, we can solve for the final velocity of the box:

mgh = (1/2)mv^2

Canceling out the mass and simplifying the equation, we get:

gh = (1/2)v^2

Solving for v, we find:

v = sqrt(2gh)

Substituting the given values of g (9.8 m/s^2) and h (3.50 m), we get:

v = sqrt(2 * 9.8 m/s^2 * 3.50 m) = 10.8 m/s

Finally, we can use the equation v = d/t to find the time it takes for the box to reach the floor. Rearranging the equation to solve for t, we have:

t = d/v

Substituting the given value of d (3.50 m) and the calculated value of v (10.8 m/s), we get:

t = 3.50 m / 10.8 m/s = 0.324 s

Therefore, it takes approximately 0.324 seconds for the box to reach the floor.

Answer 2

The velocity of the box is related to the angular velocity of the spool, which is given by the equation:

v = r * ω

where r is the radius of the spool and ω is the angular velocity of the spool. The angular velocity of the spool, in turn, is related to the torque applied to the spool by the tension in the string, which is given by the equation:

τ = I * α

where τ is the torque, I is the moment of inertia of the spool, and α is the angular acceleration of the spool.

The tension in the string is equal to the weight of the box, which is given by:

T = m * g

Putting all of these equations together, we can solve for the time it takes for the box to reach the floor. Here's how:

First, we can find the angular acceleration of the spool using the torque equation:

τ = I * α

T = m * g = τ

m * g = I * α

α = (m * g) / I

α = (35.30 kg * 9.81 m/s^2) / 4.00 kg*m^2

α = 86.53 rad/s^2

Next, we can find the angular velocity of the spool using the kinematic equation:

ω^2 = ω_0^2 + 2 * α * θ

where ω_0 is the initial angular velocity (which is zero), θ is the angle through which the spool has turned (which is equal to the distance the box has fallen divided by the radius of the spool), and ω is the final angular velocity (which is what we want to find). Solving for ω, we get:

ω^2 = 2 * α * θ

ω = sqrt(2 * α * θ)

ω = sqrt(2 * 86.53 rad/s^2 * (3.50 m / 0.10 m))

ω = 166.6 rad/s

Finally, we can find the time it takes for the box to reach the floor using the equation:

v = r * ω

v = 0.10 m * 166.6 rad/s

v = 16.66 m/s

t = d / v

t = 3.50 m / 16.66 m/s

t = 0.21 s