Question

In a charming 19th-century hotel, an old-style alevator is connected to a counterweight by a cable that passes over a rotating disk 3.50 mm in diameter (Figure 1). The elevator is raised and lowered by turning the disk, and the cable does not slip on the rim of the disk but turns with it At how many rom must the disk turn to raise the elevator at 15.0 cm/s? Express your answer in revolutions per minute.. Part B To start the elevatur maving, it must be accelerated at What must be the angular acceleration of the disk, in rad/s² Express your answer in radians per second squared Part C Through what angle (in radians) has the disk turned when it has raised the elevator 3.50 m between floors? Express your answer in radians. Part D Through what angle (in degrees) has the disk turned when it has raised the elevator 3.50 m between floors? Express your answer in degrees.

Answer 1

The tension in the elevator cable during various stages of motion can be calculated using Newton's second law, with the forces involved being the gravitational force and the force due to acceleration or deceleration. The height and final velocity also depend on the kinematics of constant acceleration.

Step-by-step explanation:

We are tasked with performing a series of calculations related to the motion of an elevator and the forces involved. These calculations are centered around tension, acceleration, and angular motion, which involve principles of classical mechanics.

Part A: Calculating Tension During Acceleration

To calculate the tension in the cable while the elevator accelerates, we can use Newton's second law (F = ma). For the elevator mass of 1700 kg accelerating at 1.20 m/s2, the tension is the sum of the force due to gravity (mg) and the force due to acceleration (ma), which is T = mg + ma. Substituting for g (9.8 m/s2), we find T = (1700 kg)(9.8 m/s2 + 1.20 m/s2).

Part B: Tension at Constant Velocity

When the elevator moves at constant velocity, acceleration is zero, so the tension in the cable is simply the force of gravity on the elevator: T = mg.

Part C: Tension During Deceleration

While decelerating, the force required is less than the gravitational force, as the elevator is slowing down. The tension is thus T = mg - ma, with a being the deceleration rate.

Part D: Height and Final Velocity

The height moved by the elevator can be calculated using the kinematic equation for constant acceleration and the final velocity can be deduced from the time spent accelerating and decelerating.