Answer to Question #193069 in Mechanics | Relativity for rhy

Question

A pulley rotating in the counterclockwise directions is attached to a mass suspended from a string. The mass causes the pulley's angular velocity to decrease with a constant angular acceleration, α = -2.10 rad/s^2. (a) if the pulley's initial angular velocity is ω = 5.40 rad/s, how long does it take for the pulley to come to rest? (b) Through what angle does the pulley turn during this time? (c) if the radius of the pulley is 5.0 cm, through what distance is the mass lifted? *

Answer 1

$\omega_o =5.40rad/s, \alpha=-2.10 rad/s^2$

(a) Here, final velocity $\omega=0$

As we know,

$\omega=\omega_o+\alpha t\\\Rightarrow 0=5.40-2.10t\\\Rightarrow t=\dfrac{5.40}{2.10}=2.57s$

(b) The angle pulley turn during this time

$\theta=\omega_o t\\\Rightarrow\theta = 5.40\times 2.57=13.87 rad$

(c) r=0.05 m,

The distance at which mass lifted-

$x=r\theta=0.05\times 13.87=0.6939m$