Question

A weight of mass 1.61 kg is suspended by a string wrapped around a pulley wheel, which consists of a solid disk of mass 5.96 kg and radius 0.712 m. The system is released from rest. Over what vertical distance does the hanging mass move in 3.0 seconds? Ignore friction and drag forces, and assume that the string does not slip.

Answer 1

Answer:

19.23 m

Explanation:

Given:

Mass suspended, m = 1.61 kg

Mass of the pulley, M = 5.96 kg

Radius of the pulley, r = 0.712 m

now,

let the tension in the string be 'T'

therefore,

the net force in the vertical direction

mg - T = ma    ............(1)

where,

g is the acceleration due to the gravity

a is the acceleration of the hanging mass.

also,

Torque on the pulley = Iα

where,

I is the moment of inertia of the pulley

α is the angular acceleration

the torque in the pulley is due to the movement of the string

thus,

torque = T × r

also

α = a/r

and I = (Mr²/2) for the circular disc

on substituting the values in the above relation for torque, we get

T × r = (Mr²/2) × (a/r)

or

T = (Ma)/2 ..............(2)

from 1 and 2 we have

mg - [(Ma)/2] = ma

or

$mg=a(\frac{M}{2}+m)$

on substituting the values, we get

$1.61\times9.8=a(\frac{5.96}{2}+0.712)$

or

a = 4.27 m/s²

Now, we know

$s=ut+\frac{1}{2}at^2$

where, s is the distance covered

u is the initial speed = 0 (for this case)

t is the time = 3 seconds

therefore on substituting the values, we get

$s=0\times3+\frac{1}{2}\times4.27\times(3.0)^2$

or

s = 19.23 m