a weight of mass 1.09 kg is suspended by a string wrapped around a pulley wheel, which consists of a solid disk of mass 4.22 kg and radius 1.04 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.

Respuesta :

a weight of mass 1.09 kg is suspended by a string wrapped around a pulley wheel, which consists of a solid disk of mass 4.22 kg and radius 1.04 m. the system is released from rest. vertical distance does the hanging mass move in 3.0 seconds is 18.27 m

distance can be calculate as follows:

Suspended mass, m = 1.09 kg

Pulley mass, M=4.22 kg

pulley radius, r = 1.04 m

now,

Let's put some tension on the "T" string

For this,

vertical net force

mg - T = ma …………(1)

Where,

g is the gravitational acceleration

a is the acceleration of the suspended mass.

Pulley torque = Iα

I is the moment of inertia of the pulley

α is the angular acceleration

Pulley torque is due to string movement

therefore,

Torque = T x r

Also

α = a/r

and I = (Mr²/2) (for circular discs)

Substituting the values ​​of the above relationship for torque, we get

T × r = (Mr²/2) × (a/r)

T = (Ma)/2 …………(2)

from 1 and 2

mg - [(Ma)/2] = ra

mg= a(M + r)

          2

(1.09x9.8) =  a( 4.22 + 1.04)

                         2

a= 4.06

now we know that

s= ut+1/2 at²

where s is the distance traveled

u is the initial velocity = 0 (in this case)

t is time = 3 seconds

So if you replace the value

s= (0x3)+ 1/2 4.06 (3)²

s = 18.27 m

learn more about distance at https://brainly.com/question/13077188

#SPJ4