Question

f R = 12 cm, M = 360 g, and m = 70 g (below), find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles. (Treat the pulley as a uniform disk.) Solve the problem using energy conservation principles. (b) Repeat (a) with R = 5.0 cm.

Answer 1

The speed of the block is 1.65 m/s.

Step-by-step explanation:

Given that,

Radius = 12 cm

Mass of pulley= 360 g

Mass of block = 70 g

Distance = 50 cm

(a). We need to calculate the speed

Using energy conservation

`P.E=K.E`

`P.E=mgh`

`K.E=(1)/(2)mv^2+(1)/(2)I\omega^2`

`K.E=(1)/(2)mv^2+(1)/(2)I*((v)/(r))^2`

`K.E=(1)/(2)mv^2+(1)/(2)*0.5Mr^2*((v)/(r))^2`

`K.E=(1)/(2)mv^2+(1)/(2)*0.5M* v^2`

`K.E=(1)/(2)v^2(m+0.5M)`

Put the value into the formula

`mgh=(1)/(2)v^2(m+0.5M)`

`v^2=(2mgh)/(m+0.5M)`

`v=\sqrt{(2mgh)/(m+0.5M)}`

`v=\sqrt{(2*70*10^(-3)*9.8*50*10^(-2))/(70*10^(-3)+0.5*360*10^(-3))}`

`v=1.65\ m/s`

(b), We need to calculate the speed of the block

When r = 5.0 cm

Here, The speed of the block is independent of radius of pulley.

Hence, The speed of the block is 1.65 m/s.

Answer 2

a)v= 1.6573 m/s

Step-by-step explanation:

a) Considering center of the disc as our reference point. The potential energy as well as the kinetic energy are both zero.

let initially the block is at a distance h from the reference point.So its potential energy is -mgh as its initial KE is zero.

let the block descends from h to h'

During this descend

PE of the block = -mgh' {- sign indicates that the block is descending}

KE= 1/2 mv^2

rotation KE of the disc= 1/2Iω^2

Now applying the law of conservation of energy we have

`-mgh = (1)/(2)mv^2+(1)/(2)I\omega^2-mgh'`

`mg(h'-h) = (1)/(2)mv^2+(1)/(2)I\omega^2` ................i

Rotational inertia of the disc =
`(1)/(2)MR^2`

Angular speed ω =
`(v)/(R)`

by putting vales of ω and I we get

so,
`(1)/(2)I\omega^2= (1)/(4)Mv^2`

Now, put this value of rotational KE in the equation i

`mg(h'-h) = (1)/(4)(2m+M)v^2`

⇒
`v= \sqrt{(4mg(h'-h))/(2m+M) }`

Given that (h'-h)= 0.5 m M= 360 g m= 70 g

`v= \sqrt{(4*70* 9.81* 0.5)/(140+360) }`

v= 1.6573 m/s\

b) The rotational Kinetic energy of the disc is independent of its radius hence on changing the radius there is no change in speed of the block.