Answer :
Answer:
a)v= 1.6573 m/s
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
[tex]-mgh = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2-mgh'[/tex]
[tex]mg(h'-h) = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2[/tex] ................i
Rotational inertia of the disc = [tex]\frac{1}{2}MR^2[/tex]
Angular speed ω =[tex]\frac{v}{R}[/tex]
by putting vales of ω and I we get
so, [tex]\frac{1}{2}I\omega^2= \frac{1}{4}Mv^2[/tex]
Now, put this value of rotational KE in the equation i
[tex]mg(h'-h) = \frac{1}{4}(2m+M)v^2[/tex]
⇒[tex]v= \sqrt{\frac{4mg(h'-h)}{2m+M} }[/tex]
Given that (h'-h)= 0.5 m M= 360 g m= 70 g
[tex]v= \sqrt{\frac{4\times70\times 9.81\times 0.5}{140+360} }[/tex]
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.
Answer:
The speed of the block is 1.65 m/s.
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
[tex]P.E=K.E[/tex]
[tex]P.E=mgh[/tex]
[tex]K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2[/tex]
[tex]K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\times(\dfrac{v}{r})^2[/tex]
[tex]K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times0.5Mr^2\times(\dfrac{v}{r})^2[/tex]
[tex]K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times0.5M\times v^2[/tex]
[tex] K.E=\dfrac{1}{2}v^2(m+0.5M)[/tex]
Put the value into the formula
[tex]mgh=\dfrac{1}{2}v^2(m+0.5M)[/tex]
[tex]v^2=\dfrac{2mgh}{m+0.5M}[/tex]
[tex]v=\sqrt{\dfrac{2mgh}{m+0.5M}}[/tex]
[tex]v=\sqrt{\dfrac{2\times70\times10^{-3}\times9.8\times50\times10^{-2}}{70\times10^{-3}+0.5\times360\times10^{-3}}}[/tex]
[tex]v=1.65\ m/s[/tex]
(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.