The speed of the block after it has descended 50 cm starting from rest is 1.6573 m/s.
a) Using the disc's center as a point of reference. Both the kinetic energy and the potential energy are zero.
Let's assume that the block's starting distance from the reference point is h.
Since its initial KE is zero, its potential energy is -mgh.
The block should fall from h to h'.
Throughout this descent
The block's PE is -mgh', and the - symbol denotes that it 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]\begin{aligned}&-m g h=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2-m g h^{\prime} \\&m g\left(h^{\prime}-h\right)=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2 \\&\end{aligned}[/tex]
Rotational inertia of the disc = 1/2 MR²
Angular speed ω =v/R
by putting vales of ω and I we get
so, 1/2 IW² = 1/4 MV²
Now, put this value of rotational KE in the equation i
[tex]\begin{aligned}&m g\left(h^{\prime}-h\right)=\frac{1}{4}(2 m+M) v^2 \\&\Rightarrow^v=\sqrt{\frac{4 m g\left(h^{\prime}-h\right)}{2 m+M}}\end{aligned}[/tex]
Given that (h'-h)= 0.5 m M= 360 g m= 70 g
V =[tex]=\sqrt{\frac{4 \times 70 \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.
The complete question is : 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.
To learn more about energy conservation principle refer to:
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