Answer :
Answer: 2.86 m
Explanation:
To solve this question, we will use the law of conservation of kinetic and potential energy, which is given by the equation,
ΔPE(i) + ΔKE(i) = ΔPE(f) + ΔKE(f)
In this question, it is safe to say there is no kinetic energy in the initial state, and neither is there potential energy in the end, so we have
mgh + 0 = 0 + KE(f)
To calculate the final kinetic energy, we must consider the energy contributed by the Inertia, so that we then have
mgh = 1/2mv² + 1/2Iw²
To get the inertia of the bodies, we use the formula
I = [m(R1² + R2²) / 2]
I = [2(0.2² + 0.1²) / 2]
I = 0.04 + 0.01
I = 0.05 kgm²
Also, the angular velocity is given by
w = v / R2
w = 4 / (1/5)
w = 20 rad/s
If we then substitute these values in the equation we have,
0.5 * 9.8 * h = (1/2 * 0.5 * 4²) + (1/2 * 0.05 * 20²)
4.9h = 4 + 10
4.9h = 14
h = 14 / 4.9
h = 2.86 m
Answer:
The value of h is 2.86 m
Explanation:
Applying the energy conservation:
[tex]E_{p,i} +E_{k,i} =E_{p,f} +E_{k,f} \\mgh+0=0+\frac{1}{2} mv^{2} +\frac{1}{2} Iw^{2} \\I=\frac{m(R_{1}^{2}+R_{2}^{2}) }{2}[/tex]
Where
m = 2 kg
R₁ = 0.2 m
R₂ = 0.1 m
[tex]I=\frac{2*(0.2^{2}+0.1^{2}) }{2} =0.05kgm^{2}[/tex]
The angular speed is:
w = v/R₂ = 4/0.2 = 20 rad/s
[tex]h=\frac{\frac{1}{2}mv^{2}+\frac{1}{2}Iw^{2} }{mg} =\frac{\frac{1}{2}*2*4^{2}+\frac{1}{2}*0.05*20^{2} }{2*9.8} =2.86m[/tex]