Respuesta :
Explanation:
Given that,
Mass of the hamster, m = 0.139 kg
Diameter of the wheel, d = 20.8 cm
Radius, r = 10.4 m
Frequency of the wheel, f = 1 Hz
Time, t = 0.823 s
(a) There exist a relationship between the tangential and angular acceleration of the wheel. It is given by :
[tex]a=\alpha \times r[/tex]
Since, [tex]\alpha =\dfrac{\omega}{t}[/tex]
[tex]a=\dfrac{\omega}{t} \times r[/tex]
[tex]a=\dfrac{2\pi f}{t} \times r[/tex]
[tex]a=\dfrac{2\pi \times 1}{ 0.823} \times 10.4\times 10^{-2}[/tex]
[tex]a=0.793\ m/s^2[/tex]
(b) In radial direction, applying the second law of motion as :
[tex]N-mg=ma[/tex]
a is the radial acceleration, [tex]a=\dfrac{v^2}{r}[/tex]
[tex]N=mg+ma[/tex]
[tex]N=mg+m(\dfrac{v^2}{r})[/tex]
[tex]N=mg+m(\dfrac{(r\omega)^2}{r})[/tex]
[tex]N=mg+m\omega^2 r[/tex]
[tex]N=m(g+\omega^2 r)[/tex]
[tex]N=m(g+(2\pi f)^2 r)[/tex]
[tex]N=0.139\times (9.8+(2\pi 1)^2\times 10.4\times 10^{-2})[/tex]
[tex]N=1.93\ N[/tex]
Hence, this is the required solution.
