Question

A hamster of mass 0.139 kg gets onto his 20.8−cm-diameter exercise wheel and runs along inside the wheel for 0.823 s until its frequency is 1.00 Hz.
(a) What is the tangential acceleration of the wheel, assuming it is constant?
(b) What is the normal force on the hamster just before he stops? The hamster is at the bottom of the wheel during the entire 0.823 s.

Answer 1

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 :

$a=\alpha \times r$

Since, $\alpha =\dfrac{\omega}{t}$

$a=\dfrac{\omega}{t} \times r$

$a=\dfrac{2\pi f}{t} \times r$

$a=\dfrac{2\pi \times 1}{ 0.823} \times 10.4\times 10^{-2}$

$a=0.793\ m/s^2$

(b) In radial direction, applying the second law of motion as :

$N-mg=ma$

a is the radial acceleration, $a=\dfrac{v^2}{r}$

$N=mg+ma$

$N=mg+m(\dfrac{v^2}{r})$

$N=mg+m(\dfrac{(r\omega)^2}{r})$

$N=mg+m\omega^2 r$

$N=m(g+\omega^2 r)$

$N=m(g+(2\pi f)^2 r)$

$N=0.139\times (9.8+(2\pi 1)^2\times 10.4\times 10^{-2})$

$N=1.93\ N$

Hence, this is the required solution.