Answer to Question #112863 in Mechanics | Relativity for Clara

Question

A set of keys has a mass of 0.2kg. The keys are swung on a light lanyard of an initial length of 0.48m. Each time the lanyard is swung around the person's fingers it loses 0.03m of length. After how many swings will the keys be moving twice as fast as they were initially. Your answer will be a whole number of swings.

Answer 1

As per the given question,

mass of the set of keys =0.2 kg

Initial length of the key (l)=0.48 m

let the initial angular speed of the key was $=\omega$

After each rotation length of the key is getting reduce $(\Delta l)=0.03m$

As per the question, final angular speed $(\omega_2)=2\omega$

Now, applying the conservation of the angular momentum,

$I\omega=I_2\omega_2$

$I_2=\dfrac{I\omega}{2\omega}=\dfrac{I}{2}=\dfrac{ml^2}{3\times 2}=\dfrac{0.2\times 0.48^2}{6}$

$l_2=\sqrt{\dfrac{0.48^2}{2}}=0.3394 m$

so, time =$\dfrac{0.48-0.3394}{0.03}=4.686sec$ or t= 4 sec