Answer to Question #118531 in Mechanics | Relativity for Jeshneel Kumar

Question

A flywheel has a moment of inertia of 1.6 × 10−3 kg.m2
. When a constant torque is
applied, it reaches an angular velocity of 1200 rev/min in 15 s. Assuming it started from
rest, find:
i) the angular acceleration
ii) the unbalanced torque applied
iii) the angle turned through 15 s
iv) the work done on the flywheel by the torque

Answer 1

Solution.

$I=1.6\sdot10^{-3}kg.m^2;$

$n=1200rev/min=20rev/s;$

$t=15s;$

w_0=0;

i)$\alpha=\dfrac{\Delta w}{t}; \Delta w=w-w_0;$

$w=2\pi n;$

$w=2\sdot3.14rad\sdot20rew/s=125.6rad/s$ ;

$\alpha=\dfrac{125.6rad/s}{15s}=8.37rad/s^2;$

ii) $\alpha=\dfrac{T}{I};\implies T=\alpha I;$

$T=8.37rad/s^2\sdot1.6\sdot10^{-3} kg.m^2=0.013kg.m^2/s^2;$

iii)$\theta=\theta_0+w_0t+\dfrac{\alpha t^2}{2};$

$\theta=\dfrac{8.37rad/s^2\sdot(15s)^2}{2}=941.6rad;$

iv)$W=\dfrac{1}{2}Iw^2-\dfrac{1}{2}Iw_0^2;$

$W=\dfrac{1}{2}\sdot1.6\sdot10^{-3}kg.m^2\sdot(125.6rad/s)^2=$

$=12.62J;$

Answer: i)$\alpha=8.37rad/s^2;$

ii)$T=0.013kg.m^2/s^2;$

iii)$\theta=941.6rad;$

iv)$W=12.62J.$