Answer to Question #123778 in Mechanics | Relativity for Ojugbele Daniel

Question

A torque of 4Nm is applied to accelerate a wheel from rest to 6rev/sec in a time of 10sec. Determine
(i) angular acceleration
(ii) angular velocity
(iii) angular momentum
(iv) angular impulse
(v) power applied

Answer 1

Solution.

$M=4Nm;$

$\omega_0=0rev/s;$

$\omega=6rev/s;$

$t=10s;$

$\epsilon=\dfrac{\omega-\omega_0}{t};$

$\epsilon=\dfrac{6rev/s}{10}=0.6rew/s^2=3.768rad/s^2;$

$\omega=6rev/s=37.68rad/s;$

$M=I\epsilon\implies I=\dfrac{M}{\epsilon};$

$I=\dfrac{4Nm}{3.768rad/s^2}=1.06kgm^2;$

$L=I\omega;$

$L=1.06kgm^2\sdot37.68rad/s=40kgm^2/s;$

$P=M\omega;$

$P=4Nm\sdot37.68rad/s=150.72W$ ;

Answer: $i)\epsilon=3.768rad/s;$

$ii)\omega=37.68rad/s;$

$iii)I=1.06kgm^2;$

$iv)L=40kgm^2/s;$

$v)P=150.72W;$