Question #123778
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
1
Expert's answer
2020-06-24T11:21:04-0400

Solution.

M=4Nm;M=4Nm;

ω0=0rev/s;\omega_0=0rev/s;

ω=6rev/s;\omega=6rev/s;

t=10s;t=10s;

ϵ=ωω0t;\epsilon=\dfrac{\omega-\omega_0}{t};

ϵ=6rev/s10=0.6rew/s2=3.768rad/s2;\epsilon=\dfrac{6rev/s}{10}=0.6rew/s^2=3.768rad/s^2;

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

M=Iϵ    I=Mϵ;M=I\epsilon\implies I=\dfrac{M}{\epsilon};

I=4Nm3.768rad/s2=1.06kgm2;I=\dfrac{4Nm}{3.768rad/s^2}=1.06kgm^2;

L=Iω;L=I\omega;

L=1.06kgm237.68rad/s=40kgm2/s;L=1.06kgm^2\sdot37.68rad/s=40kgm^2/s;

P=Mω;P=M\omega;

P=4Nm37.68rad/s=150.72WP=4Nm\sdot37.68rad/s=150.72W ;

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

ii)ω=37.68rad/s;ii)\omega=37.68rad/s;

iii)I=1.06kgm2;iii)I=1.06kgm^2;

iv)L=40kgm2/s;iv)L=40kgm^2/s;

v)P=150.72W;v)P=150.72W;




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