The rate of rotation of a solid disk with a radius of $4 m$ $4 m$ and mass of $5 k g$ $5 kg$ constantly changes from $12 H z$ $12 Hz$ to $5 H z$ $5 Hz$ . If the change in rotational frequency occurs over $7 s$ $7 s$ , what torque was applied to the disk?

Question

1438 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-11T11:41:00

The torque was $= 251.3 N m$ $=251.3Nm$

The torque is the rate of change of angular momentum

$τ = \frac{d L}{d t} = \frac{d (I ω)}{d t} = I \frac{d ω}{d t}$ $tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt$

where $I$ $I$ is the moment of inertia

The mass of the disc is $m = 5 k g$ $m=5kg$

The radius is $r = 4 m$ $r=4m$

For the solid disc, $I = \frac{1}{2} m r^{2}$ $I=1/2mr^2$

So, $I = \frac{1}{2} \cdot 5 \cdot {(4)}^{2} = 40 k g m^{2}$ $I=1/2*5*(4)^2=40kgm^2$

And the rate of change of angular velocity is

$(d omega)/dt=(Deltaomega)/t=(24pi-10pi)/7=2pirads^-2$

So,

The torque is

$tau=40*2pi=251.3Nm$

The rate of rotation of a solid disk with a radius of 4 m4m4 m and mass of 5 kg5kg5 kg constantly changes from 12 Hz12Hz12 Hz to 5 Hz5Hz5 Hz. If the change in rotational frequency occurs over 7 s7s7 s, what torque was applied to the disk?