Question

A circular metal disc of mass 4 kg and radius 0.4m makes 10 revolutions per second about an axis passing through its centre and perpendicular to its plane. Calculate the magnitude of torque which will increase the angular momentum by 20% in 10 sec.

Answer 1

we are given the circular metal disc and we know it is rotating about an axis passing through its centre of mass. We will take the moment of inertia for the solid disc and torque can be calculated using the relationship between torque and moment of inertia. Here we can exploit the symmetry since the disc is uniform and it is rotating around its axis passing through its centre of mass.

Complete Step by step answer: The moment of inertia of the disc about an axis passing through its centre of mass is given by $I=\dfrac{M{{r}^{2}}}{2}$, Where M is the mass of the disc and r is its radius
$I=\dfrac{4\times 0.4}{2}=0.8kg{{m}^{2}}$
frequency = 10 revolutions per second
So angular velocity, $\omega =2\pi f=20\pi rad/{{s}^{2}}$
If the angular velocity of the disc is ω, then its angular momentum can be written as, $L=I\omega$

$$L=0.8\times 20\pi \\\ \Rightarrow L=16\pi kg{{m}^{2}}{{s}^{-1}} \\\$$

Now we need to find the magnitude of torque which will increase the angular momentum by 20% in 10s

$$\tau =20\Rightarrow \tau =\dfrac{20}{100}\times 16\pi \\\ \Rightarrow \tau =3.2\pi\; kg{{m}^{2}}{{s}^{-1}} \\\$$

So final angular momentum is,

$$L =(3.2\pi +16\pi )\; kg{{m}^{2}}{{s}^{-1}} \\\ \Rightarrow L =19.2\pi\; kg{{m}^{2}}{{s}^{-1}} \\\$$

We know $\dfrac{\Delta L}{\Delta t}=\tau$

$$\tau =\dfrac{3.2\pi }{10} \\\ \Rightarrow \tau=0.32\times 3.14 \\\ \Rightarrow \tau =1.004Nm \\\$$

Thus, the magnitude of torque required is 1.004 Nm.

Note: While substituting the values all the units must be in the same notation preferably in SI. If the frequency is given in round per minute, always convert it into round per second and if we multiply it by 2 $\pi$ we get the angular frequency.