Question

What constant torque should be applied to the disc of mass 10kg and diameter 50cm so that it acquires an angular velocity of 2rad/sec in 4sec ?? The disc is initially at rest and rotates about an axis through the centre of the disc and in a plane perpendicular to the disc?

Answer 1

Torque = Momemt of inertia * Angular acceleration

Answer 2

The torque applied to the disc is 0.156 N-m.

Explanation:

It is given that,

Mass of the disc, m = 10 kg

Diameter of the disc, d = 50 cm = 0.5 m

Radius, r = 0.25 m

Initial angular velocity of the disc, $\omega_i=0$

Final angular velocity of the disc, $\omega_f=2\ rad/s$

Time, t = 4 s

The moment of inertia of the disc is given by :

$I=\dfrac{mr^2}{2}$

$I=\dfrac{10\times (0.25)^2}{2}$

$I=0.312\ kg-m^2$

Angular acceleration of the disc is given by :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

$\alpha =\dfrac{2-0}{4}$

$\alpha =0.5\ m/s^2$

The torque in case of rotational motion is given by :

$\tau=I\times \alpha$

$\tau=0.312\times 0.5$

$\tau=0.156\ N-m$

So, the torque applied to the disc is 0.156 N-m. Hence, this is the required solution.

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Rotational kinematics

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