Question

Problem: The circular blad on a radial arm saw is turning at262

Answer 1

Answer:

Net torque, $\tau=-0.033\ N-m$

Explanation:

It is given that,

Initial angular speed of the blade, $\omega_i=262\ rad/s$

Final angular speed of the blade, $\omega_f=85\ rad/s$

Time, t = 18 s

Radius of the disk, r = 0.13 m

Mass of the disk, m = 0.4 kg

We need to find the net torque applied to the blade. We know that in rotational mechanics the net torque acting on an object is equal to the product of moment of inertia and the angular acceleration such that,

$\tau=I\times \alpha$

The moment of inertia of the disk, $I=\dfrac{mr^2}{2}$

$\tau=\dfrac{mr^2}{2}\times \dfrac{\omega_f-\omega_i}{t}$

$\tau=\dfrac{0.4\times (0.13)^2}{2}\times \dfrac{85-262}{18}$

$\tau=-0.033\ N-m$

Negative sign shows that the net torque is acting in the opposite direction of its motion. Hence, this is the required solution.