Question

A rigid body of moment of inertia 0.5 kg.M^2 rotates with 2 RPM. How much torque is needed to increase the rotation to 10 RPM in 5 seconds.

Answer 1

Answer:

T = 0.084 Nm

Explanation:

First, we will calculate the angular acceleration:

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

where,

α = angular acceleration = ?

ωf = final angular speed = (10 RPM)(2π rad/1 rev)(1 min/60 s) = 1.05 rad/s

ωi = initial angular speed = (2 RPM)(2π rad/1 rev)(1 min/60 s) = 0.21 rad/s

t = time = 5 s

Therefore,

$\alpha = \frac{1.05\rad/s - 0.21\ rad/s}{5\ s}\\\\\alpha = 0.168\ rad/s^2$

Now, for the torque:

$T = I\alpha$

where,

T = torque = ?

I = moment of inertia = 0.5 kg.m²

Therefore,

$T = (0.5\ kg.m^2)(0.168\ rad/s^2)$

T = 0.084 Nm