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.

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Physics Kim Chi 4 years 2021-08-17T15:40:34+00:00 2021-08-17T15:40:34+00:00 1 Answers 10 views 0

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Captcha* Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )

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Acacia · Answer 1 · 2021-08-17T15:42:21+00:00

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