Question 126: Centre of Mass and Rotational Motion

Question

A circular disk of radius R meter and mass M kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that $	heta(t)=5 t^2-8 t$, where $	heta(t)$ is the angular position of the rotating disc as a function of time $t$.
How much power is delivered by the applied torque, when $t=2 \mathrm{~s}$ ?

Accepted Answer

Moment of Inertia and Angular Motion
For a solid circular disk, the moment of inertia is given by
$$ I = \frac{1}{2}MR^2. $$
The angular position is defined as
$$ 	heta(t) = 5t^2 - 8t. $$
Angular Velocity and Acceleration
Differentiate with respect to time to obtain the angular velocity:
$$ \omega(t) = \frac{d	heta}{dt} = 10t - 8. $$
Differentiating again, the angular acceleration is:
$$ \alpha(t) = \frac{d\omega}{dt} = 10. $$
At time $$ t = 2 \, 	ext{s} $$:
Angular velocity:
$$ \omega(2) = 10(2) - 8 = 12 \, 	ext{rad/s}. $$
Angular acceleration:
$$ \alpha(2) = 10 \, 	ext{rad/s}^2. $$
Torque Calculation
The torque applied by the external force is related to the moment of inertia and angular acceleration:
$$ 	au = I \alpha = \frac{1}{2}MR^2 	imes 10 = 5MR^2. $$
Power Delivered
Power delivered by a torque is given by:
$$ P = 	au \omega. $$
At $$ t = 2 \, 	ext{s} $$, substitute the values:
$$ P = 5MR^2 	imes 12 = 60MR^2. $$
Thus, the power delivered by the applied torque at $$ t = 2 \, 	ext{s} $$ is:
$$ \boxed{60MR^2}. $$

Answer

$60 \mathrm{MR}^2$

Answer

$72 \mathrm{MR}^2$

Answer

$8 \mathrm{MR}^2$

Answer

$108 \mathrm{MR}^2$

Centre of Mass and Rotational Motion

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